Locally Finite, Simple Groups

Ulrich Meierfrankenfeld March 30, 2011

2

Contents

1 Local Properties 5 2 P.Hall’s universal locally finite group 9 3 Kegel-covers of locally finite, simple groups 11 4 Serial Subgroups 15 5 On the Jordan Hölder Theorem for locally finite groups 19 6 Absolute Simplicity 25 7 Simple locally finite groups which are not absolutely simple 29 8 Conjugacy centralizer property 31 9 Transitive groups of finitary permutations 33 10 Linear periodic groups in characteristic 0 39 11 Inverse limits of finite sets 45 12 Ultra-products and a little bit of model theory 47 13 Periodic, simple, linear groups 55 14 The maximal subgroups of SL_{2}(K) 63 15 The generalized Fitting subgroup of periodic, linear group 73 16 Non-linear, locally finite, simple groups with p-groups as centralisers 79 17 Examples of simple and of characteristically simple groups 87 18 Finitary groups 91 4 CONTENTS 19 Elementary equivalence 99 20 Exercises 101

Chapter 1

Local Properties

local

Definition 1.1 Let C be a class of groups and G a group. Then G is called a locally locally C-group, provided that for each finite subset X of G there exists Y ≤ G with X ⊆ Y and Y ∈ C.

For example a locally abelian group is abelian as any two elements of G have to lie in an abelian subgroup of G and so commute. A less trivial observation is:

Lemma 1.2 Let G be a locally simple group. Then G is simple. ls=s

Proof: Let N be a non-trivial normal subgroup of G and 1 ≠ x ∈ N. We need to show that G = N. For this let y ∈ G. Since G is locally simple there exists a simple subgroup S of G so that x, y ∈ S. As x ∈ S ∩ N, S ∩ N is a non-trivial normal subgroup of S and so S = S ∩ N ≤ N. Thus y ∈ N. Now y was arbitrary and so G = N

Another explanation why ”simple” is a local property is as follows. A group G is simple if and only if G = ⟨x^{G}⟩ for all 1 ≠ x ∈ G. This is true if and only if y ∈ ⟨x^{G}⟩ for all y ∈ G. This in turn just means that y can be written as a product of finitely many conjugates of x and x^{−1}. But the last statement just involves finitely many elements of G and so is a local property.

The next lemma introduces a fairly general method to construct locally C-groups.

Lemma 1.3 Let C be a class of groups and let clcg

G_{1} ≤ G_{2} ≤ G_{3} ≤ … ≤ G_{i} ≤ G_{i+1} ≤ …

be an ascending chain of C-groups . Put G = ⋃_{i=1}^{∞} G_{i}. Then G is a locally C-group.

Proof: Let X be a finite subset of G. As G is the union of the G_{i}, for each x ∈ X there exists i_{x} ≥ 1 with x ∈ G_{i_{x}}. Put i = max{i_{x} | x ∈ X}. Then x ∈ G_{i_{x}} ≤ G_{i} and so X ⊆ G_{i}. Since G_{i} is a C group, this proves that G is locally a C-group.

For example the infinite cyclic group (ℤ, +) has lots of subgroups isomorphic to itself and we easily obtain a chain of groups as in 1.3 with each of the G_{i}’s isomorphic to ℤ and properly contained in G_{i+1}. The resulting group G is locally cyclic, but (as any possible element of G lies in one of the G_{i}’s and so generates a proper subgroup), G is not cyclic. A more concrete example of a locally cyclic but not cyclic group is (Q, +). The reader might also convince herself that our first example of such a locally cyclic, but not cyclic group is isomorphic to a subgroup of Q. For example if |G_{i+1}/G_{i}| = 2 for all i, then G is isomorphic to \{\frac{n}{2^{k}} | n ∈ ℤ, k ∈ N \}.

To obtain an example of an infinite, locally finite, locally cyclic, groups, let p be prime and choose G_{i} ≅ C_{p^{i}} = ℤ/p^{i}ℤ. The resulting group G is called the Prüfer p-group and will be denoted by C_{p^{∞}} or ℤ_{p^{∞}}.^{1}

Next let Ω_{i} = {1, 2, …, i}. Then we can identify Sym(Ω_{i}) with the subgroup of Sym(Ω_{i+1}) fixing i + 1. Application of 1.3 yields a locally finite group G = ⋃_{i=1}^{∞} Sym(Ω_{i}). Note that we can view G as a subgroup of Sym(Ω), where Ω = {1, 2, 3, …}. But G is not the full symmetric group, indeed

… → 8 → 6 → 4 → 2 → 1 → 3 → 5 → 7 → 9 → …

is an element of infinite order in Sym(Ω) and so cannot be contained in G. Note also that the elements in G fix all but finitely many elements of Ω. Conversely, if π ∈ Sym(Ω) fixes all but finitely many elements of Ω, then π ∈ G_{i}, where i is the largest positive integer not fixed by π. For a set Δ define the finitary symmetric group on Δ by

FSym(Δ) = {π ∈ Sym(Δ) | | Supp(π)| < ∞}.

With this notation G is FSym(Ω). The groups FSym(Δ) are always locally finite and we have found locally finite groups of arbitrary cardinality. FSym(Δ) is a normal subgroup of Sym(Δ) and has as a normal subgroup the alternating group Alt(Ω), which consists of those permutations which are product of an even number of 2-cycles. Also it might be interesting to notice that for infinite Δ, | FSym(Δ)| = |Δ| while | Sym(Δ)| = 2^{|Δ|}.

Next we use 1.3 one more time to construct one of the most fascinating locally finite groups: P.Hall’s universal locally finite group U.

For this let n_{1} = 3 and put G_{1} = Sym(3). Let n_{2} = |G_{1}| = n_{1}! and G_{2} = Sym(n_{2}) = Sym(G_{1}). View G_{1} as a subgroup of G_{2} via the regular permutation action. Inductively put n_{k+1} = |G_{k}| = n_{k}!, G_{k+1} = Sym(n_{k+1}) = Sym(G_{k}) and view G_{k} as a subgroup of G_{k+1} via the regular permutation. Put U = ⋃_{i=1}^{∞} G_{i}. Then U is a locally finite group. We will study U in more details later on but for now let us notice that as U contains symmetric groups of arbitrary large finite groups, U contains an isomorphic copy of any finite group.

1The reader may check that the Prüfer p-groups are subgroups of the multiplicative group of complex numbers with norm equal 1

8 CHAPTER 1. LOCAL PROPERTIES

P.Hall’s universal locally finite group

PHall

Definition 2.1 A locally finite group G is called universal provided that G fulfills the universal following property:

(Uni) For all finite groups F and E and embeddings α : F → G and β : F → E there exists an embedding γ : E → G with α = γβ.

The goal of this chapter is to show that P.Hall’s group U is up to isomorphism the unique countable, universal, locally finite group. Definition 2.2 Let G be a group acting on a set Ω. Then we say that G acts semi-regularly semiregular on Ω provided that C_{G}(ω) = {g ∈ G | ω^{g} = ω} = 1 for all ω ∈ Ω. G acts regularly on Ω if G acts transitively and semi-regularly on Ω.

The proofs of the next three lemmas are left to the reader. Lemma 2.3 Let G be acting on a set Ω. Then the following three statements are equiva- esr lent:

(a) G acts semi-regularly on Ω.

(b) G acts regularly on each of its orbits on Ω.

(c) Each orbit for G on Ω is isomorphic to the action of G on G by right multiplication. Lemma 2.4 Let G be a group acting semi-regularly on a set Ω and H ≤ G. Then also H ssr acts semi-regularly on Ω. Lemma 2.5 Let G be acting semi-regularly on the sets Ω_{1} and Ω_{2}. Then Ω_{1} and Ω_{2} are isr isomorphic as G-sets if and only if |Ω_{1}| = |Ω_{2}|.

Proposition 2.6 P.Hall’s group U is universal. phu

Proof: Let F and E be finite groups and α : F → U and β : F → E be embeddings. Also let G_{1} < G_{2} < … < G_{i} < … be the ascending chain of finite symmetric groups used to construct U. Since α(F) is finite, there exists j with α(F) ≤ G_{j}. Moreover, pick k, n_{k} ≥ |E| and k ≥ j. Then α(F) ≤ G_{k} and there exists an embedding γ^{∗} : E → G_{k}. Let i = k + 1 and Ω = G_{k}. Then G_{i} = Sym(Ω). By construction G_{k} acts regularly on Ω and so by 2.4 we see that both α and γ^{∗}β define semi-regular actions of F on Ω. By 2.5 these actions are isomorphic. But that just means that there exists g ∈ Sym(Ω) = G_{i} ≤ U so that α = Inn(g)γ^{∗}β. Putting γ = Inn(g)γ^{∗} we see that (Uni) in Definition 2.1 is fulfilled.

uulg Proposition 2.7 Let X and Y be countable, universal, locally finite groups. Then X and Y are isomorphic.

Proof: Since X and Y are countable, X = {x_{1}, x_{2}, x_{3}, …} and Y = {y_{1}, y_{2}, y_{3}, …}. Put X_{i} = ⟨x_{1}, x_{2}, …, x_{i}⟩ and Y_{i} = ⟨y_{1}, y_{2}, …, y_{i}⟩ and note that X_{i} and Y_{i} are finite subgroups of X and Y, respectively. Put X_{0}^{∗} = 1 = Y_{0}^{∗}. By induction we will define finite subgroups X_{i}^{∗} ≤ X and Y_{i} ≤ Y_{i}^{∗} and embeddings ϕ_{i} : X_{i}^{∗} → Y and ψ_{i} : Y_{i}^{∗} → X so that for all i > 0

1. X_{i}^{∗} = ⟨ψ_{i−1}(Y_{i−1}^{∗}), X_{i}⟩

2. ϕ_{i} ∘ ψ_{i−1} is the identity on Y_{i−1}^{∗}.

3. ϕ_{i} equals ϕ_{i−1}, when restricted to X_{i−1}^{∗}

4. Y_{i}^{∗} = ⟨ϕ_{i}(X_{i}^{∗}), Y_{i}⟩

5. ψ_{i} ∘ ϕ_{i} is the identity on X_{i}^{∗}.

6. ψ_{i} equals ψ_{i−1}, when restricted to Y_{i−1}^{∗}

For i = 0 there is nothing to do. So assume such groups and maps exist for i − 1. Then use 1. as the definition of X_{i}^{∗}. Let F = ψ_{i−1}(Y_{i−1}^{∗}) and α : F → Y_{i−1}^{∗} ≤ Y the inverse map of ψ_{i−1}. Let β be the inclusion map from F to X_{i}^{∗}.. As Y is universal there exist γ : X_{i}^{∗} → Y with γβ = α. But this just means that γ restricted to F is α and so 2. holds with ϕ_{i} = γ. In particular, ϕ_{i} is the inverse of ψ_{i−1} on X_{i−1}^{∗}. By 5. the same is true for ϕ_{i−1} and so 3. holds.

4., 5. and 6. are done in a similar way.

Define ϕ : X → Y by ϕ(x_{i}) = ϕ_{i}(x_{i}) and ψ : Y → X by ψ(y_{i}) = ψ_{i}(y_{i}). Then it is easy to check that ψ and ϕ are group homomorphisms which are inverse to each other.

Kegel-covers of locally finite, simple groups

¿From now on G will always be a locally finite group and F the set of finite subgroups of Kegel G, i.e the set of finitely generated subgroups of G.

Definition 3.1 kegelcover

(a) A set of pairs {(H_{i}, M_{i})|i ∈ I} is called a sectional cover for G if, for all i in I, H_{i} ∈ F and M_{i} is a normal subgroup of H_{i}, and if, for each H ∈ F there exists i in I with H ≤ H_{i} and H ∩ M_{i} = 1. The groups H_{i}/M_{i}, i ∈ I, are called the factors of the sectional cover.

(b) A Kegel-cover for G is a sectional cover all of whose factors are simple groups.

Definition 3.2 G^{∞} is subgroup of G generated by the all the perfect finite subgroups of G. gi G is called absolutely perfect if G = G^{∞}.

Note that G^{∞} is the smallest normal subgroup N of G so that G/N is locally solvable. Also for finite G, G^{∞} is just the last term of the derived series of G and G is perfect if and only if its is absolutely perfect.

Lemma 3.3 Let {(G_{i}, N_{i})|i ∈ I} a sectional cover for G. Suppose that I = ⋃_{t=1}^{k} I_{t}. Then psc for at least one 1 ≤ t ≤ k, {(G_{i}, N_{i})|i ∈ I_{t}} is a sectional cover for G.

Proof: Suppose not. Then for each t there exists a finite subgroup F_{t} ≤ G so that for all i ∈ I_{t} with F_{t} ≤ G_{i} we have F_{t} ∩ N_{i} ≠ 1. Define F = ⟨F_{t} | 1 ≤ t ≤ k⟩. By the definition of a sectional cover there exists i ∈ I with F ≤ G_{i} and F ∩ N_{i} = 1. But then F_{t} ≤ G_{i} and F_{t} ∩ M_{i} = 1. Hence i ∉ I_{t}, a contradiction to I = ⋃_{t=1}^{k} I_{t}.

Lemma 3.4 Let G be a locally finite, simple group and {(G_{i}, N_{i})|i ∈ I} a sectional cover elementarykegel for G.

(a) There exists a Kegel cover {(H_{j}, M_{j})|j ∈ J} such that for all j ∈ J there exists i ∈ I with N_{i} ≤ M_{j} ⊴ H_{j} ⊴ G_{i}.

(b) For i ∈ I let M_{i} be a normal subgroup of G_{i}. Then at least one of {(G_{i}, M_{i})|i ∈ I} and {(M_{i}, M_{i} ∩ N_{i})|i ∈ I} is a sectional cover for G.

(c) Suppose G is not cyclic of prime order.Then {(G_{i}^{∞}, G_{i}^{∞} ∩ N_{i})|i ∈ I} is a sectional cover for G.

(d) Suppose G is not cyclic of prime order. Let E be a class of groups such for each i ∈ I and each non-abelian composition factor K of G_{i}/N_{i} one has K ∈ E. Then there exists a Kegel cover for G all of whose factors are in E.

Proof: (a) Let E be a non trivial finite subgroup of G and 1 ≠ e ∈ E.

As a first step we show there exists a finite subgroup T of G with E ≤ T and E ≤ ⟨e^{⟨e^{T} ⟩}⟩ for all 1 ≠ e ∈ E. Indeed since G is simple, E ≤ ⟨e^{G}⟩ and since G is locally finite, E ≤ ⟨e^{F_{e}}⟩ for some finite subgroup F_{e} of G. Similarly F_{e} ≤ ⟨e^{T_{e}}⟩ for some finite subgroup T_{e} of G. Then E ≤ ⟨e^{⟨e^{Te}⟩}⟩. Now just let T be the finite subgroup of G generated by E and all the T_{e}, 1 ≠ e ∈ E and the first step is completed.

Pick i ∈ I with T ≤ G_{i} and T ∩ N_{i} = 1. Put H_{E} = ⟨E^{G_{i}}⟩N_{i}.

Our second step is to find a maximal normal subgroup M_{E} of H_{E} with N_{i} ≤ M_{E} and E ∩ M_{E} = 1. For this let R be the set of the maximal normal subgroups of H_{E} containing N_{i} and R = ⋂ R. Then G_{i} leaves R invariant and so R is a normal subgroup of G_{i} with R < H_{E}. Suppose that E ≤ R. Then H_{E} = ⟨E^{G_{i}}⟩N_{i} ≤ R, a contradiction. Thus E ≰ R. Thus there exists M_{E} ∈ R with E ≰ M_{E}. To complete the second step it remains to show that E ∩ M_{E} = 1. Otherwise pick 1 ≠ e ∈ E ∩ M_{E}. Then

E ≤ ⟨e^{⟨e^{T} ⟩}⟩ ≤ ⟨e^{H_{E}}⟩ ≤ M_{E},

a contradiction.

It follows now immediately from the second step that {(H_{E}, M_{E}) | E ∈ F} is a Kegel cover that fulfills (a).

(b) Assume that {(G_{i}, M_{i})|i ∈ I} is not a sectional cover for G. Then there exists a H ∈ F such that H ∩ M_{i} ≠ 1 for all i ∈ I with H ≤ G_{i}. Without loss H ≤ G_{i} for all i ∈ I. For 1 ≠ h ∈ H let I_{h} = {i ∈ I|h ∈ M_{i}}. Then I is the finite union of these I_{h} and so by 3.3 there exists 1 ≠ h ∈ H such that {(G_{i}, N_{i})|i ∈ I_{h}} is a sectional cover for G. Hence we may assume that I = I_{h}. Let E be any finite subgroup of G. Since G is LF S, there exists a finite subgroup T in G with E ≤ T and E ≤ ⟨h^{T} ⟩. Pick i ∈ I with T ≤ G_{i} and T ∩ N_{i} = 1. Then E ≤ ⟨h^{T} ⟩ ≤ M_{i} and E ∩ (N_{i} ∩ M_{i}) ≤ T ∩ N_{i} = 1. Thus {(M_{i}, M_{i} ∩ N_{i})|i ∈ I} is a sectional cover for G.

(c) Otherwise we conclude from (b) that {(G_{i}, G_{i}^{∞})|i ∈ I} is a sectional cover for G. Hence by (a) G has a Kegel cover all of whose factor are of prime order. Hence every finite subgroup of G is embedded into a group of prime order and so G itself has prime order.

(d) By (a) there exists a Kegel cover all of whose factors have prime order or lie in E. Thus by 3.3 there exists a Kegel cover all of whose factors have prime order or there exists a Kegel cover all of factors are in E. In the first case G itself has prime order and (d) is proved.

We remark that there exists examples of locally finite groups which do have a Kegel cover, but are not simple.

Corollary 3.5 Non-cyclic, locally finite simple groups are absolutely perfect. In particular, lss a locally solvable, locally finite, simple group is cyclic of prime order.

Proof: Observe that every locally finite simple groups has a sectional cover, for example {(F, 1) | F ∈ F}. Hence the corollary follows from 3.4c.

14 CHAPTER 3. KEGEL-COVERS OF LOCALLY FINITE, SIMPLE GROUPS

Serial Subgroups

The main goal of this section is to show that a subgroup of G is serial if and only if its Serial locally subnormal.

For the next definition recall that a directed set is a partially ordered set so that each two elements have an upper bound. For example F is a directed set under inclusion.

Definition 4.1 Let I be a directed set and for i ∈ I let L_{i} be a subgroup of G. Then localsystem {L_{i} | i ∈ I} is called a local system for G with respect to I provided:

(a) L_{i} ≤ L_{j} for all i ≤ j ∈ I.

(b) ⋃_{i∈I} L_{i} = G

By (b) each of the elements of G lie in one of the L_{i}’s and by (a) and since I is directed each finite subgroup of G lies in one of L_{i}’s.

Definition 4.2 Let R be a group and H a group acting on R. Let S a set of H-invariant series subgroups of R.

(a) S is called an H-series on R provided that:

(a) 1 ∈ S and R ∈ S.

(b) S is totally ordered with respect to inclusion. That is, if D, E ∈ S then D ≤ E or E ≤ D.

(c) S is closed with respect to arbitrary intersections and unions. That is, if D ⊆ S, then ⋂ D ∈ S and ⋃ D ∈ S.

(d) For D ∈ S put D^{−} = ⋃{E ∈ S | E < D}. Then D^{−} ⊴ D for all D ∈ S.

(b) If D ≠ D^{−} for some D ∈ S, then (D^{−}, D) is called a jump of S and D/D^{−} a factor of S.

(c) An H-composition series is an H-series so that none of the factors has a proper Hinvariant normal subgroup.

(d) A normal H-series on R is a series all of its members are normal in R and a H-chief series is a normal composition series.

(e) A series, normal series, composition series and chief series on R is a 1-series, normal 1-series, 1-composition series and 1-chief series on R.

(f) A subgroup N of R is called H-serial in R provided that N is the member of some H-series on R.

locallysubnormal Definition 4.3 Let N be a subgroup of G. Then N is called locally subnormal provided that for each F ∈ F, N ∩ F is subnormal in F.

fsn Lemma 4.4 Let G be finite, H a subgroup of G, N a subnormal subgroup of G, M a normal subgroup of G and N a set of subnormal subgroups of G. Then

(a) H ∩ N is subnormal in H. In particular, N is locally subnormal in G.

(b) NM/M is subnormal in G/M.

(c) ⋂ N and ⟨N ⟩ are subnormal in G.

(d) If H ≤ ⟨N^{H}⟩, then H ≤ N.

(e) If H is subnormal in N, then H is subnormal in G.

Proof: (a) Just intersect a subnormal series from N to G with H to obtain a subnormal series from H ∩ N to H.

(b) Just take the images in G/M of a subnormal series from N to G to obtain a subnormal series from NM/M to G/M.

(e) follows immediately from the definition of subnormal.

(c) By induction on |N | we may assume that N = {N_{1}, N_{2}}. Then by (a) N_{1} ∩ N_{2} ⊴ ⊴N_{1} ⊴ ⊴G and so ⋂ N is subnormal in G.

For the proof that ⟨N ⟩ is subnormal in G we will use induction on |G| and then on |G/N_{1}|. If |G| = 1 or |G/N_{1}| = 1, there is nothing to prove. If N_{1} ⊴G then by (b) N_{1}N_{2}/N_{1} is subnormal in G/N_{1} and so N_{1}N_{2} is subnormal in G. So we may assume that N_{1} is not normal in G. Thus there exist subnormal subgroups N_{3}, N_{4} with

N_{1} ⊲ N_{3} ⊴ ⊴N_{4} ⊲ G

Let n ∈ N_{2}. Since n normalizes N_{4}, N_{1}^{n} is a subnormal subgroup of N_{4}. Since |N_{4}| < |G|, induction on |G| implies that ⟨N_{1}^{N_{2}}⟩ = ⟨N_{1}^{n} | n ∈ N_{2}⟩ is a subnormal subgroup of N_{4} and so also a subnormal subgroup of G.

If N_{2} does not normalize N_{1}, then |⟨N_{1}^{N_{2}}⟩| > |N_{1}| and so by induction on |G/N_{1}|, ⟨N_{1}, N_{2}⟩ = ⟨⟨N_{1}^{N_{2}}⟩, N_{2}⟩ is a subnormal subgroup of G.

So we may assume that N_{2} does normalize N_{1}. Thus ⟨N_{3}, N_{2}⟩ ≤ N_{G}(N_{1}) < G. Thus by induction on |G|, ⟨N_{1}, N_{2}⟩ is subnormal in ⟨N_{3}, N_{2}⟩. Moreover by induction on |G/N_{1}|, ⟨N_{3}, N_{2}⟩ is subnormal in G. Thus ⟨N_{1}, N_{2}⟩ is subnormal in G by (e).

(d) By induction on |G|. If N = G we are done. So suppose N ⊴ ⊴M ⊲ G. Then H ≤ ⟨N^{H}⟩ ≤ ⟨M^{H}⟩ = M. As |M| < |G| and N and H are contained in M, we are done by induction.

Lemma 4.5 Let {L_{i} | i ∈ I} be a local system for G and for i ∈ I let N_{i} be a subgroup of clsn L_{i}. Suppose that

(a) For all i ∈ I, N_{i} is locally subnormal in L_{i}.

(b) For all i ≤ j ∈ I, N_{i} ≤ N_{j}.

Then N = ⋃_{i∈I} N_{i} is a locally subnormal subgroup of G.

Proof: First notice that by (b) and as I is directed, N is a subgroup of G. Let F ∈ F. Then since F and F ∩ N are finite there exists i ∈ I so that F ≤ L_{i} and F ∩ N ≤ N_{i}. Since N_{i} is locally subnormal in L_{i}, F ∩ N_{i} is subnormal in F. But

F ∩ N_{i} ≤ F ∩ N ≤ F ∩ N_{i}.

So F ∩ N_{i} = F ∩ N and F ∩ N is subnormal in F, as desired.

We are now able to generalize 4.4 to infinite locally finite groups.

Proposition 4.6 Let H a subgroup of G, N a locally subnormal subgroup of G, M a lfsn normal subgroup of G and N a set of locally subnormal subgroups of G. Then

(a) H ∩ N is locally subnormal in H.

(b) NM/M is locally subnormal in G/M.

(c) ⋂ N and ⟨N ⟩ are locally subnormal in G.

(d) If H ≤ ⟨N^{H}⟩ and H is finite, then H ≤ N.

(e) If H is locally subnormal in N, then H is locally subnormal in G.

Proof: We will prove (d) and the second part of (c). The easy proofs for the remaining parts are left to the reader.

(c) part 2: For F ∈ F put N_{F} = ⟨E ∩ F | E ∈ N ⟩. As each E ∈ N is locally subnormal E ∩ F is subnormal in F and so by 4.4c, N_{F} is subnormal in F. Also F_{1} ≤ F_{2} ∈ F clearly implies N_{F_{1}} ≤ N_{F_{2}}. Application of 4.5 (with I = F and L_{F} = F) yields that N = ⋃_{F ∈F} N_{F} is locally subnormal in G. To complete the proof of (c) it now suffice to show that N = ⟨N ⟩.

By definition of N_{F}, N_{F} ≤ ⟨N ⟩ and so also N ≤ ⟨N ⟩. Conversely, if g ∈ E ∈ N, then g ∈ N_{⟨g⟩} ≤ N and so ⟨N ⟩ ≤ N.

(d) Since H is finite and H ≤ ⟨N^{H}⟩, there exists a finite subset A of N with H ≤ ⟨A^{H}⟩. Put F = ⟨A, H⟩. Then A ⊆ F ∩ N and so H ≤ ⟨(F ∩ N)^{H}⟩. As N is locally subnormal F ∩ N is subnormal in F. So by 4.4d H ≤ F ∩ N ≤ N.

mlsnen Lemma 4.7 Let N subgroup of G that is maximal with respect to being locally subnormal. Then N is normal in G.

Proof: Suppose N is not normal in G. Then there exists g ∈ G with N ≠ N^{g}. By 4.6c, ⟨N, N^{g}⟩ is locally subnormal in G and thus by maximality of ⟨N, N^{g}⟩ = G. In particular, ⟨g⟩ ≤ ⟨N^{⟨g⟩}⟩ and so by 4.6d, g ∈ N a contradiction to N ≠ N^{g}.

selsn Theorem 4.8 Let H be a subgroup of G. Then H is serial in G if and only if it is locally subnormal in G.

Proof: Suppose first that S is a series on G containing H. Let F ∈ F. Then S ∩ F = {E ∩ F | E ∈ S} is a subnormal series of F containing F ∩ H. So H is locally subnormal.

Suppose next that H is locally subnormal in G. Let S be a maximal chain of locally subnormal subgroups of G with H ∈ S. That such a maximal chain exists follows from Zorn’s lemma. To complete the proof of the theorem it suffices to show that S is a series. Clearly 1 ∈ S and G ∈ S.

Let E be a locally subnormal subgroup of G so that for each S ∈ S, E ≤ S or S ≤ E. Then S ∪ {E} is still a chain of locally subnormal subgroups and so the maximality of S implies E ∈ S.

Let D ⊆ S. By 4.6c, ⋃ D is locally subnormal in G. Let E ∈ S. If E ≤ D for some D ∈ D, then E ≤ ⋃ D. Otherwise D ≤ E for all D ∈ D and so ⋃ D ≤ E. Hence by the previous paragraph, ⋃ D ∈ S. With a similar argument ⋂ D ∈ S.

Finally we need to show that E^{−} ⊴ E for all E ≤ S. As E^{−} ∈ S, it is locally subnormal. Now let E^{−} ≤ R ≤ E so that R is locally subnormal in E. Then R is also locally subnormal in G and so by maximality of S we get R ∈ S. But this implies that R = E^{−} or R = E. So E^{−} is a maximal locally subnormal subgroup of G. Thus by 4.7, E^{−} ⊴ E.

So S is indeed a series, H is serial in G and the theorem is proved.

On the Jordan Hölder Theorem for locally finite groups

sJordan

Definition 5.1 Let H be a groups acting on the groups X and Y. djoho

(a) X and Y are called H-isomorphic provided that there exists an isomorphism α : X → Y which commutes with the action of H.

(b) Let C and D be sets of groups acted upon by H. Then C and D are H isomorphic provided that there exists a bijection β : C → D so that C and β(C) are H-isomorphic for all C ∈ C.

(c) We say that the Jordan-Hölder theorem holds for H on X provided that the sets of factors for any two composition series for H on X are H-isomorphic.

(d) Let calE be a class of groups acted upon by H. We say that the Jordan-Hölder theorem holds for H on X with respect to E provided that the sets of calE-factors for any two composition series for H on X are H-isomorphic.

The following lemma will produces lots of groups for which the Jordan Hölder theorem fails.

Lemma 5.2 Let I be a set of groups and ⊕I ≤ H ≤ ∏ I. Then H has a normal series sersub whose set of factors is isomorphic to I.

Proof: Let ” ⪯ ” be any reversed well ordering ( that is any non empty subset has a maximal element) on I such that I has a minimal element. For I ∈ I let H_{I}^{+} = {f ∈ H | f_{J} = 1 for all I ≺ J ∈ I} and H_{I}^{−} = {f ∈ H | f_{J} = 1 for all I ⪯ J ∈ I}. Let S = {H_{I}^{+}, H_{I}^{−} | I ∈ I}. We claim that S is a series for H with jumps (H_{I}^{−}, H_{I}^{+}), I ∈ H. As ⊕I ≤ H, it is clear that H_{I}^{+}/H_{I}^{−} ≅ I, so the lemma is proved once the claim is established.

20CHAPTER 5. ON THE JORDAN HÖLDER THEOREM FOR LOCALLY FINITE GROUPS

If m is the minimal element of I, then 1 = H_{m}^{−} ∈ H and if M is the maximal element of I, then H = H^{+}M ∈ I.

If J ≺ I ∈ I, then H_{J}^{+} ≤ H_{I}^{−} ≤ H_{I}^{+} and so S is totally ordered.

Let D be a subset of S. Let R be the set of all I ∈ I with H_{I}^{ϵ} ∈ D for some ϵ ∈ {±}.

As R has a minimimal element, there exist a largest lower bound r for for R. If H_{r}^{−} ∈ R, then ⋂ D = H_{r}^{−}. If H_{r}^{−} ∉ R we will show that ⋂ D = H_{r}^{+}. Clearly H_{r}^{+} ≤ ⋂ D. Let f ∈ ⋂ D and let I ∈ I with r ≺ I. As r is the largest lower bound of R, there exists J ∈ R with J ≺ I. Then H_{J}^{ϵ} ∈ D for some ϵ and so f ∈ H^{ϵ} ≤ H_{J}^{+}. Hence f_{I} = 1 and f ∈ H_{r}^{+}. So indeed ⋂ D = H_{r}^{+}.

Let R be the maximal element of R. If H_{R}^{+} ∈ R, then ⋃ D = H_{R}^{+} and if H_{R}^{+} ∉ R then H_{R}^{−} ∈ R and ⋃ D = H_{R}^{−}.

So S is closet under union and intersection. Next let (H_{J}^{ϵ}, H_{I}^{μ}). be the a jump of S. We need to show that (H_{J}^{ϵ}, H_{I}^{μ}) = (H_{K}^{−}, H_{K}^{+}) for some K ∈ I. If μ = +, then H_{J}^{ϵ} ≤ H_{I}^{−} < H_{I}^{+} and so (H_{J}^{ϵ}, H_{I}^{μ}) = (H_{I}^{−}, H^{+}I). If ϵ = −, then H_{J}^{−} < H_{J}^{+} ≤ H_{I}^{μ} and so (H_{J}^{ϵ}, H_{I}^{μ}) = (H_{J}^{−}, H^{+}J). So suppose ϵ = + and μ = −. Then J ≺ I. Suppose there exists K ∈ I with J < K < I. Then H_{J}^{ϵ} ≤ H_{K}^{−} < H_{K}^{+} ≤ H_{I}^{−} and so (H_{J}^{ϵ}, H_{I}^{μ}) = (H_{K}^{−}, H^{+}K). So we may assume that no such K exist. But then H_{J}^{+} = H^{−}I, a contradiction.

countjor Example 5.3 There exists a elementary abelian p-groups which has a composition series countable many factors and another with uncountable many factors.

Proof: Let A = 𝔽_{p}^{ℕ}. Then by 5.2 A has a composition series with countable many factors. On the otherhand, A is a vector space over 𝔽_{p} and so A is also a direct sum of copies of 𝔽_{p}. As A is countable this has to be a uncountable sum and so again by 5.2, A has an uncountable series.

incs Proposition 5.4 Let H be acting an the group X and let C be a H-series on X. Let M be an H-invariant subgroup of X.

(a) M ∩ C = {M ∩ C | C ∈ C is a H series on X. If M is normal in X and C is an composition series, then M ∩ C is an H-series whose factors are H-isomorphic to a factors of jumps (B^{∗}, T^{∗}) of C with T^{∗} ∩ M ≠ B^{∗} ∩ M.

(b) Suppose that M is normal in X and M fulfills the descending chain condition on Hserial subgroups. Then CM/M = {CM/M | C ∈ C} is an H-series on X/M. If D is a composition series, then DM/M is a composition series whose factors H-isomorphic to a factors of jumps (B^{∗}, T^{∗}) of C with T^{∗} ∩ M = B^{∗} ∩ M

Proof: (a) It is readily verified that M ∩D contains 1, M and is closed under intersections and unions. Let (B, T) be a jump of M ∩ D. Put B^{∗} = ⋃{C ∈ C | M ∩ C = B} and T_{∗} = ⋂{C ∈ C | M ∩ C = T }. Then clearly M ∩ B^{∗} = B, M ∩ T_{∗} = T and (B^{∗}, T^{∗}) is a jump of C. Thus B^{∗} ⊴ T_{∗} and so B ⊴ T. Moreover, M ∩ B^{∗} = (M ∩ T_{∗}) ∩ B^{∗}, T/B = M ∩ T_{∗}/M ∩ B^{∗} ≅ (M ∩ T_{∗})B^{∗}/B^{∗} ≤ T_{∗}/B^{∗}. If C is a composition series the last ” ≤ ” actually is an equality and the lemma holds.

(b) It is readily verified that DM/M contains 1, X/M and is closed under unions. Let D ⊆ C. Let E = ⋂{DM | D ∈ D and R = ⋂ D.

Claim E = RD.

By the descending chain conditions, D ∩M has a minimal element. Clearly this element is R ∩ M. So we may assume without loss that R ∩ M = D ∩ M for all D ∈ D. Let D ∈ D. Then M ≤ E ≤ DM and so E = (D ∩ E)M. If D^{∗} ≤ D with D^{∗} ≤ D, then D^{∗} ∩ E ≤ D ∩ E ≤ R = (D^{∗} ∩ E)M and so D ∩ E = (D ∩ E^{∗})(D ∩ M). But D ∩ M = R ∩ M ≤ (D^{∗} ∩ M) and so D ∩ E ≤ D^{∗} ∩ E) ≤ D^{∗}. Since this is true for all D^{∗}, D ∩ E ≤ R. Clearly R ≤ E and so E = (D ∩ E)M = RM.

By the claim the intersection property holds for CM/M. Next let (B/M, T/M) be a jump of CM/M. Let B^{∗} = ⋃{C ∈ C | CM = B} and T^{∗} = ⋃{C ∈ C | CM = T }. Then B = B^{∗}M and by the claim T = T^{∗}M. In particular, B/M ⊴ T/M and CM/M is an H-series. Moreover,

T/B = T^{∗}M/B^{∗}M ≅ T^{∗}/(B^{∗}M ∩ T^{∗}) = T^{∗}/B^{∗}(M ∩ T^{∗}).

Thus T/B is isomorphic to a quotient of T^{∗}/B^{∗}. If D is a composition series then T/B ≅ T^{∗}/B^{∗}. Conversely (B^{∗}, M^{∗}) is a jump with B^{∗} ∩ M ≠ T^{∗} ∩ M, then clearly (B^{∗}M/M, T^{∗}M/M) is jump of DM/M with factor T^{∗}/B^{∗}(M ∩ T^{∗}).

Lemma 5.5 Let D be a subnormal H-series on X and that all factors of H have the mibymi minimal conditions on H-serial subgroups. Then also X has the minimal condition on H-serial subgroups.

Proof: By induction on |D| we may assume that D = {1, M, X}. Let S be a set of serial subgroups of X. By 5.4 S ∩ M and SM/M are sets of serial subgroups of M and X/M, respectively. Hence these sets of minimal elements say A and B/M. Without loss S ∩ M = A for all S ∈ S and then also SM = B for all S ∈ S. We claim that then all elements of S are minimal. Indeed, let R, S ∈ S with R ≤ S. Then S ≤ RM = B = RM and so S = R(S ∩ M) = RA = R(R ∩ M) = R.

Using 5.5 the follwing corollary follows easily from 5.4 and induction:

Corollary 5.6 Suppose H acts on X and D is subnormal H series all of whose factors jh have the minimum conditions an H-serial subgroups. Then every H-composition series on X has isomorphic factors as some composition series containing D. In particular, if the Jordan Hölder theorem holds for all factors of D it holds for X.

Proposition 5.7 Let E be a class of finite groups, which is closed with respect to sub- bacae group,quotients and extensions. Call G to E-perfect if no non-trivial quotient of G is locally E.

(a) Let O_{E}(G) be the subgroup of G generated by all the serial locally E(G) subgroup. Then O_{E}(G) is locally E (b) Let O^{E}(G) be the intersection of all normal subgroups of G whose quotient is locally

E. Then G/O^{E}(G) is locally E and G = ⟨O^{E}(F)|F ∈ F.

(c) O^{E}(G) is E-perfect.

(d) Let C be series on G and H ∈ C. If all the factors of C above H are locally E, then O^{E}(G) = O^{E}(H).

(e) If M ⊴ G, then O^{E}(G/M) = O^{E}(G)M/M

(f) If H ≤ G, then O^{E}(H) ≤ O^{E}(G)

Proof: If G is finite, all statements are readily verified. The general case follows from the finite one via the local system F.

dces Definition 5.8 Let E be a class of finite groups, which is closed with respect to subgroup, quotients and extensions.

(a) A E-series on G is a set of subgroups calS such that

(a) 1 ∈ C and O^{E}(G) ∈ C.

(b) All members of S are E-perfect.

(c) If S is a chain.

(d) If D ⊆ S, then ⋃ D ∈ S and O^{E}(⋃ D) ∈ S.

(e) For D ∈ S put D_{i} = ⋃{E ∈ S | E < D}. Then D^{−} ⊴ D for all D ∈ S.

(b) If D ≠ D^{−} for some D ∈ S, then (D^{−}, D) is called a jump of S. Let D_{−}/D^{−} = O_{E}(D/D^{−}). Then D/D^{−} a factor of S.

(c) An calE- composition series is an E-series so that none of the factors has a proper H-invariant normal subgroup.

isces Lemma 5.9 Let E be a class of finite groups, which is closed with respect to subgroups, quotients and extensions. Let S be a series on G. Then O^{E}(S) = {O^{E}(S) | S ∈ S} is a E series whose factors are O^{E}(Q)/O^{E}(Q) ∩ O_{E}(Q), Q a factor of S. If S is a composition series, then the factors of O^{E}(S) are precisely the ≠ E factors of S.

Proof: Clearly part (a),(b) and (c) of the definition of a E-series holds for O^{E}(S)hold. Let D ⊆ S.

Claim O^{E}(⋃ D) = ⋃{O^{E}(D) | D ∈ D} and O^{E}(⋂ D) = O^{E}(⋂{O^{E}(D) | D ∈ D}).

Indeed the first part is immediate from 5.7b. For the second let T = ⋂ D and B = ⋂{O^{E}(D) | D ∈ D}. Then clearly B is a normal subgroups of T and O^{E}(T) ≤ B. Thus O^{E}(T) = O^{E}(T). So also the second statement holds.

By the claim O^{E}(S) is closed under union and ”E”- intersections. Also the jumps of O^{E}(S) are clearly of the form (O^{E}(B), O^{E}(T)) for a jump (B, T) of S. The claim about the factors is now readily verified. jhce Proposition 5.10 Let E be a class of finite groups, which is closed with respect to sub-

groups, quotients and extensions. Then 5.4 and 5.6 hold for locally finite groups if H-series, H-serial, H-composition series is replaced by E-series, E-serial, E-composition series.

With minor modification the proof for 5.4 goes through.

Corollary 5.11 Let E be a class of finite groups, which is closed with respect to subgroups, fmccf quotients and extensions. Suppose that G has a absolute composition series C with only finitely many factors which are ¬E. Then the Jordan-Hölder Theorem holds for the ¬Efactors on G.

By 5.9 has a finite E-composition series. Thus the claim follows from the E version of 5.6.

Lemma 5.12 Let G be a locally finite group with a composition series with only finitely fmnac many non-abelian factors. Then LSol(G)E(G) ≠ 1.

Proof:

By 5.9 G has a finite ”solvable” series. The lowest term of the series is either locally solvable or a component.

Lemma 5.13 Let G be countable such that G has some absolute composition series with jhls only finite many non-abelian factors. Then the Jordan-Hölder theorem holds for G.

Proof: Each composition factor is non abelian or C_{p} for a prime p. By 5.11 applied to the class of finite solvable groups, the Jordan Hölder theorem holds for the non-abelian factors. Suppose that G has some composition series with only finitely many factors C_{p}. The all but finitely many composition factors are C_{q} for q ≠ p. Hence G has only finitely many ¬p′-factors. Hence the Jordan Hölder theorem holds for the ≠ p′ factors and in particular for the C_{p}-factors. If each composition series has infinitely many factors C_{p} then as G is countable, this number is countable infinite and again the Jordan Hölder theorem holds for the C_{p}-factors. .

24CHAPTER 5. ON THE JORDAN HÖLDER THEOREM FOR LOCALLY FINITE GROUPS

Absolute Simplicity

Absolute

Definition 6.1 A group R is called absolutely simple if {1, R} is the only series on R.

The main result in this section is due to R.E Phillips. It gives a sufficient conditions for a locally finite group to be absolutely simple. The importance of this conditions is that it can be verified locally.

Definition 6.2 ddefect

(a) Let N be subnormal in G. Then the defect d_{G}(N) of N in G is smallest length of a subnormal series from N to G.

(b) Let H ≤ G. Then NC_{G}^{0}(H) = G and inductively define NC_{G}^{i+1}(H) = ⟨H^{NC_{G}^{i}(H)}⟩.

(c) Let H ≤ G. Then LSC_{G}(H) is the intersection of the locally subnormal subgroups of G containing H. In other words, LSC_{G}(H) is the smallest locally subnormal subgroup of G containing H.

Lemma 6.3 Let N be subnormal in G and d = d_{G}(N). Then dgn

N = NC_{G}^{d}(N) ⊲ NC_{G}^{d−1}(N) ⊲ … ⊲ NC_{G}^{1}(N) ⊲ NC_{G}^{0}(N) = G

is a subnormal series of length d from N to G.

Proof: Let N = N_{d} ⊲ N_{d−1} ⊲ … ⊲ N_{1} ⊲ N_{0} = G be a subnormal series of length d from N to G. If d = 0 (that is G = N) there is nothing to prove. So suppose d ≥ 1. The NC_{G}^{1}(N) = ⟨N^{G}⟩ ≤ ⟨N_{1}^{G}⟩ = N_{1} and with the same argument and using induction we get that NC_{G}^{i}(N) ≤ N_{i} for all i ≥ 0. In particular, N ≤ NC_{G}^{d}(N) ≤ N_{d} = N and so NC_{G}^{d}(N) = N.

Lemma 6.4 Let N ≤ G. Then LSC_{G}(N) = ⋃_{F ∈F} LSC_{F} (F ∩ N) scgn

Proof: Let R be the group on the right hand. Let F ∈ F and F ≤ E ≤ G. Note that F ∩ N ≤ E ∩ N ≤ LSC_{E}(E ∩ N) and that the latter group is locally subnormal in E. Thus F ∩ N ≤ F ∩ LSC_{E}(E ∩ N) and F ∩ LSC_{E}(E ∩ N) is subnormal in F. Thus by definition of LSC_{F} (F ∩ N) we get LSC_{F} (F ∩ N) ≤ F ∩ LSC_{E}(E ∩ N) ≤ LSC_{E}(E ∩ N).

Choosing E = G we conclude that R ≤ LSC_{G}(N).

Choosing E to be finite we see that we can apply 4.5. Thus R is locally subnormal in G. Clearly N ≤ R and so by definition of LSC_{G}(N), LSC_{G}(N) ≤ R.

clsnd Lemma 6.5 Let {L_{i} | i ∈ I} be a local system for G, and for i ∈ I let N_{i} be a subgroup of L_{i}. Suppose that there exists a non-negative integer d so that

(a) For all i ∈ I, N_{i} is subnormal of defect at most d in L_{i}.

(b) For all i ≤ j ∈ I, N_{i} ≤ N_{j}.

Then N = ⋃_{i∈I} N_{i} is a subnormal subgroup of defect at most d in G.

Proof: An easy induction proof shows that for all t ≥ 0, NC_{G}^{t}(N) = ⋃_{i∈I} NC_{L_{i}}^{t}(N_{i}). By 6.3 NC_{L_{i}}^{d}(N_{i}) = N_{i} and so NC_{G}^{d}(N) = N. Thus N is subnormal in G and has defect at most d.

lsr Lemma 6.6 Let LSol(G) be the subgroup of G generated by all the locally solvable, locally

subnormal subgroups of G. Then LSol(G) itself is locally solvable and locally subnormal. In particular LSol(G) is the unique maximal locally solvable, normal subgroup of G.

Proof: Suppose first that G is finite. Let A and B solvable normal subgroups of G. We claim that AB is a solvable normal subgroup of G as well. Indeed let i and j be the derived length of A and B respectively. Then (AB)^{(j)} ≤ AB^{(j)} ≤ A and so (AB)^{(j+i)} ≤ A^{(i)} = 1. Thus AB is solvable and clearly also normal in G. Hence a finite group G has a unique largest solvable normal subgroup Sol(G). We claim that Sol(G) = LSol(G). For this let A be any solvable, subnormal subgroup of G and A ⊴ N_{1} ⊲ N_{2} ⊲ … ⊲ N_{d} = G a subnormal series from A to G. Then A ≤ Sol(N_{1}). Since Sol(N_{1}) is a characteristic subgroup of N_{1}, Sol(N_{1}) is normal in N_{2} and

A ≤ Sol(N_{1}) ≤ Sol(N_{2}) ≤ Sol(N_{3}) ≤ … ≤ Sol(N_{d}) = Sol(G).

In the general case let N be the set of locally solvable, locally subnormal subgroups of G. Then LSol(G) = ⟨N ⟩. Let F ∈ F. Then N ∩ F is a set of solvable, subnormal subgroups of F and so by the finite case ⟨N ∩ F ⟩ is a solvable subnormal subgroup of F. Since {⟨N ∩ F ⟩|F ∈ F} is a local system for LSol(G), LSol(G) is indeed a locally solvable, locally subnormal subgroup of G.

bdas Theorem 6.7 Let G be a locally finite, simple group and L a local system for G. Suppose

that there exists a positive integer d with the following property:

If L ∈ L and M is an absolutely perfect, locally subnormal subgroup of L, then M is subnormal in L with defect at most d.

Then G is absolutely simple.

Proof: Let N be a non-trivial serial subgroup of G. We need to show that N = G. By 4.8 N is locally subnormal in G.

Suppose first that N is locally solvable. Then 1 ≠ N ≤ LSol(G). Since G is simple we conclude that G = LSol(G) and so by 6.6 G is locally solvable. But then by 3.4c G ≅ C_{p}. Clearly N = G in this case.

Next suppose that N is not locally solvable. Then N^{∞} ≠ 1. Moreover, it is easy to see that {(N ∩ L)^{∞} | L ∈ L} is a local system for N^{∞}. By assumption (N ∩ L)^{∞} is subnormal of defect at most d in L and so by 6.5, 1 ≠ N^{∞} is subnormal of defect at most d in G. Since G is simple, G has no proper subnormal subgroups. Thus G = N^{∞} = N.

28 CHAPTER 6. ABSOLUTE SIMPLICITY

Simple locally finite groups which are not absolutely simple

Simple

In this section we will construct the only known examples of simple locally finite groups which are not absolutely simple.

Lemma 7.1 Let H be a group, S a group acting faithfully and transitively on a set I, nsowp K = H ≀_{I} S, where ≀ stands for the restricted wreath product. Then any normal subgroup of K either is contained in the base group B(K) or it contains B(K)′.

Proof: Let N be a normal subgroup of K not contained in B(K) = ∏_{i∈I} H_{i}. Pick n ∈ N \ B(K). Since K/B(K) ≅ S, K acts on I with kernel B(K) and H_{i}^{n} = H_{i^{n}}. Pick i ∈ I with i ≠ i^{n}. Let a, b ∈ H_{i}. Then aa^{−n} = [a^{−1}, n] ≤ [K, N] ≤ N. Since a^{−n} ≤ H_{i}^{n} we have [a^{−n}, b] = 1 and so [a, b] = [aa^{−n}, b] ≤ [N, K] ≤ N.

Since a and b are arbitrary we conclude H′_{i} ≤ N. And since K is transitive on I we get H′_{j} ≤ N for all j ∈ I. Thus B(K)′ ≤ N.

Lemma 7.2 Let H be a perfect finite group. Then there exists a perfect finite group H^{∗} containing H and function X which associates to each subgroup A of H a subgroup X(A) of H^{∗} such that

(a) H ≤ ⟨h^{H^{∗}}⟩ for all 1 ≠ h ∈ H.

(b) X(A) ∩ H = A for all A ≤ H.

(c) If A ≤ B ≤ H, then A ⊴ B if and only if X(A) ⊴ X(B).

(d) X(H) ⊴ ⟨X(H)^{H^{∗}}⟩.

Proof: Let S be any finite simple group such that there exists a monomorphism α : H → S and let T be any non trivial finite perfect group. Furthermore, let S and T act transitively and non-trivially on the sets I and J, respectively. We assume that 0 ∈ I and {0, 1} ⊆ J. Let K = H≀_{I}S. For i ∈ I let β_{i} : H → K be the canonical isomorphism between H and the i′th component of the base group of K and let β be the canonical 30CHAPTER 7. SIMPLE LOCALLY FINITE GROUPS WHICH ARE NOT ABSOLUTELY SIMPLE

monomorphism from S to K. Let H^{∗} = K≀_{J}T and for j ∈ J let γ_{j} : K → H^{∗} be the canonical isomorphism between K and the j′th component of the base group of H^{∗}. Note that as H, S and T are perfect also K and H^{∗} are perfect. Define ρ : H → H^{∗} by ρ(h) = γ_{0}(β_{0}(h))γ_{1}(β(α(h))). Then ρ is clearly a monomorphism. For A ≤ H let X(A) be the set of elements in the base group of H^{∗} such that the projection onto the 0′th-component is contained in γ_{0}(∏_{i∈I} β_{i}(A)). Identifying H with ρ(H) we see immediately that (b) and

(c) hold. Now ⟨X(H)^{H^{∗}}⟩ is the base group of H^{∗} and so (d) holds. To prove (a) let h ∈ H^{#}. Put N = [h, γ_{1}(K)]. Then N is a normal subgroup of γ_{1}(K) and is not contained in the base group of γ_{1}(K). Thus by 7.1 N contains γ_{1}(B(K)). Furthermore, K/B(K) ≅ S is simple and so N = γ_{1}(K). Since N ≤ ⟨h^{H^{∗}}⟩ and H^{∗} acts transitively on J, we get γ_{j}(K) ≤ ⟨h^{H^{∗}}⟩ for all j ∈ J. Hence ⟨h^{H^{∗}}⟩ is the base group of H^{∗} and so contains H.

enaslfsg Theorem 7.3 There exists a locally finite, simple group G with an ascending chain

1 ⊲ M_{1} ⊲ M_{2} ⊲ M_{3} …

of subgroups such that G = ⋃_{i=1}^{∞} M_{i}. In particular, G is not absolutely simple.

Proof: Let G_{1} be any nontrivial perfect finite group, and inductively let G_{i+1} = G_{i}^{∗} and X_{i} any function from the subgroups of G_{i} to the subgroups of G_{i+1} which fulfills 7.2. Let G = ⋃_{i=1}^{∞} G_{i}. Then G is a locally finite group and we claim that G is simple. Indeed, let x, y ∈ G with x ≠ 1. Pick i with x, y ∈ G_{i}. Then by (a) in 7.2

y ∈ G_{i} ≤ ⟨x^{G_{i+1}}⟩ ≤ ⟨x^{G}⟩.

Thus ⟨x^{G}⟩ = G and G is simple.

Put M_{1,1} = 1, M_{1,2} = G_{1} and inductively, M_{n+1,j} = X_{n}(M_{n,j}), for 1 ≤ j ≤ 2n, M_{n+1,2n+1} = ⟨X(G_{n})^{G_{n+1}}⟩ and M_{n+1,2n+2} = G_{n+1}. Then by induction and 7.2, M_{n+1,i} ⊲ M_{n+1,i+1} for all 1 ≤ i ≤ 2n + 1 and M_{n+1,i} ∩ G_{n} = M_{n,i} for all 1 ≤ i ≤ 2n. Put M_{i} = ⋃_{n≥ \frac{i}{2}} M_{n,i}. Then G_{n} ≤ M_{2n}, G_{n} ∩ M_{i} = M_{n,i} for all i ≤ 2n, M_{i} ⊲ M_{i+1} and G = ⋃_{i=1}^{∞} M_{i}.

Conjugacy centralizer property

Conjugacy

Definition 8.1 Let N ≤ G. ccp (a) N has the conjugacy centralizer property if for all F ∈ F there exists n ∈ N with

[F, F^{n}] = 1.

(b) N has the centralizer property if F ≤ C_{G}(F)N for all F ∈ F. Lemma 8.2 Let N be a normal subgroup of G with the conjugacy centralizer property. ccpcp Then N has the centralizer property.

Proof: Let f ∈ F ∈ F and pick n ∈ N with [F, F^{n}] = 1. Then

f = n^{−1}fnn^{−1}f^{−1}nf = f^{n}n^{−1}n^{f} ≤ F^{n}N ≤ C_{G}(F)N. Lemma 8.3 Let N ≤ G have the centralizer property. Then cp (a) If M ≤ G is N-invariant, then M ⊴ G.

(b) N is normal in G.

(c) If A ⊴ N, then A ⊴ G.

(d) G′ ≤ N.

Proof: (a) Let m ∈ M and g ∈ G. Then ⟨g, m⟩ ≤ C_{G}(⟨g, m⟩)N and so we can write g = cn with c ∈ C_{G}(m) and n ∈ N. Then m^{g} = m^{n} ∈ M and so M ⊴ G.

(b) and (c) follows from (a).

(d) Let F ∈ F. Then F N/N ≤ C_{G}(F)N/N ≤ C_{G/N}(F N/N). Thus G/N is locally abelian and so abelian.

32 CHAPTER 8. CONJUGACY CENTRALIZER PROPERTY

Transitive groups of finitary permutations

Groups

The purpose of this section is to describe all transitive groups of finitary permutations. Definition 9.1 Let G be acting on a set Ω and g ∈ G. finitaryp (a) Supp_{Ω}(g) = {ω ∈ Ω | ω ≠ ω^{g}}.

(b) The degree deg_{Ω}(g) of g on Ω is defined as | Supp_{Ω}(g)|.

(c) G acts finitarily on Ω provided that all elements in g have finite degree on Ω.

(d) A block system (also called system of imprimitivity) for G on Ω is a G-invariant proper partition D of G. (here G-invariant means that Δ^{g} ∈ D for all g ∈ G and proper means D ≠ Ω and D ≠ {Ω}.

(e) A block (also called a set of imprimitivity) for G, is a proper subset Δ of Ω such that for each g ∈ G, Δ^{g} = Δ or Δ ∩ Δ^{g} = ∅ holds.

(f) If G has a block system on Ω, G is called imprimitive. Otherwise G is primitive.

(g) G acts almost primitively on Ω if G acts primitively on Ω or acts transitively on Ω and primitively on some block system for G on Ω.

(h) G acts totally imprimitively on Ω if there exists an ascending chain of finite blocks

for G on Ω with

Δ_{1} ⊂ Δ_{2} ⊂ Δ_{2} ⊂ …

such that Ω = ⋃_{i=1}^{∞} Δ_{i}.

(i) Let D be a partition of Ω. Then m(D) := inf_{Δ∈D} |Δ|.

(j) Let Δ ⊆ Ω. G_{Δ} = C_{G}(Δ) = {g ∈ G | ω^{g} = ω for all ω ∈ Δ}. G^{Δ} = N_{G}(Δ)/G_{Δ} ≤ Sym(Δ).

We remark that if G acts on Ω, G also acts on the set of subsets of Ω. In particular, if D is a set of subsets of Ω, then C_{G}(D) denotes the group {g ∈ G | D^{g} = D for all D ∈ D}. The easy proofs of the next three lemmas are left to the reader.

ubeb Lemma 9.2 Let G be acting on the set Ω and let B be a chain of blocks. Then ⋃ B is a block or equals Ω.

premx Lemma 9.3 Let G be acting transitively on the set Ω and fix ω ∈ Ω. Then the map H → ω^{H} is a one-to-one correspondence between subgroups G_{ω} < H < G and blocks containing ω.

In particular, G is primitive if and only if G_{ω} is a maximal subgroup of G.

tiec Lemma 9.4 Totally imprimitive groups of finitary permutations are countable.

sip Lemma 9.5 Let G be transitive and finitary on Ω. Let D be a block system for G on Ω and g ∈ G.

(a) If g acts non-trivially on D, then m(D) ≤ \frac{1}{2} deg(g).

(b) Put N = ⟨g ∈ G | deg(g) < 2 m (D)⟩. Then N ≤ C_{G}(D).

(c) All blocks for G on Ω are finite.

Proof: (b) and (c) follow immediately from (a). For (a) pick Δ ∈ D with Δ^{g} ≠ Δ. As Δ is a block Δ ∩ Δ^{g} = ∅. Thus g fixes none of the elements in Δ and also none of the elements of Δ^{g}. Hence Δ ∪ Δ^{g} ⊂ Supp(g) and so deg(g) ≥ 2|Δ|. As G is transitive on Ω and on D, |Δ| = m(D) and so also (a) is proved.

apoti Theorem 9.6 Let G be a transitive group of finitary permutations. Then G is either almost primitive or totally imprimitive.

Proof: We may assume that G is imprimitive and does not have a maximal block. Then each block is properly contained in a larger block and so there exists an infinite chain

Δ_{1} ⊂ Δ_{2} ⊂ Δ_{3} ⊂ …

of blocks for G on Ω. By 9.5c, all Δ_{i}’s are finite. By 9.2, ⋃_{i=1}^{∞} Δ_{i} is a block or equals Ω. The first case is impossible as ⋃_{i=1}^{∞} Δ_{i} is infinite. So ⋃_{i=1}^{∞} Δ_{i} = Ω and G is totally imprimitive.

The next lemma follows easily from 9.5 and we leave the details of the proof to the reader. dti Proposition 9.7 Let G be a totally imprimitive and transitive subgroup of FSym(Ω) and

let Ξ be a chain of blocks for G on Ω with Ω = ⋃ Ξ.

(a) Let A be a finite subset of Ω. Then there exists a Δ ∈ Ξ with A ⊆ Δ.

(b) For a block Δ put N(Δ) = C_{G}(Δ^{G}). Then G = ⋃_{Δ∈Ξ} N(Δ).

(c) Let Δ be a block for G. Then N(Δ) is a sub-direct (restricted) product of the isomorphic finite groups N(Δ)/C_{N(Δ)}(Λ), Λ ∈ Δ^{G}.

We remark that if G is a non-primitive, transitive group of finitary permutations either each block lies in a maximal block, or G is totally imprimitive and has no maximal blocks. Indeed this follow from 9.7 and the proof of 9.6.

Lemma 9.8 Let X be a group acting on the set Ω. Let A and B be finite subsets of Ω and disjoint suppose that each orbit of X on Ω has size larger than |A||B|. Then there exists a g ∈ X with A^{g} ∩ B = ∅.

Proof: Case 1 All orbits of X have infinite length.

The prove is by induction on |A|. If A = ∅, there is nothing to prove. So assume the result for any set A′ with |A′| < |A|. For a contradiction suppose that no element g with the required property exists. That is, each conjugate of A meets B (non-trivially).

Claim For any subset C of Ω with |C| ≤ |A|, only finitely many conjugates of A contain C.

The claim is proved by induction on |A| − |C|. It starts trivially when |A| − |C| = 0. So suppose that |C| < |A| and that the claim holds for all |C′| with |C| < |C′| ≤ |A|. By the first induction we may assume (conjugating C if necessary) that C ∩ B = ∅. By the second induction for each of the finitely many points b ∈ B, only finitely many conjugates of A contain C ∪ {b}. So only finitely many conjugates of A contain C and meets B. By assumption each conjugate of A meets B and so the claim is proved.

Now taking C = ∅ in the claim we see that A has only finitely many conjugates. But this is a clear contradiction to the hypothesis in Case 1 that X has no finite orbits.

Case 2 All orbits of X on Ω are finite.

Let Ω_{1}, …, Ω_{r} be the orbits that meet A. Then

∑_{i=1}^{r} |A ∩ Ω_{i}| = |A|

|Ω_{i}| > |A| · |B|

Since X acts transitively on Ω_{i} each element of Ω_{i} lies in the same number (say u_{i}) of conjugates of A. Also each conjugate of A meet Ω_{i} in the same number of elements as A, that is |A ∩ Ω_{i}|. Counting the pairs (ω, D) with ω ∈ D and D ∈ A^{X} in two different ways we get |Ω_{i}| · u_{i} = |A^{X}||A ∩ Ω_{i}|. Thus u_{i} = \frac{|A^{X}||A∩Ω_{i}|}{|Ω_{i}|}. This means that there exist at

most u_{i}|Ω_{i} ∩ B| = \frac{|A^{X}||A∩Ω_{i}||B∩Ω_{i}|}{|Ω_{i}|} conjugates of A which meet B ∩ Ω_{i} non-trivially. Since

|B ∩ Ω_{i}| ≤ |B| and |Ω_{i}| > |A| · |B| we have \frac{|A^{X}||A∩Ω_{i}||B∩Ω_{i}|}{|Ω_{i}|} < \frac{|A^{X}||A∩Ω_{i}|}{|A|}. Summing over all i we see that there exist less than

∑_{i=1}^{r} \frac{|A^{X}||A ∩ Ω_{i}|}{|A|} = \frac{|A^{X}|}{|A|} ∑_{i=1}^{r} |A ∩ Ω_{i}| = \frac{|A^{X}|}{|A|}|A| = |A^{X}|

conjugates of A that meet B non-trivially. But this means that some conjugate of A meets B trivially. This completes the proof in Case 2.

General Case

Let Ω_{0} be the union of the infinite X-orbits and Ω_{1} the union of the finite X orbits. Let A_{i} and B_{i} be the intersection of Ω_{i} with A and B, respectively. By Case 2 applied to Ω_{1} there exists x ∈ X with A_{1}^{x} ∩ B_{1} = ∅. Let Y = C_{X}(A_{1}^{x}). Since A_{1}^{x} is finite and a^{X} is finite for all a ∈ A_{1}^{x}, Y has finite index in X. Thus the orbits for Y on Ω_{0} are still infinite and we can apply Case 1 to Y and Ω_{0} to get y ∈ Y with A_{0}^{xy} ∩ B_{0} = ∅. But A_{1}^{xy} = A_{1}^{x} and so A_{1}^{xy} ∩ B_{1} = ∅ and A^{xy} ∩ B = ∅. This completes the proof of the proposition also in the general case.

We remark that the bound |A||B| in the previous lemma is the best possible. To see this, let C and D be any finite sets and X = Sym(C) × Sym(D) ≤ Sym(C × D). Let c ∈ C, d ∈ D, A = C × {d} and B = {c} × D. Then the conjugates of A under X are of the form C × {d′} with d′ ∈ D, so they meet B in (c, d′).

ntnsg Lemma 9.9 Let G be transitive and finitary on the infinite set Ω.

(a) G is neither abelian nor finite.

(b) Let N be an intransitive normal subgroup of G. Then all orbits for N on Ω are finite. Moreover G/N acts transitively and finitarily on the infinite set of orbits of N on Ω. In particular, G/N is neither finite nor abelian.

Proof: (a) G cannot be finite. If G is abelian, let g be an element acting non-trivially on Ω. Then Supp_{Ω}(g) is a finite G-invariant set, a contradiction.

(b) Since the orbits of N form a system of imprimitivity, each of the orbits is finite. As Ω is infinite, N has an infinite number of orbits. The remaining assertions now are readily verified.

trnope Theorem 9.10 Let G be transitive and finitary on the infinite set Ω.

(a) G′ acts transitively on Ω.

(b) Let N be a transitive, subnormal subgroup of G. Then G′ ≤ N and and N is normal.

(c) G′ is perfect and has no proper subgroups of finite index.

Proof: (a) Since G/G′ is abelian, 9.9(b) implies that G′ is transitive.

(b) By induction on the defect of N in G we may assume that N is normal in G. Thus in view of 8.2 and 8.3 it remains to verify that N has the conjugacy centralizer property. For this let F ∈ F and put Δ = Supp(F). As G is finitary, Δ is finite. So by 9.8 applied with X = N and A = B = Δ, there exists n ∈ N with Δ ∩ Δ^{n} = ∅. Since Supp(F^{n}) = Δ^{n}, this clearly implies [F, F^{n}] = 1. Thus N has the conjugacy centralizer property and (b) holds.

(c) By (a) G′′ is transitive and so by (b) G′ ≤ G′′. Thus G′ is perfect. Let N be a subgroup of finite index in G′. Without loss, N is normal in G′. Then N is transitive and again by 9.9b G′ ≤ N.

Lemma 9.11 Let X be a group acting primitively on a set Ω and let a ∈ Ω. pj

(a) Suppose that X_{a} has k orbits on Ω and X_{a} has an orbit of length m on Ω − a. If k and m are finite, then |Ω| ≤ ∑_{i=0}^{k−1} m^{i}.

(b) If g ∈ X acts non-trivially on Ω then the number of orbits of X_{a} on Ω is at most deg(g).

Proof: (a) Let a ≠ b ∈ Ω with |b^{X_{a}}| = m. Define a directed graph on Ω by α → β if and only if (α, β) = (a^{x}, b^{x}) for some x ∈ X. Let Δ(a) be the set of elements in Ω that can be reached from a via a directed path. Let Δ_{i}(a) be the set of elements in Δ(a) that are of distance i from a. Then |Δ_{0}(a)| = |{a}| = 1, |Δ_{1}(a)| = |b^{X_{a}}| = m and by induction |Δ_{i+1}(a)| ≤ m|Δ_{i}(a)| ≤ m^{i+1}. Clearly each Δ_{i}(a) is a union of X_{a}-orbits and so at most k of the Δ_{i}(a)’s are not empty. It follows that

|Δ(a)| = ∑_{i=0}^{k−1} |Δ_{i}(a)| ≤ ∑_{i=0}^{k−1} m^{i}.

In particular, Δ(a) is finite. If c ∈ Δ(a) then clearly Δ(c) ⊆ Δ(a). Since X is transitive, |Δ(a)| = |Δ(c)| and, since Δ(a) is finite, we conclude Δ(a) = Δ(c). But this implies that Δ(a) is a block and, as X is primitive, we conclude Δ(a) = Ω. Thus (a) holds.

(b) Without loss g does not fix a. Since X is primitive, X_{a} is by 9.3 a maximal subgroup of X. Thus X = ⟨X_{a}, g⟩ and so g does not normalize any of the orbits of X_{a} on Ω. But then each orbit of X_{a} meets Supp(g) non-trivially and so there are at most deg(g) orbits.

Theorem 9.12 Let X be a primitive subgroup of Sym(Ω) and suppose that X contains a jordan non-trivial element of finite degree d and that

|Ω| > ∑_{i=0}^{d−1}(d − 1)^{2i}.

Then Alt(Ω) ≤ X.

Proof: Let 1 ≠ g ∈ X with deg(g) = d. By assumption |Ω| > d and so there exists a ∈ Ω with a = a^{g}. By 9.11b, X_{a} has at most d orbits on Ω. If X_{a} has an orbit of length less or equal to (d − 1)^{2} on Ω − a, then 9.11a implies |Ω| ≤ ∑_{i=0}^{d−1}(d − 1)^{2i}, a contradiction.

Hence all orbits of X_{a} on Ω − a have length larger than (d − 1)^{2}. Let t be a conjugate of g in X with a ≠ a^{t}, that is a ∈ Supp(t). Then we can apply 9.8 to X = X_{a}, Ω = Ω − a and A = B = Supp(t) − a to obtain h ∈ X_{a} with (Supp(t) − a) ∩ (Supp(t) − a)^{h} = ∅. As h fixes a, Supp(t) ∩ Supp(t)^{h} = {a}. A straight-forward calculation now shows that [t, t^{h}] is a 3-cycle. Alt(Ω) ≤ X is now a consequence of exercise 14.

psio Corollary 9.13 Let G be a primitive subgroup of FSym(Ω) with Ω infinite. Then G = Alt(Ω) or G = FSym(Ω).

sifipu Corollary 9.14 Let G be an infinite, simple group of finitary permutations. Then G is an alternating group.

Proof: Let G ≤ FSym(Omega). Since G is simple, it acts faithfully on each of its orbits. So we may assume that G is transitive. Suppose that G is is totally imprimitive. Then by 9.7b, G is the union of proper normal subgroups, a contradcition to the simplicity of G. Hence by 9.6 G is almost primitive on Ω. Thus G is primitive or acts primitively on some bock system D. In the latter case, the simplicity forces G to be faithful on D and so in any case G is a primitive group of finitary permutations. As G is infinite we can apply 9.13 to conclude that G is an alternating or a symmetric group. As G is simple, G is alternating.

Proposition 9.15 Let G be an almost primitive subgroup of FSym(Ω) with Ω infinite. Let B be a maximal block for G on Ω and D := B^{G}. Put W = G^{B} ≀ FSym(D) and view W as a subgroup of FSym(Ω) with G ≤ W. Then

W′ ≤ G ≤ W.

Proof: The goal is to show that G has the centralizer property in W. The proposition then follows from 8.3.

Let E be any finite subset of D and Δ = ⋃ E.

Claim: G^{Δ} = W^{Δ}

Let E ∈ E. Then by definition of W, G^{E} = W^{E}. Hence there exists a finite subgroup F of N_{G}(E) with F^{E} = W^{E}. Pick a finite subset S of D with Supp(F) ⊆ ⋃ S. As G induces at least Alt(D) on D, there exists a g in N_{G}(E) with E ∩ S^{g} = {E}. Put H(E) = F^{g}. Then H(E) induces W^{E} on E and centralizes all other E′ ∈ E. Put H = ⟨H(E) | E ∈ E⟩. Then H^{Δ} ≅ ×_{E∈E}W^{E} ≅ C_{W} (E)^{Δ}. Moreover, G^{E} = Sym(E) = W^{E} and so the claim holds.

Next let F be any finite subgroup of W. Since W is finitary we can choose our E as above with Supp(F) ⊆ Δ. Then by the claim F ≤ C_{W} (Δ)N_{G}(Δ). Clearly C_{W} (Δ) ≤ C_{W} (F) and so F ≤ GC_{W} (F). So G indeed has the centralizer property and the proposition is proved.

Linear periodic groups in characteristic 0

LinChar0

Definition 10.1 Let X be a group, n a positive integer and K a field. Then X is linear dlin of degree n over K if X is isomorphic to a subgroup of GL_{n}(K). X is linear if it is linear of some degree over some field.

Lemma 10.2 Let X ≤ GL_{n}(K). Then there exists a finitely generated subgroup H of X, locirr such that the images of KX and KH in End_{n}(K) are equal. In particular, X and H have the same submodules in K^{n}, and X is completely reducible if and only if H is.

Proof: The image of KX in End_{n}(K) is a vector space of dimension at most n^{2} over K and spanned by the images of X. So it is spanned by the images of at most n^{2} elements. Take H to be the group generated by these elements.

Lemma 10.3 Suppose that G ≤ GL_{n}(K) and gcd(|G|, operatornamecharK) = 1. Then masch G is completely reducible.

Proof: By 10.2 we may assume that G is finite. Let W be proper KG-subspace of V = K^{n}. Then there exits a K-subspace U of V with V = W ⊕U. Define a K-linear map π : V → W by π(w + u) = w for all w ∈ W, u ∈ U. Then define ϕ : V → W by ϕ(v) = ∑_{g∈G} π(v^{g}). Since ϕ_{W} is just multiplication by |G| and since gcd(|G|, operatornamecharK) = 1, ϕ_{W} is an isomorphism. Thus V = W ⊕ ker ϕ. Let h ∈ G. Then ϕ(v^{h}) = ϕ(v) and so ker ϕ is a KG-module. By induction on the dimension both W and ker ϕ are direct sum of irreducible KG-modules. Hence V is completely irreduicble as KG-module.

Proposition 10.4 Periodic linear groups are locally finite. perloc

Proof: TO BE CONTINUED Definition 10.5 Let V be a vector space over the field F and σ an automorphism of F of duni order 2. Let f : V × V → K be a map.

(a) f is called a unitary form provided that

(a) f is linear in the first coordinate.

(b) f is σ-linear in the second coordinate.

(c) f(u, w) = f(w, u)^{σ} for all u, w ∈ V.

GU(V, f) denotes the group of invertible linear transformations fixing f).

(b) f is called a hermitian form if f is unitary with F = ℂ and σ is the complex conjugation.

(c) f is a positive definite hermitian form provide that f is a hermitian form and f(u, u) > 0 for all 0 ≠ u ∈ V.

The next lemma is easily proved by induction on dim U and we will leave the details to the reader.

ortsum Lemma 10.6 Let f be a positive definite hermitian form on V and U a finite dimensional subspace of V. Then V = U ⊕ U^{⊥}, where U^{⊥} = {v ∈ V | f(u, v) = 0 for all u ∈ U}.

invher Lemma 10.7 Let G be acting on the finite dimensional vector space V over ℂ. Then there exists a positive definite G-invariant hermitian form V.

Proof: Suppose first that G is finite. Let f be any positive definite hermitian form on G. For g ∈ G define f^{g}(u, v) = f(u^{g}, v^{g}). Then ∑_{g∈G} f^{g} does the trick.

For the general case we may assume without loss that G is irreducible. By 10.2 there exists a finite, irreducible subgroup F of G. By the finite case there exists such an F - invariant positive definite form f on V. We claim that f is already G-invariant.

For this let be F ≤ H ∈ F and h be an H-invariant positive definite hermitian form on V. Let V^{∗} = Hom_{ℂ}(V, ℂ) be the dual of V^{∗}. Define

ϕ_{f} : V → V^{∗}, ϕ_{f}(w)(v) = f(v, w)

.

Then ϕ_{f}(w)^{g}(v) = ϕ_{f}(w)(v^{g^{−1}}) = f(v^{g^{−}1}, w) = f(v, w^{g}) = ϕ_{f}(w^{g})(v). Thus ϕ_{f} ∈ Hom_{ℂG}(V, V^{∗}). By Schur’s lemma and since ℂ is algebraically closed Hom_{ℂG}(V, V^{∗}) = ℂ·ϕ_{f}. Now H is also an F -invariant hermitian form and so ϕ_{f} = λ · ϕ_{h} for some λ ∈ ℂ. Thus f(u, v) = λh(u, v). Since h is H-invariant this implies that f is H-invariant.

This is true for any finite H containing F and so f is indeed G-invariant.

norm Definition 10.8 Let f be a positive definite hermitian form on V. For v ∈ V put ||v|| = \sqrt{f(v, v)}. For A ∈ End_{ℂ}(V) put ||A|| = sup\{\frac{||Av||}{||v||} | 0 ≠ v ∈ V \}. trie Lemma 10.9 Let f be a positive definite hermitian form on V.

(a) f(v, w) + f(w, v) ≤ 2||v|| · ||w||.

(b) ||v + w|| ≤ ||v|| + ||w||.

(c) ||A + B|| ≤ ||A|| + ||B||.

(d) If A ∈ GU(V, f) then ||A|| = 1 and ||A − 1|| ≤ 2.

(e) Suppose that A ∈ GU(V, f) and there is a 0 ≠ v ∈ V with f(Av, v) = 0. Then ||A − 1|| ≥ \sqrt{2}.

(f)

||AB|| ≤ ||A|| · ||B||

Proof: Let v, w ∈ V and put a = ||v||, b = ||w|| and t = f(v, w) + f(w, v).

(a) Let r ∈ ℝ. Then 0 ≤ f(v+rw, v+rw) = a^{2}+rt+r^{2}b^{2}. Choosing r = −\frac{a}{b} we conclude that a^{2}− \frac{a}{b}t+a^{2} ≥ 0.

Thus t ≤ 2ab and a holds.

(b) f(v + w, v + w) = a^{2} + t + b^{2} ≤ a^{2} + 2ab + b^{2} = (a + b)^{2}.

(c) Follows from (b) and the definition of ||A||.

(d) As f is A-invariant, ||Av|| = ||v||. Moreover, ||(A−1)v|| = ||Av−v|| ≤ ||Av||+||v|| = 2 · ||v||.

(e) If f(Av, v) = 0, then f(Av − v, Av − v) = 2f(v, v).

(f) follows immediately from the definition of ||A||.

Lemma 10.10 Let f be a positive definite hermitian form on the finite dimensional K- norcom vectorspace V and A, B ∈ GU(V, f).

(a) V is the orthogonal sum of the eigenspaces of A.

(b)

||AC|| = ||C|| = ||CA||

for all C ∈ End_{ℂ}(V).

(c) ||[A, B] − 1|| ≤ 2 · ||A − 1|| · ||B − 1||.

(d) If ||B − 1|| < \sqrt{2}, ||A − 1|| ≤ \frac{1}{2} and [B, A, k] = 1, ^{1} for some k ≤ 1, then [B, A] = 1.

Proof: (a) As ℂ is algebraically closed, there exists a non-zero eigen-vector v. Then by 10.6 V = Kv ⊕ v^{⊥}, and (a) follows by induction on dim V.

(b) is clear.

(c) ||[A, B]−1|| = ||A^{−1}B^{−1}AB−1|| = ||AB−BA|| = ||(A−1)(B−1)−(B−1)(A−1)|| ≤≤ ||(A − 1)(B − 1)|| + ||(B − 1)(A − 1)|| ≤ 2||A − 1|| · ||B − 1||.

1[B, A, 2] := [[B, A], A] and [B, A, n + 1] := [[B, A], A]

(d) By (c) and induction, ||[B, A, k] − 1|| ≤ \sqrt{2} for all k. So by induction on k we may assume that k = 2. So suppose that [B, A] ≠ 1. Then A ≠ A^{B}. Since A commutes with [B, A] = A^{−B}A, A also commutes with A^{B}. As A ≠ A^{B} there exists some eigenspace D for A on V with D ≠ D^{B}. Since A^{B} commutes with A, A^{B} leaves D invariant and so D is the orthogonal sum of the eigenspaces for A^{B} on D. Hence D = (D ∩ D^{A}) ⊕ (D ∩ D^{B⊥}). So there exists 0 ≠ v ∈ (D ∩ D^{B⊥}). But then v^{B} ∈ D^{A} and v ⊥ v^{B}. Thus 10.9e implies ||B − 1|| ≥ \sqrt{2}.

linjor Theorem 10.11 There exists a function J : ℕ → ℕ so that whenever G is a periodic subgroup of GL_{n}(ℂ), then G has an abelian normal subgroup of index at most J(n).

Proof: Let V = ℂ^{n}. By 10.7 there exists a G-invariant hermitian form on V. Let A = \{g ∈ G | ||g − 1|| < \frac{1}{2}\} and H = ⟨A⟩.

Step 1 H is abelian.

Let a, b ∈ A and put a_{k} = [a, b, k]. By 10.4 ⟨a, b⟩ is finite. Hence there exists a k with ||a_{k} − 1|| minimal. Suppose that a_{k} ≠ 1. Then ||a_{k} − 1|| ≠ 0 and so by 10.10c

||a_{k+1} − 1|| = ||[a_{k}, b] − 1|| ≤ 2 · ||a_{k} − 1|| · ||b − 1|| < ||a_{k} − 1||,

a contradiction to the minimal choice of a_{k}. Thus a_{k} = 1 and so [a, b] = 1 by 10.10d. Step 2 |G/H| is bounded by a function of n. Let R be a transversal to H in G and r, s ∈ R. Since Hr ≠ Hs, rs^{−1} ∉ H and so

||s − r|| = ||1 − rs^{−1}|| ≥ \frac{1}{2}.

It follows that the spheres of radius \frac{1}{5} around the members of R are disjoint. As the elements of R have norm 1, all these spheres are contained in a sphere of radius \frac{6}{5}. Hence |G/H| = R is bounded by the fraction of the volumes of spheres of radius \frac{6}{5} and \frac{1}{5} in the 2n^{2} dimensional metric space End_{ℂ}(V).

Corollary 10.12 Let G be a periodic subgroup of GL_{n}(K), where K is a field of characteristic 0. Then G is an abelian normal subgroup of index less or equal to J(n).

Proof: We contend our self with a sketch of a proof. Suppose first that G is finite. Then it can be show that there exists an F ≤ K with G ≤ GL_{n}(F) and F is algebraic over ℚ. But then F ≤ ℂ, and so G ≤ GL_{n}(ℂ). So the theorem holds by 10.11.

In the general case we see that all finite subgroups of G have an abelian normal subgroup of index at most J(n). Let I be the set of finite subgroups of G. For H ∈ I let T_{H} be the set of all abelian normal subgroups A of H with |H/A| ≤ J(n). Then T_{H} is not empty. Let L ≤ H ∈ I and A ∈ T_{H}. Then A ∩ L ⊴ L and |L/A ∩ L| = |LA/A| ≤ |H/A| ≤ J(n). Thus A ∩ L ∈ T_{L} and we obtain a map, ϕ_{HL} : T_{H} → T_{L}, A → A ∩ L. Clearly ϕ_{HH} is the identity function on T_{H}. Also if M ≤ L and A ∈ T_{H}, then M ∩ A = M ∩ (L ∩ A) and so ϕ_{LM} ∘ ϕ_{HL} = ϕ_{HM}. So the T_{H} and ϕ_{HL} form an inverse limit system. Since each T_{H} is finite, 11.2 shows that that the inverse limit system has an inverse limit, (A_{H})_{H∈I}. Put A = ⋃_{H∈I} A_{H}. We will show that A is an abelian normal subgroups of A and |G/A| ≤ J(n).

Let H ∈ I. We claim that A_{H} = A ∩ H. Clearly, A_{H} ≤ A ∩ H. Let a ∈ A ∩ H. Then there exists L ∈ I with a ∈ A_{L}. Put M = ⟨H, L⟩. Then M ∈ I, A_{H} = A_{M} ∩ H and A_{L} = A_{M} ∩ L. Since a ∈ A_{L}, a ∈ A_{M} and so a ∈ A_{M} ∩ H = A_{H}. So A ∩ H ≤ A_{L}.

Let a, b ∈ A and g ∈ G. Put H = ⟨a, b, g⟩. Then H ∈ I and so {a, b} ∈ A ∩ H = A_{H}. Since A_{H} is a normal abelian subgroup of H we have ⟨a, b⟩ ≤ A_{H} ≤ A, a^{g} ∈ A_{H} ≤ A and [a, b] = 1. Thus A is an abelian normal subgroups of G.

Let R be a transversal to A in G and S a finite subset of R. Put H = ⟨S⟩. Then H ∈ I. Since As ≠ At for all s ≠ t ∈ S and since A_{H} = A ∩ H, A_{H}s ≠ A_{H}t for all s ≠ t ∈ S. Thus |S| ≤ |H/A_{H}| ≤ J(n). Since this holds for all finite subsets of R, |R| ≤ J(n) and so |G/A| ≤ J(n).

44 CHAPTER 10. LINEAR PERIODIC GROUPS IN CHARACTERISTIC 0

Inverse limits of finite sets

In this section we will show that inverse limits of finite sets are not empty.

Definition 11.1 Let I be a directed set. Then an inverse limit system based on I is a dinvlim tuple (T_{i} | i ∈ I, ϕ_{ij} | i ≥ j ∈ I) such that

(a) For i ∈ I, T_{i} is a non-empty set.

(b) For i ≥ j ∈ I, ϕ_{ij} : T_{i} → T_{j} is a function.

(c) For i ∈ I, ϕ_{ii} is the identity on T_{i}.

(d) For i ≥ j ≥ k, ϕ_{jk} ∘ ϕ_{ij} = ϕik.

The inverse limit of a inverse limits system is the set of all tuples (t_{i} | i ∈ I) such that t_{i} ∈ T_{i} and ϕ_{ij}(t_{i}) = t_{j}) for all i ≥ j ∈ I.

Proposition 11.2 The inverse limit of an inverse limits system of finite sets is not empty. inlifi

Proof: Let A be the set of all tuples (J, t_{j} | j ∈ J) such J ⊆ I, t_{i} ∈ T_{i} and

(*) For all finite subsets F of J and all F ≤ s ∈ I there exists t_{s} ∈ T_{s} so that ϕ_{sf}(t_{s}) = t_{f} for all f ∈ F.

Order A by (J, t_{j} | j ∈ J) ≤ (K, s_{k} | k ∈ K) if J ⊂ K and t_{j} = s_{j} for all j ∈ J. It is easy to verify that the assumptions of Zorn’s lemma are fulfilled and so there exists a maximal elements (J, t_{j} | j ∈ J) in A. If J = I then (*) easily implies that (t_{j} | j ∈ J) is in the inverse limit. So we may assume that J ≠ I. Pick i ∈ I \ J and put

X = ⋂ {ϕ_{si}( ⋂_{f∈F} ϕ_{sf}^{−1}(t_{f})) | F ⊆ J, |F | < ∞, F ≤ s ∈ I}

We claim that X is non empty. Since T_{i} is finite, there exists finite subsets F_{1}, F_{2}, … F_{l} of J and elements s_{1}, s_{2}, … s_{l} with F_{k} ≤ s_{k}

X = ⋂ {ϕ_{s_{k}i}( ⋂_{f∈F_{k}} ϕ_{s_{k}f}^{−1}(t_{f})) | 1 ≤ k ≤ l}

Let s be an upper bound for the s′_{i}s and F = ⋂_{k=1}^{l} F_{k}. Then by (*) there exists t_{s} ∈ T_{s} with ϕ_{sf}(t_{s}) = t_{f} for all f ∈ F. Put x = ϕ_{si}(t_{s}). Then for all k and all f ∈ F_{k} x = ϕ_{s_{k}i}(ϕ_{ss_{k}}(t_{s})) and ϕ_{s_{k}f}(ϕ_{ss_{k}}(t_{s})) = ϕ_{sf}(t_{s}) = t_{f}. Hence ϕ_{ss_{k}}(t_{s}) ∈ ⋂_{f∈F_{k}} ϕ_{s_{k}f}^{−1}(t_{f}) and x ∈ X. Thus X is indeed not empty.

Now let t_{i} ∈ X and put J^{∗} = J ∪ {i}. It’s clear from the definition of X that (J^{∗}, t_{j} | j ∈ J^{∗}) fulfills (*), a contradiction to the maximality of (J, t_{j} | j ∈ J).

Ultra-products and a little bit of model theory

sUltra

In this section we will prove Malcev’s Theorem which asserts that a group with a local system of linear groups of bounded degree is itself linear. The main tool in th proof are ultra-products. The most natural setting for ultra products is model theory. So we will use the occasion to introduce the basic definitions in model theory.

Definition 12.1 dmodth

(a) A first-order language is a set FL together with a partition (FS, RS, CS, LS) of FL. The elements of FL, FS, RS, CS and LS are referred to as symbols, function symbols, relations symbols, constant symbols and logical symbols Also LS = {), (, ∧, ¬, ≡ , ∀, v_{1}, v_{2}, v_{3} …, v_{n}, …}. The v_{i}’s are called variables and ∀ a quantifier. The functions symbols are assumed to be partitioned into 1-placed,2-placed, 3-placed, … function symbols and the relation symbols are partitioned into 1-placed, 2-placed,… relation symbols. V is the set of variables.

(b) A model A for the language ∞L is a set A together with a function which assigns to each m-placed functions symbol s an function A^{m} ← A, and to each m-placed relation symbol an m-placed relation on A and to each constant symbol an element of A. A is called the universe of the model. The object assigned to a symbol is called the interpretation of the symbol in A.

(c) A term is any sequence of symbols obtain in finitely many steps by the following procedure:

(a) A variable is a term.

(b) A constant symbol is a term.

(c) If f is an m-placed function symbol and t_{1}, t_{2}, …, t_{m} are terms, then f(t_{1}t_{2} … t_{m}) is a term.

48 CHAPTER 12. ULTRA-PRODUCTS AND A LITTLE BIT OF MODEL THEORY

(d) An atomic formula is a sequence of symbols of one of the following two forms: (a) t_{1} ≡ t_{2}, where t_{1} and t_{2} are terms.

(b) R(t_{1}t_{2} … t_{m}), where R is an m-placed relation symbol and t_{1}, … t_{m} are terms.

(e) A formula is any sequence of symbols obtain in finitely many steps by the following procedure:

(a) An atomic formula is a formula.

(b) If ϕ and ψ are formulas, then (ϕ ∧ ψ) and (¬ϕ) are formulas.

(c) If v is a variable and ϕ is an formula then (∀v)ϕ is a formula.

(f) Let v be a variable and ϕ a formula. If v is in the sequence ϕ, we say that v is a variable of ϕ. If v is a variable of ϕ and the last occurrence of v is of the form ∀v then v is called a bound variable of ϕ. A variable of ϕ which is not bound is called a free variable of ϕ. Otherwise v is a bounded variable of ϕ.

dquant Definition 12.2 Given formulas ϕ and ψ and a variable v. (a) ϕ ∨ psi) denotes the formula (¬((¬ϕ) ∧ (¬ψ))).

(b) (ϕ ← ψ) denotes the formula ((¬ψ) ∧ ϕ)

(c) (ψ ↔ ψ) denotes the formula ((ϕ ← ψ) ∧ (ψ ← phi)).

(d) (∃v)ϕ denotes the formula (≠ (∀v)(¬ψ))).

dsatmod Definition 12.3 Let ∞L be a first order language and A a model for ∞L with universe A. Let E be a function from V → A.

(a) Let t be a term. The the value t |_{E} of t at E is defined as follows: (a) If t is a variable, then t |_{E}= E(t).

(b) If t is a constant, then t |_{E} is the interpretation of t in A.

(c) If t = F (t_{1} … t_{m}), for a function symbol F and terms t_{1}, …, t_{m}, then

t |_{E}= F^{∗}(t_{1} |_{E}, …, t_{m} |_{E}).

where F^{∗} is the interpretation of F in A.

(b) Let ϕ be a formula. Then the statement E satisfies ϕ in A (written A ⊧ ϕ |_{E}) is defined as follows.

(a) If ϕ is the atomic formula t_{1} ≡ t_{2} for terms t_{1}, t_{2} when A ⊧ ϕ |_{E} if and only if t_{1} | E = t_{2} | E.

(b) If ϕ is the R(t_{1} … t_{m}) for a relation symbol R and terms t_{1}, … t_{m}. Then A ⊧ ϕ |_{E} if and only if (t_{1} |_{E}, …, t_{m} |_{E}) fulfills the interpretation ofR in A.

(c) If ϕ = ϕ_{1} ∧ phi_{2}, then A ⊧ ϕ |_{E} if and only if both A ⊧ ϕ_{1} |_{E} and A ⊧ ϕ_{2} |_{E} hold.

(d) If ϕ = ¬ψ, then A ⊧ ϕ |_{E} if an only if not A ⊧ ψ |_{E}.

(e) If ϕ = (∀v)ψ, then A ⊧ ϕ |_{E} if and only if A ⊧ ϕ |_{E∗}, for all functions

E^{∗} : V → A,which agree with E except maybe on v. The easy proof of the following lemma is left to the reader.

Lemma 12.4 Let ∞L be a first order language, A a model for ∞L and E, F : V → A. dov (a) If t is a term and E and F agree on the variables of t, then t | E = t | F.

(b) If ϕ is a formula and E and F agree on the free variable of ϕ then A ⊧ ϕ |_{E} if and

only if A ⊧ ϕ |_{F}. In view of the previous lemma we can talk about t |_{E}, whenever t is a term and E is

a function from the variables of t to A and about A ⊧ ϕ |_{E} whenever ϕ is a formula and E a function defined on the free variables of ϕ. In particular, if ϕ has no free variable, the statement A ⊧ ϕ already makes sense. Definition 12.5 Let FL be a first order language and A a model for A. dtheory (a) ϕ is a sentence if ϕ has no free variables.

AIb Let ϕ be a sentence then we say that A is a model of ϕ if A ⊧ ϕ.

(c) A theory for A is a set of sentences.

(d) A model for a theory Σ is a model for FL which is a model for each of the sentences in Σ.

(e) Two models A and B for FL are called elementary equivalent if for each sentence ϕ of FL, A ⊧ ϕ if and only if B ⊧ ϕ.

Choose FL = {·, 1} ∪ LS, where ” · ” as 2-placed function symbol and ”1” as a constant symbol. Then groups are exactly the models of the theory with the following sentences: (where we write a · b for ·(ab)

• (∀v_{1})(∀v_{2})(∀v_{3})v_{1} · (v_{2} · v_{3}) ≡ (v_{1} · v_{2}) · v_{3}.

• (∀v_{1})v_{1} · 1 ≡ v_{1}

50 CHAPTER 12. ULTRA-PRODUCTS AND A LITTLE BIT OF MODEL THEORY

• (∀v_{1})(∃v_{2})v_{1} · v_{2} = 1 Definition 12.6 Let I be a set. dfilter

(a) Let P(I) is a power set of I, i.e the set of all subsets of I.

(b) A filter D on I is a subset of P(I) such that

(a) I ∈ D.

(b) If D, E ∈ D, then D ∩ E ∈ D.

(c) If D ∈ D and D ⊆ E ⊆ I, then E ∈ D.

(c) An ultra filter on I is a filter D on I so that for each D ⊆ I exactly one of D and I ⊂ D is in D.

Examples for filter on a set I are:

The improper filter P(I)

The trivial filter {I}.

For D ⊆ I the principal filter for D on I: {E | D ⊂ E ⊆ I}.

The cofinite or Frechet filter : {E ⊂ I | |I \ E| < ∞.

A proper filter is any filter other than the improper filter

Its an easy exercise to show that on a finite set each filter is a principal filter. Also arbitrary intersections of filters are filters and unions of chains of filters are filters. In particular, if E is any subset of P(I) than there exists a smallest filter containing E, namely the intersection of all the filters containing E. This filter is called the filter generated by E. We say that E has the finite intersection property if any intersection of finitely many members of E is not empty. The following proposition follows immediately from the definitions:

genfil Lemma 12.7 Let I be a set and E ⊆ P(I). Then the filter generated by E is the filter

{D | ⋂ J ⊂ D ⊂ I for some finite subset J ofE.

In particular, the filter generated by E is proper if and only if E has the finite intersection property.

ult=max Lemma 12.8 A filter is an ultra-filter if and only if its a maximal proper filter.

Proof: Clearly a ultra-filter is a maximal proper filter. So let D be maximal proper filter and E ⊂ I with E ∉ E. We need to show that I \ E ∈ D. Since E is maximal, the filter generated by D + E is improper. By 12.7 this means that D + E does not have the finite intersection property. Since finite intersection of members of D are still in D we conclude that there exists D ∈ D with D ∩ E =. Hence D ⊂ I \ E and so I \ E ∈ D.

Corollary 12.9 Every proper filter, and so also every subset of D with the finite intersection property, lies in an ultra filter.

Proof: By Zorn’s lemma every filter lies in a maximal filter. By 12.8 this maximal filter is an ultra filter.

dultprod Definition 12.10 Let I be a set, D a filter on I and for i ∈ I let A_{i} be a non-empty set.

Let A = ∏_{i∈I} A_{i} = {⟨a_{i}|mida_{i} ∈ A_{i}, i ∈ I⟩} be the cartesian product of the A_{i}’s. The equivalence relation ≡_{D} on A is defined by

⟨a_{i} | i ∈ I⟩ ≡_{D} ⟨b_{i} | i ∈ I⟩ if and only if {i ∈ I | a_{i} = b_{i}} ∈ D

For a ∈ A let a_{D} be the equivalence class of ≡_{D} containing a. Let ∏_{D} A_{i} be the set of equivalence classes of ≡_{D}.

∏_{D} A_{i} is called the reduced product of A_{i} modulo D A reduced product modulo D is called an ultra-product of A_{i} modulo D if D is an ultra filter. If A_{i} = B for all i ∈ I, then ∏_{D} B is called an ultra-power of B.

Since over-sets of filter sets are filter sets ⟨a_{i} | i ∈ I⟩ ≡ ⟨b_{i} | i ∈ I⟩ if and only if there exists D ∈ D with a_{i} = b_{i} for all i ∈ D. This observation will be usefully in any concrete calculation since in it might be rather difficult to determine the exact sets on which a_{i} and b_{i} agree

Definition 12.11 Let I be a set, FL a first order language and A_{i}, i ∈ I a model for FL. Let A_{i} be the universe of A_{i} and for a non-logical symbol s let s_{i}^{∗} be the interpretation of s in A_{i}. Then the reduced product ∏_{D} A_{I} is the model for FL describes as follows.

(a) The universe A of ∏_{D} A_{i} is ∏_{D} A_{i}.

(b) If R is an m-placed relation symbol, when the interpretation R^{∗} for R in ∏_{D} A_{I} is defined by

R^{∗}(a^{1}, … a_{m}) if and only if {i ∈ I | R_{i}^{∗}(a^{1}(i), … a^{m}(i)} ∈ D

where a^{k} = ⟨a^{k}(i) | i ∈ I⟩_{D}

(b) If F is an m-placed function, symbol then the interpretation F^{∗} for F in ∏_{D} A_{i} is defined by

F^{∗}(a^{1}, …, a^{m}) = ⟨R_{i}^{∗}(a_{i}^{1}, …, a_{i}^{m}) | i ∈ I⟩_{D}

where a^{k} = ⟨a_{i}^{k}) | i ∈ I⟩_{D}.

(c) If c is an constant symbol then the interpretation c^{∗} for c in ∏_{D} A_{i} is defined by

c = ⟨c∗_{i}⟩ | i ∈ I⟨_{D}

.

The reader should convince herself that this is well defined, i.e. does not depend on the particular choice of the representatives ⟨a^{k}(i) | i ∈ I⟩ for a^{k}.

fundult Theorem 12.12 Let D be an ultra filter on I and A = ∏_{D} A_{i} be an ultra product of models of the first order language FL. Let E_{i} : V ← A_{i} be a functions and E = ∏_{D} E_{i} : V ← A, v ← ⟨E_{i}(v) | i ∈ I⟩_{D}.

(a) Let t be a term of FL, then

t_{E} = ⟨t_{E_{i}} | i ∈ I⟩_{D}

(b) Let ϕ be a formula in FL. Then

A ⊧ ϕ_{E} if and only {i | A_{i} ⊧ ϕ |_{E_{i}}} ∈ D

(c) Let Σ be a finite theory. Then Ais a model for Σ if and only if there exists D ∈ D so that for all i ∈ D, A_{i} is a model for Σ.

Proof: (a) This is merely an unwinding of the definition and the details are left to the reader.

(b) As usual this is proved using the inductive definition of a formula. We will carry out this analysis for formulas of the form (∀v)ϕ and ≠ ϕ an d leave the remaining easier cases to the reader. So suppose that (b) holds for the formula ϕ.

Let v be a variable. We wish to show that (b) also holds for (∀v)ϕ.

By definition A ⊧ ((∀v)ϕ) |_{E}

if and only if for all Ê : V ← A such that Ê agrees with E on V − v one has

A ⊧ ϕÊ.

As (b) holds for ϕ this is equivalent to

(∗) {i | A_{i} ⊧ ϕ |Ê_{i}} ∈ D

for all such Ê and where we may choose the Ê_{i} to agree with E_{i} on V − v.

Let D = {i | A_{i} ⊧ (∀vϕ) |_{E_{i}}} ∈ D. Then for any Ê, D is a subset of the set in (*). So if D ∈ D, (*) holds.

Conversely suppose that D ∉ D. Then by definition of ⊧ (∀v)ϕ |_{E_{i}}, for each i ∈ I \ D there exists Ê_{i} so that Ê_{i} agrees with E_{i} on V − v and we do not have A_{i} ⊧ ϕ |Ê_{i}. Put Ê_{d} == E_{d} for d ∈ D and put Ê = ∏_{D} Ê_{i}. For this choice of Ê the set in (*) is a subset of D and so as D ∉ D (*) does not hold. This completes the proof of (b) for formulas of the form (∀v)ϕ.

Next we consider the formula ≠ ϕ.

By definition A ⊧≠ ϕ |_{E} if and only if not A ⊧ ϕ_{E}. As (b) holds for ϕ, this is equivalent to

{i | A_{i} ⊧ ϕ) |_{E_{i}}} ∉ D

As D is an ultra-filter (this is the only place there the ultra filter assumption is used) this holds if and only if

{i | not A_{i} ⊧ ϕ) |_{E_{i}}} ∈ D

By definition of ⊧ ¬ϕ |_{E} this is the same as

{i | A_{i} ⊧ (¬ϕ) |_{E_{i}}} ∈ D

(c) By (b), (c) holds a single sentence, As a D is closed under finite intersection (c) holds for arbitrary finite theory (or more general for any theory with a finite set of axioms). bigskip

As an immediate consequence of 12.12 we obtain that an ultra products of groups is again a group and an ultra-product of fields is a field. Also if n is a given positive integer and each A_{i} is a set of size at most n, then any ultra-product of A_{i} is a set of size n. Indeed, a set has size at most n if and only if it satisfies the sentence

(∃v_{1} … v_{n})(∀v_{n+1})(v_{1} ≡ v_{n+1} ∨ … ∨ v_{n} ≡ v_{n+1})

To be able to talk about groups,fields and vector spaces at the same time we add an additional set Ξ to a first order language ,where each element of Ξ stands for one of the structures. The universe of now is of the form (A_{ξ}|ξ ∈ Ξ) where each A_{ξ} is a set. The function symbols are now (m_{ξ}| | ξ ∈ Ξ) − ξ_{0})-placed (with ∑_{ξ∈Xi} m_{ξ}) finite) where such a symbol is interpreted by a function ∏_{ξ∈Ξ} A_{ξ}^{m_{ξ}} → A_{ξ_{0}}. The relation symbols are treated similarly. The variables are of the form v_{n}^{ξ}, n ∈ ℕ, ξ ∈ Xi. To evaluate terms v_{n}^{ξ} is only allowed to be replaced to element of A_{ξ}, that is the only consider functions E : ⋃ V ⋃ A_{ξ} with E(v_{k}^{ξ}) ∈ A_{ξ}. All the assertions about first order language we made so far are still true in this more general set-up. Indeed, it is not difficult to see that this set can be expressed in terms of the old set up.

If V_{i} is vector space over the field K_{i}, we now obtain ∏_{D} A_{i} is a vector space over the field ∏_{D} K_{i}. Now if each A_{i} has dimension at most n, then ∏_{D} A_{i} has dimension at most n over ∏_{D} K_{i}. Indeed a vector space over a field has dimension at most n if the following sentence is fulfilled:

(∃v_{1} … v_{n})(∀v_{n+1})(∃k_{1} … k_{n})v_{n+1} ≡ k_{1} · v_{1} + k_{2} · v_{2} + … + k_{n}v_{n})

where the v_{i}’s are vector space variables and the k_{i}’s a field variables.

If g ∈ GL_{K}(V) define deg(g) = deg_{V} (g) = dim[V, g] = dim V/C_{V} (g). Furthermore pdeg_{V} (g) = inf{deg k · g | o ≠ k ∈ K}. The statements deg_{V} (x) ≤ n and pdeg_{V} (g) ≤ n can be expressed in the first order language of groups acting on vector-space . For example pdeg_{(} g) ≤ n reads as (∃v_{1} … v_{n})(∃k_{n+1})(¬(k_{n+1} ≡ 0)∧((∀v_{n+1})(∃k_{1} … k_{n})k_{n+1}v_{n+1}^{g} ≡ k_{1}·v_{1}+k_{2}·v_{2}+…+k_{n}v_{n}+v_{n+1})

We also have a natural embedding ∏_{D} GL_{K_{i}}(V_{i}) → GL_{∏_{D} K_{i}}(∏_{D} V_{i}. This induces an embedding ∏_{D} P GL_{K_{i}}(V_{i}) → P GL_{∏_{D} K_{i}}(∏_{D} V_{i}. Indeed, ⟨g_{i}⟩_{D} acts as a scalar ⟨k_{i}⟩D on ∏_{D} V_{i} if and only if there exists D ∈ D with k_{i} = g_{i} for all i ∈ D. But this is the same as saying that ⟨g_{i}⟩_{D} is in the kernel of the map from ∏_{D} GL_{K_{i}}(V_{i}) to ∏_{D} P GL_{K_{i}}(V_{i}). If pdeg V_{i}(g_{i}) ≤ n for all i ∈ D ∈ D, then we get pdeg_{∏ V_{i}}(⟨g_{i}⟩_{D}) ≤ n.

Let X be a group, (L_{i} | i ∈ I) a local system for X and for each i ∈ I, let ϕ_{i} : L_{i} → H_{i} be a group homomorphism. For i ∈ I ≤ I^{≥I} = {j ∈ I | j ≥ i}. Since I is directed, {I^{≥i} | i ∈ I} as the finite intersection property. Hence there exists an ultra-filter D on I so that I^{≥i} ∈ D for all i ∈ I.

Let H = ∏_{D} H_{i}. Extend ϕ_{i} to a map (but not a homomorphism) X → H_{i} by ϕ_{i}(x) = 1 if x ∈ X \ L_{i}. Define

ϕ : X → H, x → ⟨ϕ_{i}(x)⟨_{D}.

Let x, y ∈ X and pick i with ⟨x, y⟩ ≤ L_{i}. Then ϕ_{j}(xy) = ϕ_{j}(x)ϕ_{j}(y) for all j ∈ I^{≥i}. Since I^{≥i} ∈ D this implies ϕ(xy) = ϕ(x)ϕ(y). Hence ϕ is a homomorphism. In general it seems to a non-trivial task to determine the kernel of ϕ and in general the kernel depends on the choice of the ultra filter. But there is an easy sufficient condition which guarantees ϕ(x) ≠ 1. Namely if there exists i ∈ I so that ϕ_{j}(x) ≠ 1 for all j ≥ i then ϕ(x) ≠ 1. In particular if all ϕ_{i} are one to one, then

For locally finite X (or more generally for arbitrary groups if instead of finite groups, finitely generated groups are considered) we give a similar conditions which ensures that ϕ is one to one:

Suppose that {(L_{i}, M_{i})|i ∈ I} is a sectional cover and that ker ϕ_{i} = M_{i}. For i ∈ I let I^{←i} = {j ∈ I | L_{i} ≤ L_{j}, L_{i} ∩ M_{j} = 1}. The definition of a sectional cover implies that {L^{←i} | i ∈ I as the finite intersection property. Hence we can choose an ultra-filter D with L^{←i} ∈ D for all i ∈ I. Then clearly ϕ is one to one.

malcev Theorem 12.13 Let X be a group with a local system of linear groups of degree n. Then X is linear of degree n.

Proof: Let L_{⟩} | ⟩ ∈ I be the local system such that there exist field K_{i} and embeddings ϕ_{i} → GL_{n}(K_{i}). By the preceding discussion X can be embedded into ∏_{D} → GL_{n}(K_{i}) which in turn is embedded into GL_{n}(∏_{D} K_{i}).

With the same sort of argument we obtain:

joncev Theorem 12.14 Let X be a group and {(L_{i}, M_{i}) | i ∈ I} be a sectional cover for G. For i ∈ I let V_{i} be a vector space over the field K_{i} and ϕ_{i} : L_{i}/M_{i} → P GL_{K_{i}}(V_{i}) an embedding. Furthermore let Y ⊂ X so that for all y ∈ Y there exists an integer d_{y} and a finitely generated subgroup F_{y} so that for all i ∈ I with F_{y} ≤ L_{i} and F_{y} ∩ M_{i} = 1 one has pdeg_{V_{i}}(ϕ_{i}(y)) ≤ d_{y}.

Then there exists a vector space V over the field K and an embedding ϕ : X → P GL_{K}(V) so that pdeg_{V} (ϕ(y)) ≤ d_{y} for all y ∈ Y.

Periodic, simple, linear groups

slinear

In this chapter we present and partly prove the classification of the periodic, simple, linear groups.

Definition 13.1 A projective representation of the group X is homomorphism of X into dpr P GL_{K}(V), where V is a vector space over the field K. dim_{K} V is called the degree of the projective representation.

For example P SL_{n}(K) has projective representation of degree n. This indicates that a projective representation can be a more natural object to look at as a linear representation. Also we remark that any group with a faithful projective representation of finite degree is linear (but maybe of larger degree). Indeed if V is a projective representation for X, then X ⊗ X^{∗} provides a linear representation, where X^{∗} is the dual of X.

Definition 13.2 Let H be a finite group. Then mdeg(H) is the minimal positive integer md d so that there exists a field K and finite groups B ⊴ T ≤ GL_{d}(K) such that T ∩ K ≤ B and M ≅ H.

Lemma 13.3 Let q be a prime, M a finite q-group with 1 ≠ M′ ≤ Φ(M) ≤ Z(M). Let irrext |M/Z(M)| = q^{n}.

(a) If Z(M) is cyclic, then M has an abelian subgroup of rank at least \frac{n}{2}.

(b) Let K be a field and V faithful irreducible KM-module. Then V has dimension at least q^{\frac{n}{2}} .

Proof: (a) Let x ∈ M. Then the commutator map induces defines an isomorphism from M/C_{M}(x) into [M, x] ≤ Z(M). Since M/C_{M}(x) is elementary abelian and Z(M) is cyclic this implies |M/C_{M}(x) ≤ q. An easy induction argument shows that |M/C_{M}(A)| ≤ |A/A ∩ Z(M)| for all A ≤ M. Since C_{M}(A) ≤ A for any maximal abelian subgroup of M,

(a) holds.

(b) Since M is irreducible, Z(M) is cyclic Also if 1 ≠ z ∈ Z(M), then C_{V} (z) is an M submodule and so C_{V} (z) = 0. Without loss K is algebraically closed.

Let A be a maximal abelian subgroups of M. Then Z(M) ≤ A and A is normal in M. Let W be a Wedderburn component for A on V and let B = N_{M}(W). Then [A, B] ≤ M′ ≤ Z(M). As A is abelian and K is algebraically closed, A has an eigenvector on W and so acts as scalars on W. Thus [A, B] centralizes W. Put C_{Z(M)}(W) = 1 and so [A, B] = 1. The maximality of A implies B ≤ A. Hence N_{M}(W) = A and dim V = dim W ·M/A|. It remains to show that |M/A| ≥ |A/Z(M)|. For this pick B ≤ M with M = AB and A ∩ B = Z(M). Then as seen above |M/C_{M}(B)| ≤ |B/Z(M)| = |M/A|. Thus also |A/C_{A}(B)| ≤ |M/A|. Put C_{A}(B) ≤ C_{M}(AB) = Z(M) and so |A/Z(M)| ≤ M.

minsim Lemma 13.4 Let H be a non-cyclic, finite, simple group, d = mdeg(H) and B ⊴ T ≤ GL_{d}(K) finite groups with H ∩ K ≤ B and T/B ≅ H. Then T acts primitively on K^{d} and B ≤ K.

Proof: Suppose that T normalizes a proper subspace W in V = K^{d}. By minimality of d neither T/C_{T} (W) not T/C_{T} (V/W) have composition factor isomorphic to H. But neither does the abelian group C_{T} (W) ∩ C_{T} (V/W).

Hence T is irreducible on V. Suppose next that Δ is a system of imprimitivity for T on V. Since C_{T} (Δ) is not irreducible it cannot have composition factor isomorphic to H. Thus T^{Δ} does. But T^{δ} ≤ Sym(Δ) ≤ Sym(n) and Sym(n) is isomorphic to a subgroup of GL_{d−1}(K) via the even permutation module, a contradiction.

So T is indeed primitive on V.

Suppose now that B K̸ and pick a counter example with |T | minimal. Without loss K is algebraically closed. Put Z = T ∩ K and let q be a prime dividing the order of B/Z. Let Q be a Sylow q-subgroup of B. By the Frattini argument T = BN_{T} (Q) and so N_{T} (Q)/N_{M}(Q) ≅ H. Thus by minimality of |B|, Q is normal in T. Let M be a subgroup of Q minimal with respect to M ≰ Z and M ⊴ T. As T is primitive Clifford’s theorem implies that V is the direct sum of isomorphic irreducible M-modules. If M would be abelian then as K is algebraically closed we get M ≤ Z, a contradiction.

Thus M is not abelian. By minimality of M. Φ(M) ≤ Z and Z(M) ≤ Z. Put M̄ = M/M ∩Z.By minimality of T and as M ≤ C_{T} (M̄) the latter group has no composition factor isomorphic to H. Hence T/C_{T} (M̄) has a composition factor isomorphic to H. Let M̄ = q^{m}. Note that M̄ is a vector space over GF (q) and so by minimality of d, d ≤ m. On the other hand by 13.3, d ≥ q^{fracm2}. So \frac{q^{m}}{≤} m^{2}. As m > 1, q = 2 and m ≤ 4. But then (Aut(M)/Inn(M))∞ ≤ Sp_{4}(2)^{∞} ≅ P SL_{2}(9). Hence d ≤ 2, m = 2 and Aut(M) is solvable, a contradiction.

boundq Lemma 13.5 Let A be a elementary abelian r subgroup of X = GL_{n}(K) with r ≠ operatornamecharK. Then A has rank at most n and |N_{X}(A)/C_{X}(A)| ≤ n!.

Proof: Without loss K is algebraically closed. Let E be the set of eigenspaces for A on K^{n}. Then |E| ≤ n. Let E ∈ E. Then A/C_{A}(E) is cyclic and so A has rank at most |E|. Since A acts as scalars on E we get that [N_{X}(E), A] centralizes E and so [C_{X}(E), A] centralizes K^{=} ⨁ E. Hence N_{X}(A)/C_{X}(A) is is isomorphic to a section of Sym(E) and so has order at most |E|!.

Proposition 13.6 Let G be a infinite, periodic,linear, simple group and d the minimal lsls degree of a projective representation for G.

(a) Let K be any Kegel cover for G. Let J consists of whose K ∈ K with K̄ perfect and mdeg(K̄) = d. Then J is Kegel cover for G, K \ J is not a Kegel cover and H_{J} is simple for all J ∈ J.

(b) G is the union of an ascending chain G_{1} ≤ G_{2} ≤ G_{3} … of finite simple subgroups with mdeg(G_{i}) = d for all i.

Proof: (a) Without loss K has non cyclic factors.Suppose that K \ J is a Kegel-cover for G. Then by 13.4 all the factors of this Kegel cover have a projective representation of degree less than d. By Malcev’s theorem 12.13 we conclude that also G has a projective representation of degree less then d. A contradiction.

So K \ J is not a Kegel cover and thus J is a Kegel cover. Let J ∈ J. Since H_{J} is has a projective representation of degree d and d = mdeg J̄. Hence H_{J} is simple by 13.4

(b) By exercise 16. G is countable. Thus (b) follows from (a).

In order to proceed we proudly present:

The classification of finite simple groups

Any finite simple groups is isomorphic to one of the following groups where p is a prime, q = p^{k} and n, k are positive integers.

• C_{p}

• Alt(n), n ≥ 5

• A_{n}(q) = P SL_{n+1}(q), (n, q) ≠ (1, 2), (1, 3).

• ^{2}A_{n}(q^{2}) = P SU_{n+1}(q^{2}), n ≤ 3, (n, q) ≠ (3, 2)

• B_{n}(q) = P Ω_{2n+1}(q), n ≥ 3, p odd.

• ^{2}B_{2}(q) = Sz(q), p = 2, k > 1 odd.

• C_{n}(q) = P Sp_{2n}(q), n ≥ 2, (n, q) ≠ (4, 2).

• D_{n}(q) = P Ω_{2n}^{+}(q), n ≥ 4

• ^{2}D_{n}(q^{2}) = P Ω_{2n}^{−}(q), n ≥ 4.

• ^{3}D_{4}(q^{3})

• E_{6}(q), E_{7}(q), E_{8}(q)

• ^{2}E_{6}(q^{2})

• F_{4}(q)

• ^{2}F_{4}(q), p = 2, k > 1 odd, ^{2}F_{4}(q)′

• G_{2}(q), q ≠ 2.

• ^{2}G_{2}(q), p = 3, k > 1.

• M_{11}, M_{12}, M_{22}, M_{23}, M_{24}, J_{1}, J_{2}, J_{3}, J_{4}, HS, McL, Suz, Ly, He, Ru,

O′N, Co_{3}, C0_{2}, C0_{1}, F i_{22}, F i_{23}, F i′_{24}, F_{3}, F_{5}, F_{3}, F_{1}

• DDELNP P RST

All of these groups are simple. And they are pairwise non-isomorphic except for Alt(5) ≅ P SL_{2}(4) ≅ P SL_{2}(5), P SL_{2}(9) ≅ Alt(6), P SL_{3}(2) ≅ P SL_{2}(9) and P Sp_{4}(3) ≅ P SU_{4}(2^{2}).

lslt Proposition 13.7 Let G be a infinite, periodic,simple linear group.

(a) There exists a prime p, an positive integer n, a Lie-type X and an ascending chain G_{1} < G_{2} < G_{3} < … of finite simple subgroups of G such that G = ⋂_{i=1}^{∞} G_{i} and G_{i} ≅ X_{n}(p^{k_{n}}). Moreover, any linear representation of G has to be over a field in characteristic p.

(b) If G is a linear representation over the field K, then operatornamecharK = p.

Proof: Let G_{1} ≤ G_{2} ≤ ldots be as 13.6. Using 3.3 we will select a subsequence which fulfills (a). (b) will be proved along the way. First of all we may assume that none. of the G_{i}’s is one of the sporadic groups. Also suppose that K is a field so that G has a faithful projective representation of degree d over K. Let p = operatornamecharK.

Let r be any prime with r ≠ p and let L be any simple section of one of the G_{i}’s. By 13.4 L has faithful projective representation of degree d. L is linear of degree d^{2} and so L has no elementary abelian subgroups of rank larger than d^{2}. In particular there is bound on the size of the alternating sections. Hence only finitely many of the G_{i}’s can be alternating groups. So we may assume that G_{i} ≅ X(i)_{n_{i}}(q_{i}), q_{i} = p_{i}^{k_{i}}.

We will use the following facts about groups of Lie type without proof:

1. X_{n}(q) has a section isomorphic to Alt(n)

2. X_{n}(q) has an elementary abelian subgroup A with |A| = q and |N(A)/C(A)| ≥ \frac{q−1}{2}

¿From 1. we conclude that there are only finitely many distinct n_{i}’s. Also there are only finitely many X(i)’s and so by 3.3 we may assume that for all i and some X and n, X(i) = X and n_{i} = n. By 2. and 13.5 there is a bound on the q_{i}’s with p_{i} ≠ operatornamecharK. Thus, p = operatornamecharK is a prime and p_{i} = p for all but finitely many i.

bglv Lemma 13.8 Let K be a field of characteristic p ≥ 0, V a finite dimensional vector space over K, B a basis for V, G = GL_{V} (K) and S = SL_{H}(V). Let W be a K-subspace of V and put Q_{W} = C_{G}(W) ∩ C_{G}(V/W)

(a) Q_{W} ≤ SL_{K}(W) and Q_{W} ∩ K = 1.

(b) N_{G}(W)/Q_{W} ≅ GL_{K}(W) × GL_{K}(W),

N_{S}(W)/W ≅ {(a, b)|a ∈ GL_{K}(W), b ∈ GL_{K}(V/W), det(a) det(b) = 1}.

(c) Q_{W} is an elementary abelian p-group (where an elementary abelian 0-group is defines to be a torsion free abelian group). As ℤN_{G}(W)/Q_{W} -module Q_{W} is isomorphic to Hom_{K}(V/W, W).

(d) If W is a proper subspace of V, then C_{G}(Q_{W}) = Q_{W} K^{#} and C_{P SL_{K}(V)}(Q_{W}) = Q_{W} K^{#}/K^{#}

(e) SL_{K}(V) = ⟨Q_{Kb} | b ∈ B⟩.

(f) Q_{B} acts transitively on the points of V \ B.

(g) Let V = A ⊕ B with dim A = 1. Then SL_{K}(V) = ⟨Q_{A}, Q_{B}⟩.

(h) Suppose that dim_{K} V ≥ 3 or |K| ≥ 4. Then every proper normal subgroup of SL_{K}(V) is contained in K and P SL_{K}(V) is simple.

Proof: (a) Obvious.

(b) There is an obvious homomorphism from ϕ : N_{G}(W) → GL_{K}(W) × GL_{K}((V/W) and Q_{W} is the kernel of this homomorphism. Also if V = W ⊕U,then GL_{K}(W)×GL_{K}(U) ≤ GL_{K}(V), and ϕ restricted to this subgroup is an isomorphism. Hence ϕ is onto. This proves the first part of (b). The second is obvious.

(c) For q ∈ Q_{W} define q^{∗} ∈ Hom(V/W, W) by q ∗ (v + W) = [v, q]. As q centralizes V/W, [v, q] indeed lies in W and since q centralizes W this well defined. Conversely if α ∈ Hom(V/W, W) define α^{∗} ∈ Q_{W} be α^{∗}(v) = v + α(v + W).

(d) Let g ∈ C_{G}(Q_{W}). Let H be a hyper-plane (that is a subspace of codimension 1) of V/W and U a 1-space in W. Pick α ∈ Hom(V/W, W) with ker α = H and ℑα = U. Then ker α^{g} == H^{g} and ℑα^{g} = U^{g}. Since α^{∗} = α^{g} we also have α = α^{g} and so H^{g} = H and U^{g} = U. Since this is true for all such H and U it is easy to see that g acts as an scalar on W and on U. αg = gα implies that these scalars are the same. Thus the first part of (d) holds. The second follows from the first and (a).

(e) Let n = dim KV. If n = 1, there is nothing to prove. So suppose n ≥ 2. Let H = ⟨Q_{Kb} | b ∈ B. Let a ∈ B, B_{0} = B − a, H_{0} = ⟨Q_{Ka} ∩ C_{G}(a) | a ∈ B_{0}⟩ and V_{0} = K⟨B_{0}⟩. Let b ∈ B. By induction on n, H_{0} = SL_{K}(V_{0}).

Step 1 If v + Ka = w + Ka, then v^{q} = w for some q ∈ Q_{Ka}.

Pick α ∈ Hom_{V/Kb,Kb} with α(v + Ka) = w − v. Then v^{α^{∗}} = w.

Step 2 H acts transitively on V^{#}.

By step 1 every 1-space in V \ K_{a} is conjugate under Q_{Ka} to am element of V_{O}. Since H_{0} acts transitively one the 1-spaces of V_{0}, we conclude that Q_{Ka}H_{0} acts transitively on the 1-spaces different from 1. Moreover, Q_{Kb} does not fix Ka and so H is transitive one the 1-spaces of V. It remains to show that for all λ ∈ K^{#}, a and λ · a are conjugate under H. By step 1 (sometimes applies to b in place of a), a and a + b are conjugate under Q_{Kb}, a + b and λ · a + b are conjugate under Q_{Ka}, and λ · a + b and λ · a are conjugate under Q_{Kb}.

Step 3 Q_{Ka} acts transitively on the complements to Ka in V.

Let W be any complement to Ka in V. Let β(v_{0} + Ka) be the projection of v_{0} onto Ka via V = Ka ⊕ W. Then β ∈ Hom_{V/Kb,Kb} and v_{O} + β(v_{0}) ∈ W for all v_{0} ∈ V_{0}. Thus V_{0}^{β^{∗}} = W and Step 3 is established.

Let s ∈ SL_{V} (K). By Step 2 and Step 3 there exists h ∈ H so that sh centralizes a and normalizes V_{0}. Thus sh ∈ SL_{K}(V_{0}) ≤ H_{0} ≤ H and s ∈ H.

(f) This is just the dual statement to Step 3 in (e).

(g) By (f) V = ⟨A^{Q_{B}}⟩ and so A^{Q_{B}} contains a basis. Thus by (e) ⟨Q_{A}^{Q_{B}}⟩ = SL_{K}(V).

(g) Let N be a normal subgroup of S. Let A be a 1-space and M = N_{S}(A). Clearly SL_{K}(V) acts double transitive on the 1-spaces and so M is a maximal subgroup of G. If N ≤ M, then as N is normal in S and S is transitive, S normalizes all the 1-spaces in V. Thus N ≤ K. So suppose N ≰ M. Since M normalizes Q_{A} we conclude that MQ_{A} ⊴ S. Thus S = NQ_{A} and [Q_{A}, M] ≤ N. Suppose that [Q_{A}, M] ≠ Q_{A}. Since C_{G}(A) acts irreducibly on Q_{A}^{#} we conclude that [Q_{A}, M] = 1. Thus by (d) M ≤ Q_{A}K. This implies dim V = 2 and |K| ≤ 3, a contradiction to the assumptions. Hence Q_{A} = [Q_{A}, M] ≤ N and S = N.

mdegsl Lemma 13.9 Let K be a field in positive characteristic p. Let n > 1 an integer. Let ϕ : P SL_{n}(K) ← P GL_{F} (W) be a embedding with operatornamecharK = operatornamecharF and dim_{F} W ≤ n. Then dim F W = n. Moreover if A is point or a hyper-plane in V, then ϕ(Q_{A}) ≤ Q_{B} for a unique point or hyper-plane B in W.

Proof: By induction on n. Let p = operatornamecharF. Since Q_{A} is a p-group, Q_{A} centralizes proper subspace B in W. If n = 2, B has to be a point and we are done. So suppose n > 2. Note that N_{SL_{n}(K)}(A)/Q has a subgroup isomorphic to SL_{n−1}(K). Induction and the simplicity of P SL_{n−1}(K) (respectively the structure of P SL_{2}(2) and P SL_{2}(3) implies either B or W/B has dimension at least n − 1. Thus B is a point or a hyper-plane and Q_{A} ≤ Q_{B}. pslslt Theorem 13.10 Let G and G_{1} ≤ G_{2} ≤ … be as in 13.7(a) with X = A. Let F be the

subfield of 𝔽̄_{p} generated by the 𝔽_{p^{k_{i}}}. Then G ≅ A_{n}(F).

Proof: Let d = n = 1. Let F_{i} = 𝔽_{p^{k_{i}}} and V_{i} = F_{i}^{d}. Let V_{∞} and F_{∞} be a suitable ultra product of the V_{i} and F_{i}, respectively. Then V_{∞} is a projective G-module of degree d over F_{∞}. Put G_{∞} = G. Let Z_{i} = {k ∈ F_{i} | k^{d} = 1}. Then for i < ∞, G_{i} ≅ P SL_{K_{i}}(V_{i}) ≅ SL_{K_{i}}(V_{i})/Z_{i}. |Z_{i}| ≤ d and so passing to a subsequence we may assume |Z_{i}| = z for all 1 ≤ i < ∞ and some z. Then also |Z_{∞}| = z.

Let V_{1} = A_{1} ⊕B_{1} for a point A_{1} and a hyper-plane B_{1}. Then for 1 ≤ i ≤ ∞, Q_{A_{1}} ≤ Q_{A_{i}} for some point or hyper-plane A_{i} in V_{i}. Replacing V_{i} by its dual if necessary we may assume that A_{i} is a point. Also let B_{i} be the point or hyper-plane of V_{i} with Q_{B_{1}} ≤ Q_{B_{i}}. Then G_{i} = ⟨Q_{A_{1}}, Q_{B_{1}}⟩ normalizes A_{i} + B_{i} and so V_{i} = A_{i} ⊕ B_{i}. Note also that if i ≤ j ≤ ∞, then Q_{A_{i}} ≤ Q_{A_{j}}. Fix 1 ≤ i < ∞. Let M_{i} = N_{G_{i}}(A_{i}) and M_{i}^{∗} the inverse image of M_{i} ∈ SL_{F_{∞}}(V_{∞}). Let K_{i} be the subfield of F_{∞} generated by the scalars induced by M_{i}^{∗} on A_{∞}. The p-elements in M_{i}^{∗} centralize A_{∞}. Also M_{i}/O^{p′}(M_{i}) ≅ F_{i}^{#}/Z_{i}, M_{i}^{∗}/Z_{∞} ≅ M_{i} and Z_{i} ≅ Z_{∞}. Thus K_{i} is isomorphic to a subfield of F_{i}. Let O ≠ a ∈ A_{∞}. Let W_{i} be the K_{i}G_{i} submodule of V_{∞} spanned by a. Also let t_{1}, t_{2}, … t_{d} ∈ G_{1} so that V_{1} = ⊕_{i=1^{d}}A_{1}^{t_{i}} and let a_{j} = a^{t_{j}}

Claim For all i < ∞, W_{i} = K_{i}a ⊕ (W_{i} ∩ B_{∞})

Let g ∈ G_{i}. If A_{i}^{g}B_{i}, then Q_{B_{i}} ≤ M_{i}^{g} and so A_{i}^{g} ≤ C_{W_{i}}(Q_{A_{i}}) ≤ B_{∞}. If A_{i}^{g} ≰ B_{i}, then by 13.8f, A_{i}^{g} = A_{i}^{q} for some q ∈ Q_{B_{i}}. Hence g = mq for some m ∈ M_{i}^{∗}. By definition of K_{i}, m normalizes K_{i}a and so K_{i}a^{g} = K_{i}a^{q} ≤ K_{i}a[K_{i}, q] ≤ K_{i}a(B_{∞} ∩ W_{i}).

Thus the claim is proved. It follows that W_{i} ∩ A_{∞} = K_{i}a and so [W_{i}, Q_{A_{i}}] ≤ K_{i}a. Hence G_{i} = ⟨Q_{A_{i}}^{t_{j}} | 1 ≤ j ≤ d normalizes ⊕_{j=1}^{d}K_{i}a_{j}. Hence a_{1}, a_{2}, … a_{d} is a K_{i}-basis for W_{i}. W_{i} is a faithful projective G_{i} module and so |P SL_{d}(K_{i})| ≥ |P SL_{d}(K_{i}). Since K_{i} is a isomorphic to a subfield of F_{i} this implies that K_{i} and F_{i} are isomorphic. Moreover G_{i} maps isomorphically onto P SL_{K_{i}}(W_{i}).

As G = ⋃ G_{i} we have W_{∞} = ⋃ i = 1^{∞}W_{i}, K_{∞} = ⋃ K_{i} ≅ F and G = ⋂ P SL_{K_{i}}(W_{i}) = P SL_{K_{∞}}(W_{∞}) ≅ P SL_{d}(F).

The next theorem (or in some cases parts of it) was proved independently by Belyeav [?], Borovik [?], Hartley and Shute [?] and Thomas [?]. We refer the reader to whose papers for a proof.

Theorem 13.11 Let G be an infinite, periodic,linear, simple groups. Then G is a groups BBHST of Lie type over an infinite,locally finite field. More precisely if G_{1} ≤ G_{2} ≤ … are as 13.7(a) and F is the subfield of 𝔽̄_{p} generated by the 𝔽_{p^{k_{i}}}. Then G ≅ X_{n}(F).

62 CHAPTER 13. PERIODIC, SIMPLE, LINEAR GROUPS

The maximal subgroups of SL_{2}(K)

masusl

In this section p is a prime and K is locally finite field in characterictic p, that is K is a subfield of the algebraic closure of 𝔽_{p}. Let V be a 2-dimensional vector space over K , L = SL_{K}(V), Z = Z(L) = {±1} and L̄ = L/Z(L) = PSL_{K}(V). The goal of this section is to determine the maximal subgroup of L.

Lemma 14.1 Let m = sup{i | 𝔽_{p^{2^{i}}} ≤ K}. ffson

(a) F has a field extension of degree 2 if and only if m < ∞.

(b) Let F be a field extension fo degree 2 of K. Then

(a) F : K is Galois.

(b) F is unique up to K-isomorphism. Namely F is the splitting field if x^{p^{2^{m+1}}} − 1 over K

(c) Let σ the non-trivial K-automorphism of F. Then K = {ff^{σ} | f ∈ F }.

Proof: Let E the algebraic closure of K. We start with describing the subfields of E. Let q be a power of p. Then (ab)^{q} = a^{q}b^{q} and (a + b)^{q} = a^{q} + b^{q} for all a, b ∈ E. Hence the q roots of the polynomial x^{q} − x for a subfield 𝔽_{q} of E. Conversely, if T is a subfield of order q in E, then as (T^{#}, ·) is a groups of order q − 1, t^{q−1} = 1 for all o ≠ t ∈ T and so t^{q} − t = 0 for t ∈ T. Thus 𝔽_{q} is a unique subfield of order q in E. Let T be any subfield and for a prime r let m_{r}(T) = sup{r^{i} | 𝔽_{p^{r^{i}}} ≤ T }. Let N_{K} be the formal product ∏_{r} m_{T} (r),there the product is taken of all the primes. For two such formal we can build the product, the greates comman divisor and leat commom multiple in an obviuos way. Given a formal product N = ∏ m_{r} let 𝔽_{p^{N}} = ⋃_{n∈ℕ}{𝔽_{p^{n}} | ndividesN}. Then 𝔽_{p^{N}} is a subfield of E with N_{𝔽_{p^{N}}} = N and T = 𝔽_{p^{N_{T}}}. Hence we obtain a one to one correspondence between subfield of E and the formal products. Given two formal products N, M. Then 𝔽_{p^{N}} ∩ 𝔽_{p^{M}} = 𝔽_{p^{gcd(N,M)}}, The subfield of E generated by 𝔽_{p^{N}} and 𝔽_{p^{M}} is 𝔽_{p^{lcm(N,M)}}. Also 𝔽_{p^{N}} ≤ 𝔽_{p^{M}} if and only if N divides M, in which case the degree of this extension is M/N.

64 CHAPTER 14. THE MAXIMAL SUBGROUPS OF SL_{2}(K)

By the preceeding dicussion it remains to prove (bc) Since F is the union of its finite subfields and since all finite subfields are invariant under σ we may assume that F is finite. Let q = |K|. Then x → x^{q} is non trivial K-autmorphism of F and so σ(f) = f^{q} for all f ∈ F. Since σ has order two, ff^{σ} is fixed by σ and so lies in K. The map f → ff^{σ} = f^{q+1} induces a homomorphism form the multiplicative group of F to the multiplicative group of K. Since F has at most q + 1 (q + 1)’th roots of unity, the cernel of this map has order at most q + 1. As |F^{#}| = q^{2} − 1 = (q − 1)(q + 1) and |K^{#}| = q − 1, this map must be onto.

absl Lemma 14.2 Let A be an abelian subgroup of L. Then C_{L}(A) is is the unique maximal abelian subgroup of L containing A, N_{L}(A) is solvable of derived length two and one of the following holds:

1. (a) Ā is an elementary abelian p-group.

(b) KA has exactly one proper submodule W in V.

(c) C_{L}(M) = Z(L)U where U = C_{L}(W) ≅ (K, +) is a maximal unipotent subgroup

of L. With respect to some of basis V, U = \{( \begin{matrix}1 & 0 \\ k & 1 \end{matrix} ) | k ∈ K\}.

(d) N_{L}(A) = N_{L}(W), N_{L}(W)/U ≅ (K^{#}, ·) and with respect to some basis of V,

N_{L}(A) = \{( \begin{matrix}λ & 0 \\ k & λ^{−1} \end{matrix} ) | k ∈ K, λ ∈ K^{#}\}.

2. (a) A is a locally cyclic p′ group.

(b) KA has excactly two proper submodules W_{1}, W_{2} on V

(c) C_{L}(A) = N_{L}(W_{1}) ∩ N_{L}(W_{2}) ≅ (K^{#}, ·). With respect to some basis V, C_{L}(A) =

\{( \begin{matrix}k & 0 \\ 0 & k^{−1} \end{matrix} ) | k ∈ K^{#}\}.

(d) N_{L}(A) = N_{L}({W_{1}, W_{2}}), |N_{L}(A)/C_{L}(A)| = 2 and N_{L}(A)/C_{L}(A) inverts C_{L}(A).

3. (a) A is a locally cyclic p′-group.

(b) A acts irreducible on V.

(c) Let F = End_{KA}(V). Then F is a field extension K of degree 2. Let 1 ≠ σ ∈ Gal(F, K), then C_{L}(A) = F ∩ L = {f ∈ F | ff^{σ} = 1}.

(d) N_{L}(A) = N_{L}(F), |N_{L}(A)/C_{L}(A)| = 2 and N_{L}(A)/C_{L}(A) inverts C_{L}(A).

Proof:

Case 1 A is not a p′ group.

Let 1 ≠ b ∈ B be an element of order p. Then (b − 1)^{p} = b^{p} − 1 = 0. Hence there exists v_{2} ∈ V with v_{1} = v_{2}^{(b−1)} ≠ neq0 and v_{1}(b − 1) = 0. Then W = C_{V} (b) = Kv_{1} is the only proper K∠b⟩-invarinant subspace of V. As A is abelian, A normalizes W and so 1b holds. Hence also N_{L}(A) leaves W invariant and 1d holds. A simple calculation shows that C_{L}(b) = UZ(L) where U is as in 1c. Since UZ(L) is abelian, UZ(L) = C_{L}(A) and so 1c holds. Also Ā ≤ Ū and so 1a holds.

Case 2 A is a p′ group and A is not irreducible.

Let W_{1} be a proper KA-submodule in V. Let a ∈ A \ Z(L). Then a acts as a scalar λ on W_{1}. Since det a = 1, a acts as λ^{−1} on V/W_{1}. Suppose that λ = λ^{−1}. Then λ = ±1. Put then a^{p} = ±1 and as A is a p-group, a = ±1 a contradcition to a ∉ Z(L). So λ ≠ 1. Let W_{2} be the eigenspace of a corresponding to the eigen vector λ^{−1}. Then W_{1} and W_{2} are the only proper K⟨a⟩ submodules in V and so 2b holds. Is is now easy to check that also 2c and 2d hold. Finally finite subgroups the multiplicative groups of field are cyclic and so also 2a holds.

Case 3 A is irreducible( and so also a p′ group)

By Schur lemma F is a division ring. Clearly K ≤ F and as A is abelian A ≤ F and so F ≠ A. Since V is a vector space over F we get 2 = dim_{K} V = dim_{F} V · dim_{K} F. Hence F as degree 2 over K and V is 1-dimensional over F. In particular we may and do identify the K-space V with the K-spave F. Let f ∈ F \ K. Then 1, f is a K basis for V. Let f^{2} = −k − lf for some k, l ∈ K. So m_{f}(x) = x^{2} + lx + k is the minimal polynomial for f

over K and the matrix for f with respect to the basis 1, f is ( \begin{matrix}0 & 1 \\ −k & −l \end{matrix} ). Thus det f = k.

As f ∈ SL(V) if and only if k = 1.

In particular for a ∈ A \ Z(L) we see that a and a^{−1} are the roots of m_{a}(x), F contains all roots of m_{a}(x) and m_{a}(x) does not have a double root. Thus F : K is a Galois extension.

Let σ be the non trivial element of Gal(F : K). Then f, f^{σ} are the roots of m_{f}(x) and so k = ff^{σ}. Hence f ∈ L if and only if ff^{σ} = 1. Note that σ ∈ GL_{K}(F) = GL_{K}(V). With rspect to the basis a, a^{−1} we see that det σ = −1. So in order to show that N_{L}(F) ≰ C_{L}(F) we need find f ∈ F with det f = −1, for then fσ ∈ N_{L}(F) \ C_{L}(F). This amounts to finding f ∈ F with ff^{σ} − 1. But such an f exists by 14.1. Since C_{L}(A) ≤ F, 3a holds and thus 3. holds in this case.

Lemma 14.3 Let M ≤ L. Then one of the following holds. solmn1

(a) M has an abelian normal subgroup A with A ≰ Z(L) and the structure of M ≤ N_{L}(A) is described by 14.2. In particular, M is solvable of derived length at most 2.

(b) p is odd and there exists Q ⊴ M so that Q is a quaternion group of order eigth. Moreover NL̄(Q) ≅ Alt(4) or Sym(4).

(c) M̄^{∞} is locally ”finite,simple” and so simple.

Proof: Suppose first that M has a solvable normal subgroup Q with Q ≰ Z(L). Choose Q of minimal derived length. If Q is abelian, then (a) holds. So suppose that Q is not abelian. Then by minimal choice Z(Q) = Q′ = Z(L) and p is odd. By 18.13 Q/C_{Q}(Q) ≤ 2^{2} = 4. If |Q/Z(Q)| ≤ 3, Q/Z(Q) is cyclic and so Q is abelian. So |Q/Z(Q)| = 4 and |Q| = 8. By 14.2, C_{2} × C_{2} is nor a subgroup of L and so Q is a quaternion group. It is easy to see that C_{L}(Q) = Z(L) and so NL̄(Q) is ismorphic to a subgroup of Aut(Q). Since Aut(Q) ≅ Sym(4) it rmains to verify that 3 divides N_{L}(Q). We leave this as an exercise to the reader.

Note that we have probed that all solvable subgroups of L have derived length at most 4 and so all locally solvable subgroups of L are solvable. So we may assume that LSol(M) ≤ Z(L). Let R be non trivial normal finite perfect subgroup of M and T a normal subgroup of R minimal with respect to T ≰ Z(L). Let S be some Sylow subgroup of T with S ≰ Z(L). Then by Frattini, R = T N_{R}(S). Since N_{L}(S) is solvable and R is perfect, R = T. We conclude that R̄ is simple.

BurTho Lemma 14.4 Let H be a finite group.

(a) Let r a prime and S a Sylow r-subgroup of G. If S is cyclic and N_{H}(S) ≤ C_{H}(S), then H has a normal p-complement, that is H = NS for some normal p′ subgroups N of H.

(b) Let T < S ≤ G with S a Sylow 2-subgroup of H. Let t be an involution in H and suppose that |S/T | = 2. Then either t is conjugate in H to an element of T or H has a normal subgroup of index two which does not contain t.

(a) Since S is cyclic S has a faithful 1 dimensional module over ℂ. Let W be the ℂ module induced from U. For h ∈ H let sign(h) ∈ {1, −1} be the sign of h as a permutation on H/S and let det h the determinant of h as a linear map map on W.

If h is a r′ element then no non-trivial power of h normalizes a conjugate of W under H and so as ℂ⟨h⟩-module, W is isomorphic to the permutation module on H/S. Thus det h = sign h.

So suppose now that s ∈ S with |h| = p. Let λ be scalar induced by s on U. Let g ∈ H so that s fixes Sg. Then both s ∈ S ∩ S^{g} and so both S and S^{g} are Sylow p-subgroups of C_{H}(s). Hence S^{g} = S^{c} for some c ∈ C_{H}(s). Thus gc^{−1} ∈ N_{H}(S) = C_{H}(S) ≤ C_{H}(s) and so g ∈ C_{H}(s). In particular s acts as λ on U^{g}. Let m = |C_{H}(s)/S|. We conclude that det s = λ^{m} sign h.

Define ϕ(h) = det(h) sign(h). Then ϕ is a homorphism form H into the multiplicative group of C. Also by the above ϕ(h) = 1 for all p′ elements and ϕ(s) = λ^{m} ≠ 1 as λ is a p-root of unity and m is coprime to p. Hence H = ker ϕS and ker ϕ ∩ S = 1.

(b) Consider ther action of H on Ω = H/T by right multiplication. Suppose first that H has a fixed point T h on Ω. Then T ht = T h and so t^{h_{1}} ∈ T. So suppose that t acts fixed point freely on Ω. Then t as \frac{|Ω|}{2} = \frac{|H|}{|S|} orbits on Ω. Thus t induces an odd permutaion on Ω. Thus the normal subgroup of H consisting of all the elements inducing an even permutation on Ω has index two and does no contain Ω.

ssl Lemma 14.5 Let M ≤ L so that M̄ is non-abelian, finite simple simple. The one of the following holds.

1. There exists a finite subfield K_{M} of K and 2-dimensional K_{M} subspace V_{M} with KV_{M} = V and M = SL_{K_{M}} (V_{M}).

2. p is odd and M̄ ≅ Alt(5).

Proof: Let A_{1}, A_{2}, … A_{r} be representatives for the conjucagcy classes of maximal abelian p′ subgroups of M. Also let Ā_{0} be a Sylow p subgroup of M̄.

1. Every non trivial element of M̄ is contained in exactly one conjugate of exactly on of ssl-1 the Ā_{i}’s, 0 ≤ i ≤ r.

Let x ∈ M \ Z(L) and put A = ⟨x⟩. Then A is abelain. If A is not a p′ group then by 14.2 Ā is a p-group and so A is contained in a conjugate of A_{O}. If A is a p′ group, then A is contained a maximal abelian p-prime subgroup of M, which by definition is a conjugate of one of the A_{i}’s, 1 ≤ i ≤ r. Finally if A ≤ A_{i}^{x} for some x ∈ M, then A_{i}^{x} ≤ C_{M}(A) which by 14.2 is abelian and either (modulo Z(L) a p or a p′ group. Thus A_{i}^{x} = C_{M}(A) is unique.

Let a_{i} = |Ā_{i}|, m = |M̄| and r_{i} = |N_{L}(A_{i})/A_{i}|. Let e = gcd(2, p − 1)

2. 1 − \frac{1}{m} = ∑_{i=0}^{r} \frac{1}{r_{i}} (1 − \frac{1}{a_{i}}) ssl-2

Fist note |NL̄(A_{i})| = r_{i}a_{i} and so A_{i} has |L/N_{L}(A_{i})| = \frac{m}{r_{i}a_{i}} conjugates. Each of these conjugates contains a_{i} − 1 non-trivial elements and so by 1. and as M̄ as m − 1 non-trivial elements:

m − 1 = ∑_{i=0}^{r} \frac{m}{r_{i}a_{i}} (a_{i} − 1).

Dividing boths sides by m gives 2..

3. a_{i} ≠ 1 for at least two 0 ≤ i ≤ r ssl-2.5

Suppose a_{i} ≠ 1 for exactly one i. Then as a_{i}r_{i} ≤ m, 2.

1 − \frac{1}{a_{i}r_{i}} ≤ 1 − \frac{1}{m} = \frac{1}{r_{i}} (1 − \frac{1}{a_{i}}) = \frac{1}{r_{i}} − \frac{1}{a_{i}} r_{i} ≤ 1 − \frac{1}{a_{i}r_{i}}

Hence all inequalites are inequalities and we conclude, a_{i}r_{i} = m, r_{i} = 1 and a_{i} = m. Thus M̄ = Ā_{i} is abelian a contradiction.

4. ssl-3

(a) Let 1 ≤ i ≤ r, then r_{i} ≤ 2.

(b) If a_{0} ≠ 1, then r_{0} = a_{j} for some 1 ≤ j ≤ r.

(a) follows directly from 14.2. For (b) suppose that a_{0} ≠ 1. Suppose that r_{0} = 1. If a_{0} ≤ 3, then Ā_{0} is a cyclic Sylow p-subgroup of M̄. Since r_{0} = 1, N_{M}(A_{0}) = A_{0} is abelian, a contradiction to 14.4. Thus a_{0} ≥ 4 and so \frac{1}{r_{0}} (1 − \frac{1}{a_{0}} ≥ \frac{3}{4}. By 3. r ≥ 1 and by (a) \frac{1}{r_{1}} (1 − \frac{1}{a_{1}} ≥ \frac{1}{2}(1\frac{1}{2}) = \frac{1}{4}.

Thus 2. provides the contradiction

1 − \frac{1}{m} ≥ \frac{3}{4} + \frac{1}{4} = 1

Hence r_{0} ≠ 1. Let r be a prime divisor of r_{0} and T be a Sylow r subgroup of N_{M}(A_{0}). Let A = C_{M}(T). Then be 2., A = A_{i}^{g} for some i and g in M. By 14.2(1) there exists a unique 1-space Y invariant under A_{0}. In particular also T normalises Y and by 14.2(2), T fixes exactly two 1-spaces and both of these are also invarinat under A. Thus A normalizes Y and so also A_{0}. Thus A = N_{N_{M} (A_{0})}(T). Since N_{M}(A_{O})/A_{0} is cyclic, A_{0}T is normal in N_{M}(A_{0}) and so by a Frattini argument, N_{M}(A_{0}) = A_{0}A. Thus r_{0} = |Ā| = a_{i}.

ssl-4 5. Let r be a primes divisor of M̄.

(a) r divides exactly one of the a_{i}’s.

(b) Let S_{r} be a Sylow r subgroup of M. Then exactly one of the following holds:

1. r is odd or p = 2, and S_{r} ≤ A_{i}^{g} for some g ∈ M.

2. r = 2, p is odd and for some g ∈ M, S_{r} ∩ A_{i}^{g} is a Sylow r-subgroup of A_{i}^{g} and |S_{r}/S_{r} ∩ A_{i}^{g}| = 2.

(c) r_{i} = 2 for all i ≥ 1.

Let ā be an element of order r in M̄. Then a ∈ A_{i}^{g} for some i and some g ∈ M. Without loss a ∈ A_{i}. Then r divides a_{i}. Let T_{r} be the Sylow r-subgroup of A_{i}. Let t ∈ N_{M}(T_{r}). Then T_{r} = T_{r}^{g} ≤ A_{i} ∩ A_{i}^{g} and so by 1., A_{i} = A_{i}^{g}. Thus N_{M}(T_{r}) = N_{M}(A_{i}).

Supppose first that r is odd or p = 2. Without loss T_{r} ≤ S_{r}. By 4., r does not divide r_{i}. So T_{r} is a Sylow p-subgroup of N_{M}(T_{r}) and hence N_{S_{r}}(T_{r}) = T_{r}. Hence T_{r} = S_{r}. So (b) holds. Assume that r also divides a_{j}. Then also A_{j} contains a Sylow r-subgroup of M and so 2. implies i = j. Thus (a) holds. Suppose i ≥ 1 and r_{i} = 1, then N_{M}(T_{r}) = A_{i} is abelian. Since T_{r} is a cyclic Sylow r subbgroup of M, this contradicts 14.4. So (c) holds for this i.

Suppose next that r = 2 and p is odd. Without loss ā ≤ Z(S̄). Then T_{r} = C_{S_{r}}(a). S_{r} acts on the coset ā = aZ(L) and aZ(L) contains only two elements. Thus |S_{r}/T_{r}| = 2 and S_{r} ≤ N_{M}(T_{r}). If S_{r} = T_{r} we obtain the same contradiction as at the end of the last paragraph. Thus S_{r} ≠ T_{r} and r_{i} = 2. So (b) and (c) hold. Assume that r = 2 also divides a_{j}. Then Ā_{j} contains an involution b̄. As T_{r} has indes two on S_{r}, 14.4 implies that b^{g} ∈ T_{r} ≤ A_{i} for some g. Thus b^{g} ∈ A_{j}^{g} ∩ A_{i} and so by 2., j = i. So also (a) holds and all parts of 5. are proved.

ssl-5 6. m = e ∏_{i=0}^{r} a_{i}.

Clearly m = ∏ |S̄_{r}|, where the product is taken over all prime divisors r of m and S_{r} is a Sylow rp-subgroup of M. Thus 6. follows from 5.(b).

7. One of the following holds ssl-6

(a) r = 3

(b) r = 2 and a_{0} ≠ 1.

Since r_{i} ≤ 2 for all i ≥ 1 2. implies

1 − \frac{1}{m} ≥ r \frac{1}{2} \frac{1}{2}

hence 1 < \frac{r}{4} and r < 4. Suppose that neither (a) nor (b) hold. Then r ≤ 2 and a_{0} = 1, or r ≤ 1. So by 3. there

exists exactly two a_{i}’s ( say i and j) with a_{i} ≠ 1. Then r_{i} ≥ 2 and r_{j} ≥ 2 and so by 2.

1 − \frac{1}{m} = \frac{1}{r_{i}} (1 − \frac{1}{a_{i}} + \frac{1}{r_{j}} (1 − \frac{1}{a_{j}} ≤ \frac{1}{2}(1 − \frac{1}{a_{i}} + 1 − \frac{1}{a_{j}})

Thus

\frac{1}{a_{i}} + \frac{1}{a_{j}} ≤ \frac{1}{m}

By 6., m = ea_{i}a_{j} so multiplication with m implies

e(a_{j} + a_{i}) ≤ 1

A contradiction, as e, a_{i} and a_{j} ar all larger or equal to a.

8. Suppose that r = 3. Then r_{0} = 1, p > 5, {a_{1}, a_{2}, a_{3}} = {2, 3, 5} and m = 60. In ssl-7 particular, M̄ ≅ Alt(5).

Since r = 3, 2. implies:

\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} = 1 + \frac{2}{m} + \frac{2}{r_{0}} (1 − \frac{1}{a_{0}} ) > 1

Suppose that {a_{1}, a_{2}, a_{3}} ≠ {2, 3, 5}. Then since the a_{i}’s are relatively prime

\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} ≤ \frac{1}{2} + \frac{1}{3} + \frac{1}{7} = \frac{21 + 14 + 6}{42} = \frac{41}{42} < 1

a contradiction.

Thus {a_{1}, a_{2}, a_{3}} = {2, 3, 5}. Since p does not divide any a_{i} for i ≥ 1, p > 5. In particualar, p is odd and m = 2 · 2 · 3 · 5 · a_{0} and so m = 60a_{0}. Also

\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} = \frac{15 + 10 + 6}{30} = \frac{31}{30}

\frac{1}{60} = \frac{1}{60a_{0}} + frac1r_{0}(1 − \frac{1}{a_{0}} )

But this clearly implies, a_{0} = 1 and so m = 60.

ssl-8 9. Suppose that r = 2 and a_{0} ≠ 1. Put q = a_{0} and choose notation so that r_{0} = a_{1}. Then one of the follwing holds:

(a) ea_{1} = q − 1, ea_{2} = q + 1, Z(M) = Z(L) and |M| = q(q + 1)(q − 1).

(b) q = p = 3, a_{1} = 2, a_{2} = 5 and m = 60. In particular, M̄ ≅ Alt(5).

Without loss A_{1} ≤ N_{M}(A_{0}. By 1. CM̄ ≤ Ā_{0} for all 1 ≠ a ∈ Ā_{0} and so all orbits of Ā_{1} on Ā_{0}^{#} have length a_{1}. Thus

a_{1} dividesq − 1  (1)

By 2.

1 − \frac{1}{eqa_{1}a_{2}} = \frac{1}{a_{1}} (1 − \frac{1}{q}) + \frac{1}{2}(1 − \frac{1}{a_{1}} ) + \frac{1}{2}((1 − \frac{1}{a_{2}} )

Thus

\frac{1}{2a_{2}} − \frac{1}{eqa_{1}a_{2}} = \frac{1}{2a_{1}} − \frac{1}{a_{1}q}

Put d = \frac{2}{e} = gcd(2, p) and multiply the previous equation with 2a_{1}a_{2}q. We obtain:

qa_{1} − d = qa_{2} − 2a_{2} = (q − 2)a_{2}  (2)

Thus q − 2 divides qa_{1} − d and so also qa_{1} − d − a_{1}(q − 2) = 2a_{1} − d. Since a_{1} > 1 and d ≤ 2, 2a_{2} − d > 0 and thus q − 2 ≤ 2a_{1} − d and so

a_{1} ≥ \frac{q − 1}{2} + \frac{d − 1}{2}  (3)

Suppose first that p = 2. Then d = 2 and (3) and (1) imply that a_{1} = q − 1 and by (2), a_{2} = \frac{q(q−1)−2}{q−2} = \frac{(q+1)(q−2)}{q−2} = q + 1. Thus 9. (a) holds in this case. Soppose next that p is odd. Then d = 1 by (3) and (1), a_{1} = \frac{q−1}{2} or a_{1} = q − 1. If a_{1} = \frac{q−1}{2} then by (2)

a_{2} = \frac{q \frac{q−1}{2} −1}{q−2} = \frac{1}{2} \frac{q(q−2)}{q−2} = \frac{q+1}{2} and again 9. holds. So suppose a_{1} = \frac{q−1}{2}. Thus by (2) a_{2} = q + 1 + \frac{1}{q−2}. This implies q = 3, a_{1} = 2 and a_{2} = 5. Hence 9. is proved.

ssl-9 10. Suppose that 9.(a) holds. Then there exists a subfield K of F and a 2-dimensional F subspace W of V with V = KW such that M = SL_{2}(W), where SL_{F} (W) is (in the canonical way) viewed as a subgroup of SL_{K}(V).

Let U be the 1-dimensional subspace normalizes by A_{0} and choose notation so that N_{M}(A_{0}) = N_{M}(U) = A_{0}A_{1}. For a ∈ A_{1} let λ(a) ∈ K be defined by u^{a} = λ(a)u for all u ∈ U. Let F = λ(a) | a ∈ A_{1} ∪ {0}. Since |A_{1}| = q − 1 and λ is a homorphism, F^{#} is cyclic of order q. Thus f^{q} = f for all f ∈ F and as q is a power of p, F ≅ 𝔽_{q} is a subfield of U. Let w ∈ N_{M}(A_{1}) \ A_{1}. Then N_{M}(A_{O}) ∩ N_{M}(A_{0}^{w}) = N_{M}(U) ∩ N_{M}(U^{w}) = A_{1}. In particular, |A_{0}^{wA_{0}}| = |A_{0}| = q. On the otherhand, |A_{O}^{M}| = |M/N_{M}(A_{0})| = |q(q − 1)(q + 1)/q(q − 1)| = q + 1 and so A_{0}^{wA_{0}} ∪ {A_{0}} = A_{O}^{M}. Hence {1} ∪ wA_{0} is a transversal to A_{O}A_{1}.Thus

M = A_{0}A_{1} ∪ A_{0}A_{1}wA_{1}  (4)

Let Y be a 1-dimensional F subspace of U. Let W = ⟨Y^{M}⟩. Then W as an F Msubmodule of V. Since Y is A_{0}A_{1} invariant,(4) implies W = ⟨Y +Y^{wA_{0}}⟩ = Y +Y^{w}+[Y^{w}, A_{0}]

Since [V, A_{0}] ≤ U and Y ≤ U we conclude that W = (W ∩ U) + Y^{w}. Thus W ∩ U^{w} = (W ∩ U ∩ U^{w}) + Y^{w} = Y^{w} and so W ∩ U = Y and W = Y + Y^{w}. Hence W is 2-dimensioal over F and M ≤ SL_{2}(W). As |SL_{2}(W) = q(q − 1)(q + 1) = |M| we conclude M = SL_{2}(W) and is proved. ssl-9

Lemma 14.6 Let p = 3 or p > 5 and M ≤ L with M̄ ≅ Alt(5). Then M = N_{L}(M). altsl

Proof: Note that Z(L) ≤ M. Let R = N_{M}(L) and let T be the kernel of the actions of R on the five Sylow 2-subgroups of M. If T Z̸(L) we conclude that N_{L}(T)/T is solvable, a contradcition to M ≤ N_{L}(T) and M ∩ T = Z(L). Thus T = Z(L) and so R̄ is isomorphic to a subgroup of Sym(5). Thus R = M or R̄ ≅ Sym(5). Let A be a Sylow 5-subgroup of M. Then A is abelian and as p ≠ 5 we conclude from 14.2 that |N_{R}(A)/C_{R}(A)| ≤ 2. This excludes the case R̄ ≅ Sym(5). Thus R = M.

Lemma 14.7 Let M ≤ SL_{2}(V) so that M̄ is infinite simple. Then there exists an infinite issl subfield F of K and a 2-dimensional F -subspace W of V with M = SL_{2}(W) and V = KW.

Proof: By 14.3 M̄ is locally ” finite ,simple”. As M is countbale we conclude that M is an ascending union of finite subgroups

M_{1} < M_{2} < M_{3} < M_{4} < … M_{k} <

such that M̄_{k} is simple. Without loss M̄_{k} is non-abelian and has order larger than 60. Thus by 14.5 M_{k} = SL_{2}(F_{k}) for some finite subfield F_{k} of K. Let k ≤ j. Then M_{k} ≤ M_{j} and so by 14.5 applied to K = M_{j}, F_{k} is a subfield of F_{j}. Let F = ⋃_{k=1}^{∞} F_{k}. Then F is an infinite subfield of K. Let a be an element of order p in M_{1} and 0 ≠∈ u ∈ C_{V} (a). Put W_{k} = ⟨F_{k}u^{M_{k}}⟩. Then as seen in the proof of 14.5 ??, W_{k} is 2-dimensional over F_{k} and M_{k} = SL_{F_{k}}(W_{k}). Put W = ⟨F u^{M} = ⋃_{k=1}^{∞} W_{k}. Then W is 2 dimensional over F and M = ⋃_{k=1}^{∞} M_{k} = ⋃_{k=1}^{∞} SL_{F_{k}}(W_{k}) = ⋃_{k=1}^{∞} SL_{F} (W).

Lemma 14.8 Let F be a subfield of K and W a 2-dimensional subspace F subspace of V nslf with V = KW. Let M = SL_{K}(W)

(a) N_{L}(M) = GL_{K}(W)F^{#} ∩ L = {λb | b ∈ GL_{F} (W), λ ∈ F^{#}, λ^{2} det b = 1}.

(b) Put F^{\frac{1}{2}} = \{k ∈ K | k^{2} ∈ F \}. Then

|N_{L}(M)/M| = |F^{\frac{1}{2} #}/F^{#}| ≤ 2

Proof: Let M = SL_{F} (W) and a ∈ GL_{K}(V). Then M^{a} = SL_{F} (W^{a}). Hence M = M^{a} if and only of M normalizes W^{a}. Let u be a non-trivial unipotent element in M, 0 ≠ w ∈ C_{W} (u) and 0 ≠ y ∈ C_{W^{a}}(u). Then y = λw for some λ ∈ K. Then W^{a} and λW are both 2-dimensional F M-submodules of V and containe y = λw. Thus W^{a} = λW. Hence aλ^{−1} ∈ N_{GL_{K}(V)}(W) = GL_{F} (W) and a = bλ for some b ∈ GL_{K}(W). Moreover a ∈ L if and only if 1 = det a = λ^{2} det b. Thus (a) holds.

To prove (b) define ϕ : N_{L}(M) → F^{\frac{1}{2} #}/F^{#}, λb → λ. We claim that ϕ is a well defined homomorphism, is onto and has kernel L. Indeed, suppose that λbμc. Then λμ^{−1} = cb^{−1} ∈ GL_{F} (W) ∩ K^{#} = F^{#} and so λF^{#} = μF^{#}. Also as λ^{2} det b = 1, λ ∈ F^{\frac{1}{2}} . So ϕ is well defined. Clearly ϕ is a homomorphism. Let λ ∈ F^{\frac{1}{2} #}. Then there exists b ∈ GL_{F} (W) with det b = λ^{−2} and so λb ∈ N_{L}(M) and ϕ is onto. If λb is in the kernel of ϕ if and only if λ ∈ F. And that is the case if and only if λb ∈ GL_{F} (W) ∩ L = SL_{F} (W). So the kernel of ϕ is M. Note that F^{\frac{1}{2} #} isa multiplicative subgroup of K^{#} and so is locally cyclic. It folloes that F^{\frac{1}{2} #}/F is a locally cyclic group, in which the square of every element is trivial. Thus F^{\frac{1}{2} #}/F has order at most two, and the proof of (b) is completed.

masusl Theorem 14.9 (Dickson’s List) Let M be a maximal subgroup of L. Then one of the follwing holds.

(a) M = N_{L}(U), U a 1-dimensional K subspace of W.

(b) M = N_{L}({U_{1}, U_{2}}), U_{1}, U_{2} distinct 1-dimensional K-subspaces of V.

(c) M = N_{L}(F), where F is field extension of degree teo of K.

(d) M = N_{L}(SL_{F} (W)), where F is a subfield of K, and W is a 2-dimensional F -subspace of V with V = F W.

(e) p is odd and M = N_{L}(Q), where Q ≤ L with Q ≅ Q_{8}.

(f) p = 3 or p > 5, and M̄ ≅ Alt(5).

Proof: This merely summarizes the results of this chapte. We also refer the reader to the lemmas in this section for a more detailed descriptions of structure of M.

The generalized Fitting subgroup of periodic, linear group

This section is devoted to the general structure of periodic linear groups. Our aim goal is to show that modulo the unipotent radical, the generalized Fitting subgroup of a periodic linear group has finite index. As an application we show that Jordan Hoelder Theorem holds for countable, periodic, linear groups.

Definition 15.1 Let X be a group. dgf

(a) X is called quasi simple if X is perfect, X ≠ 1 and X/Z(X) is simple.

(b) X is (absolutely semisimple if X is a direct sum of (absolutely) simple groups.

(c) A component for X is a quasi simple, subnormal subgroup of X.

(d) The layer E(X) is the subgroup of X generated by the components of X.

(e) The Hirsch-Plotkin radical LN(X) is the subgroup of X generated by all the locally nilpotent normal subgroup of X.

(f) The generalized Fitting subgroup F^{∗}(X) of X is the group LN(X) E(X)

We remark that the Hirsch-Plotkin radical is always locally nilpotent so LN(X) is the largest locally nilpotent normal subgroups of X. A proof for this non-trivial fact can be found in [?]. On the other hand for locally finite groups this is readily verified and we leave the details to the reader.

Lemma 15.2 Let X be group. comp

(a) Let Y be a subnormal subgroup of X and K a component. Then K ≤ Y or [Y, K] = 1.

(b) Distinct components of X commute. In particular E(X)/Z(E(G) ≅ ⨁{K/Z(K) | K a component of X.

74CHAPTER 15. THE GENERALIZED FITTING SUBGROUP OF PERIODIC, LINEAR GROUP

Proof: Note first that (a) implies (b). Indeed let K_{1} and K_{2} be components of X with [K_{1}, K_{2}] ≠ 1. Then by (a) K_{1} ≤ K_{2} and K_{2} ≤ K_{1}.

We prove (a) by induction on the defects of K and Y. If K = G, then either Y ≤ Z(K) or Y Z(K)/Z(K) is a subnormal subgroup of the simple group K/Z(K). The latter implies K = Y Z(K) Thus K = K′ ≤ Y. So we may assume that K ⊴ ⊴L ⊲ X

If Y = G we are done. So Y ⊴ ⊴N ⊲ G. If K ≤ N, we are done by induction. Hence [K, N] ≤ L ∩ N and K ≰ L ∩ N. Hence by induction [L ∩ N, K] = 1. Thus [N, K, K] = 1 and the three subgroup lemma implies, [N, K] = 1 and so [Y, K] = 1.

ff*g Proposition 15.3 Let G be a finite group. Then C_{G}(F^{∗}(G) ≤ F^{∗}(G).

Proof: Let D = C_{G}(F^{∗}(G)). Clearly a component of D is a component of G and F (D) ≤ F (G). Thus F^{∗}(D) ≤ F^{∗}(G) ≤ C_{G}(D) and F^{∗}(D) = Z(D). If D = Z(D) we are done. So let E/Z(D) be a minimal subnormal subgroup of D. Then E/Z(D) is simple and so either E/Z(D) is abelian or perfect. In the first case E is nilpotent and in the second case E = E′Z(D) and E′ is a component of G. Hence E ≤ F^{∗}(D) ≤ Z(D), a contradiction.

siass Lemma 15.4 Let S be a set of perfect, simple groups and X = ⊕S.

(a) If N ⊴ ⊴X, then N = ⊕N for some N ⊂ S.

(b) If N is serial in X and X is absolutely semisimple, then N = ⊕N for some N ⊂ S.

Proof: Let C be a (subnormal) series on X with N ∈ S. Let S ∈ S and C ∈ S. Then C ∩ S is a subnormal (serial) subgroup of S and as S is (absolutely) simple. C ∩ S = 1 or C ∩ S = S. In the latter case S ≤ S. Let T_{S} the intersection of the elements of S containing S and B_{S} the union of the elements which meet S trivially. Then (B_{S}, T_{S}) is a jump, T_{S} ≤ B_{S}, B_{S} ∩ S = 1 and since SB_{S}/B_{S} is a normal subgroup of T_{S}/B_{S}, T_{S} = SB_{S}. It T_{S} = T_{S^{∗}} for some other S^{∗} ∈ S we conclude [T_{S}, T_{S}] ≤ [S, S^{∗}]B_{S} ≤ B_{S},a contradiction. Let 1 ≠ n ∈ N and write s = s_{1} · . . . · s_{n} with 1 ≠ s_{i} ∈ S_{i} and distinct S_{1}, … S_{n} ∈ S. To complete the proof of the lemma it suffices to show that S_{i} ≤ N for all i. Without loss T_{S_{i}} ≤ B_{S_{n}} for all i < n. Then n ∉ B_{S_{n}} and so N ≰ B_{S_{n}}. But then T_{S_{n}} ≤ N and the lemma is proved.

sas Proposition 15.5 Let K be a quasi- absolutely simple, serial subgroup of G and Y a serial subgroup of G. Then Y ≤ K or [K, Y] = 1. In particular K is a component.

Proof: Let D be a composition series for G with Y ∈ D.

Suppose first that G = K. Completing the chain {D^{∞}Z(K)/Z(K) | D ∈ D} with respect to intersections we obtain a series for K/Z(K). As K/Z(K) is absolutely simple we conclude that Y = K or Z(K) ≤ Y.

In the general case we conclude K ≤ D or D ∩ K ≤ Z(K) for all D_{∈}D. So there exists exists a jump (B, T) of C with K ≤ T and T ∩ B ≤ Z(K). If T ≤ Y, K ≤ Y. So suppose Y ≤ B. Let F be a finite subgroup of B. Then K ∩ F is subnormal in F and so [F, K ∩F, r] ≤ K, where is the defect for K ∩F in F Hence [F ∩B, K ∩F, r ≤ B ∩K ≤ Z(K) and the three subgroup lemma implies [F ∩ B, (F ∩ K)^{∞} = 1]. Thus K = K^{∞} centralizes B. It follows that K commutes with its distinct conjugates. Hence K is normal in ⟨K^{G}⟩

Lemma 15.6 Let X be linear of degree n then X has at most ⌊\frac{n}{2}⌋ components. nclg

Proof: Without loss X = E(X) and say X ≤ GL_{K}(V) with dim_{K} V = n and K algebraically closed. If n = 1, GL_{K}(V) is abelian and X = 1. So suppose n ≥ 2If W is a proper KX submodule of V of dimension m. Clearly no component can centralize W and V/W. Moreover by induction at most \frac{m}{2} components act non-trivially on V and at most \frac{n−m}{2} on V/W. So we may assume that X acts irreducible. Let L be a component of X. Let l ∈ L so that l does not act as a scalar on V and let W be an eigenspace for l on V. Any components R other than L centralize l and so normalize W. Moreover, since R is normal in X and so C_{V} (R) = 0 and [V, R] = V. But this implies C_{W} (R) = 0 and [V/W, R] = V/W. Thus R acts non-trivially on W and V/W. But at least one of W and V/W has dimension at most \frac{n}{2}. Thus by induction X has at most ⌊\frac{n}{4}⌋ + 1 components.

Lemma 15.7 Let d be a positive integer so that each F ∈ F has at most d non-abelian fcbc composition factors. Then any composition series for G has at most d non abelian factors.

Proof: Let C be a composition series for G and (B_{i}, T_{i}), 1 ≤ i ≤ k by jumps of C with nonabelian factors. Then B_{i}/T_{i} is an non-abelian locally finite simple groups and thus is not locally solvable. Hence there exists finite subgroup F_{i} ≤ B_{i} so that F_{i}B_{i}/B_{i} is not solvable. Let F = ⟨F_{i} | 1 ≤ i ≤ k. Then F ∩ T_{i}/F ∩ B_{i} has a subgroup isomorphic to F_{i}B_{i}/B_{i} and so is not solvable. Hence F ∩ T_{i}/F ∩ B_{i} has at least one non-abelian composition factor and so F has at least k non-abelian composition factors. Hence k ≤ d.

The following examples show that absolutely semisimple by finite groups can fail the Jordan Hölder theorem. For n ∈ ℕ let K_{n} be an absolutely simple group and 1 ≠ k_{n} ∈ K_{n}. Let B = ⊕K_{n} and define α_{1} ∈ Aut(X) by (r_{n})^{α} = (r_{n}^{k_{n}^{−1}}). Let T = X ⋉⟨α_{1}⟩ and inductively let α_{i+1} = α_{i}k_{i} ∈ T. Then α_{i} centralizes K_{j} for all j < i and so T_{i} = oplus_{t≥i}⟨α_{i}⟩ is a normal subgroups of T. Moreover T_{i}/T_{i+1} ≅ K_{i} and ⋃ T_{i} = 1. Hence T has a composition series without abelian composition factor. If the K_{i} are locally finite and the k_{i} have bounded order then T is locally finite.

Definition 15.8 Let H be acting acting on a group X. dnilact

(a) H acts unipotently on X provided that [X, H, k] = 1 for some positive integer k.

(b) H acts locally unipotently on X provided that for each finitely generated subgroups F of H and each finitely generated subgroup L of X, F acts unipotently on ⟨X^{F} ⟩.

Lemma 15.9 Let G = MH with M ⊴ G, H ≤ G and C_{G}(M) ≤ M. nilbynil

(a) If M is nilpotent of class l and H acts unipotently of degree k on M , then G/M is nilpotent of class at most kl − 1 and G is nilpotent of class at most kl + k − 1.

(b) If M is locally nilpotent and H acts locally unipotently on M, then G is locally nilpotent.

Proof: Clearly (a) implies (b).

(a) Let 1 = M_{l} ≤ M_{l−1} … ≤ M_{1} ≤ M_{0} = M be the upper central series for M. Define L_{ik+r} = [M_{i}, H, r] for all O ≤ r < k and all O ≤ o ≤ l. Then G = HM centralizes the factors of the series.

1 = L_{lk} ≤ L_{lk−1} ≤ … L_{1} ≤ L_{0} = M.

Let G_{s} = [G, G, s − 1] be the s’th term of upper-central series of G. We claim that [L_{u}, G_{s}] ≤ L_{u+s+1}. If s = 0, then G_{0} = G and [L_{u}, G] ≤ L_{u+1}. Assume by induction that [L_{u}, G_{s−1}] ≤ L_{u+s}, for all u. Then

[L_{u}, G_{s−1}, G]] ≤ L_{u+s+1} and[L_{u}, G, G_{s−1}] ≤ [L_{u+1}, G_{s−1}] ≤ L_{u+s+1}.

Thus by the 3-subgroup lemma [L_{u}, [G, G_{s_{1}}]] ≤ L_{u+s+1}. But [G, G_{s−1}] = G_{s} and the claim is proved.

By the claim [M, G_{kl−1}] = 1. Since C_{G}(M) ≤ M we get G_{kl−1} ≤ Z(M). Hence G/M is nilpotent of class at most kl − 1. Moreover, G_{kl+k−1} = [G_{kl−1}, G, k] ≤ [Z(M), H, k] = 1 and so G is nilpotent of class at most kl + k − 1.

abchar Lemma 15.10 Suppose that X has an abelian normal subgroup of finite index n. Then X has an abelian, characteristic subgroup of finite index bounded by a function of n.

Proof: . Let A be a maximal abelian normal subgroup of index at most n in X. Let α ∈ Aut(X). Then also A^{α} is a maximal abelian normal subgroup of index at most n. Then C_{A}(A^{α})A^{α} is an abelian normal subgroup of X and the maximality of A^{α} implies A ∩ A^{α} = C_{A}(A^{α}). But C_{A}(A^{α}) = C_{A}(AA^{α}). Since |G/A| ≤ n, the number of possibly AA^{α} is bounded by a function of n. Hence there are only boundedly many intersection A ∩ A^{α}, α ∈ Aut(X). It follows that the index of the characteristic subgroup ⋂_{α∈Aut(X)} A^{α} is bounded by a function of n.

npabf Lemma 15.11 Let G be a locally nilpotent, periodic, linear group of degree n with trivial unipotent radical. Then G has an characteristic, abelian normal subgroup of index bounded by a function of n. In particular, G is solvable of derived length bounded by a function of n.

Proof: By 15.10 it suffices to find a normal abelian subgroup of bounded index. By an inverse limit argument we may assume that G is finite. Let A be a maximal abelian normal subgroup of G. Since G is nilpotent, C_{G}(A) = A. Thus |G/C_{G}(A)| ≤ n! by 13.5

F*gp Theorem 15.12 Let G be a periodic, linear group of degree n over a field in characteristic p with trivial unipotent radical.

(a) C_{G}(F^{∗}(G) ≤ F^{∗}(G).

(b) G has an abelian normal subgroup and components L_{1}, L_{2} … L_{r} so that each L_{i} is a group of Lie type over a locally finite field in characteristic p and the index AL_{1}L_{2} ⋯ L_{r} in G is bounded by an function of n.

(c) The index of F^{∗}(G) in G is bounded by a function of n.

(d) The number of non-abelian composition factors (in any composition series) of G is bounded by a function of n.

Proof: Suppose for the moment that we established that all four statement are equivalent. As (a) is true for finite groups, we conclude that (d) is true for finite groups. But then by 15.7 (d) is true in general and the theorem is established.

Suppose now that (a) holds. In several steps we show that (b) is true. In the following bounded always means bounded by a finite function of n

Step 1 The number of components of G is bounded

Thus is true by 15.6.

Step 2 Let H the the largest subgroup of G normalizing all the components of G. Then the index of H in G is bounded.

G/H acts faithfully in the set of components so G. Thus Step 2 follows from Step 1. Step 3 Let I the the largest subgroup of G inducing inner automorphism on each of

the components of G, (that is I = C_{G}(E(G)E(G)). Then the index of I in H is bounded.

By Step 1 it suffices to bound N_{H}(L)/C_{H}(L)L for each component L in G. Here we use the classification of simple periodic linear groups. We conclude that up to boundedly many finite exceptions L is a group of Lie type X_{m}(p^{k}), where is the characteristic of the underlying vector space and m is bounded. Next we use one more property of groups of Lie Types

• There exists an abelian p′ groups C (the Cartan subgroup) in X = X_{m}(p^{k}) such that Aut(X) = XN_{Aut(X)}(C) and |C_{Aut(X)}(X)X/X| is bounded by a function of m.

Put R = N_{G}(L). Then we get R = N_{R}(C)L and C_{R}(C)L/LC_{R}(L) is bounded. By 13.5 also R/C_{R}(C)L| = N_{R}(C)L/C_{R}(C)L ≤ |N_{R}(C)/C_{R}(C) is bounded.

Step 4 Then there exists a characteristic, abelian subgroup of G which is bounded index in C_{G}(E(G)).

Let J = C_{G}(E(G) and L = LN(G). Then L ≤ J and C_{J}(L) ≤ C_{J}(LE(G)) ≤ C_{G}(F^{∗}(G)) ≤ L. By 15.11, there exists a characteristic, abelian subgroup A of L of bounded index in L. Then C_{J}(L/A) has bounded index in J. Also by 13.5 C_{J}(A) has bounded index in J. Put P = C_{J}(L/A) ∩ C_{J}(A). Then P has bounded index in J. Since [L, P] ≤ A, [L, P, P] = 1. Thus P acts unipotently on L and by 15.9 LP is locally nilpotent. Thus P ≤ L and so A has bounded index in P and so also in J. So Step 4 holds.

(b) now follows form Steps 2,3 and 4.

As LN(G) contains any abelian normal subgroup of G, (b) implies (c). As G has boundedly many components (c) implies (d).

It remains to show that (d) implies (a). This proof is very similar to the one in 15.3 But note first that as (a) and so also (b) holds for finite groups , any finite solvable subgroup of G is solvable of bounded derived length. Thus also any locally solvable subgroup of G is solvable (of bounded derived length).

Put D = C_{G}(F^{∗}(G)).Then F^{∗}(D) = Z(D). Suppose D ≠ Z(D). By assumption D/Z(D) has only finitely many non-abelian composition factors and so by 5.12, D/Z(D) either has a non-trivial locally solvable normal subgroup, or a component. In the first case the previous paragraph implies that c that D/Z(D) has non-trivial abelian normal subgroup. Thus in either case we get the same contradiction as in 15.3.

jhpl Corollary 15.13 The Jordan Hölder theorem holds for all countable, periodic, linear groups. In particular it holds for all periodic linear groups with trivial unipotent radical.

Proof: 15.12d and 5.13

Non-linear, locally finite, simple groups with p-groups as centralisers

Definition 16.1 Let X be a group, Y a subgroup of X and W a Y -set. Then dcoin

W ⇑_{Y}^{X}= {f : X → W | f(xy) = f(x)^{y} for all x ∈ X, y ∈ Y }.

W ⇑_{Y}^{X} is called the set coinduced from W to X.

Note Y acts on X by right multiplication and W ⇑_{Y}^{X} consists of all Y -equivariant maps from X to W. Lemma 16.2 Let X be a group, Y a subgroup of X and W a Y -set. Put V = W ⇑_{Y}^{X}. bcoin Then

(a) X acts on V by f^{h}(x) = f(hx) for all x, h ∈ X.

(b) Let I be a left transversal to Y in X. Then the map ρ_{I} : V → W^{I}, f → f_{I} is a bijection. In particular, V and W^{X/Y} are isomorphic as sets.

(c) Define

π : V → W, π(f) = f(1).

Then π is an onto Y -equivariant map.

(d) Suppose that t is a fixed-point for Y on W. For w ∈ W define

κ_{t}(w) : X → W, x → \{ \begin{matrix}w^{x} & if x ∈ Y \\ t & if x ∉ Y \end{matrix}

80CHAPTER 16. NON-LINEAR, LOCALLY FINITE, SIMPLE GROUPS WITH P -GROUPS AS CENTRALISERS

Then (a) κ_{t}(w) ∈ V, and κ_{t} : W → V is a one to one Y -equivariant map.

(b) π(κ_{t}(w)) = w for all w ∈ W.

(c) π(κ_{t}(w)^{x})) = t for all w ∈ W and x ∈ X \ Y.

(e) Suppose in addition that W is a Y -group, that is Y is a group and for each y ∈ Y the map W → W, w → w^{x} is a homomorphism of group. Then the maps ρ_{I}, π and κ_{1} all are homomorphism of groups.

Proof: (a) We need to verify that f^{h} ∈ V and f^{hl} = (f^{h})^{l} for all f ∈ V, h, l ∈ X. Let x ∈ X and y ∈ Y. Since f^{h}(xy) = f(h(xy)) = f((hx)y) = f(hx)^{y} = (f^{h}(x))^{y}, f^{h} ∈ V. Also f^{hl}(x) = f((hl)x) = f(h(lx)) = f^{h}(lx) = (f^{h})^{l}(x) and so (a) holds.

(b) Let x ∈ X. Then x = iy for some unique i ∈ I, y ∈ Y.

Let f ∈ V. Then f(x) = f(iy) = f(i)^{y} and so f is uniquely determined by f_{I}. Thus f → f_{I} is one to one.

Let g ∈ W^{I}. Define f : X → W by f(x) = f(i)^{y}, It is easy to verify that f ∈ V and f_{I} = g. Hence f → f_{I} is onto.

(c) Let f ∈ V, y ∈ Y. Then π(f^{y}) = f^{y}(1) = f(y · 1) = f(1 · y) = f(1)^{y} = π(f)^{y}. So π is Y -equivariant. By the axiom of choice, 1 is contained in some left transversal. Hence (b) implies that π is onto.

(d) Let w ∈ W, y ∈ Y and x ∈ X. Then x ∈ Y if and only if xy ∈ Y. Also by assumption t = t^{y} and so

κ_{t}(w)(xy) = \{ \begin{matrix}w^{xy} & if xy ∈ Y \\ t & if xy ∉ Y \end{matrix} = \{ \begin{matrix}w^{x}^{y} & if x ∈ Y \\ t^{y} & if x ∉ Y \end{matrix} = (κ_{t}(w)(x))^{y}

Thus κ_{t}(w) ∈ V.

Also yx ∈ Y if an only if x ∈ Y and so

κ_{t}(w^{y})(x) = \{ \begin{matrix}(w^{y})^{x} & if x ∈ Y \\ t & if x ∉ Y \end{matrix} = \{ \begin{matrix}w^{yx} & if yx ∈ Y \\ t & if yx ∉ Y \end{matrix} = κ_{t}(w)(yx) = (κ_{t}(w)^{y}(x)

Thus κ_{t}(w^{y}) = (κ_{t}(w))^{y} and (da) holds.

Since 1 ∈ Y, π(κ_{t}(w)) = κ_{t}(w)(1) = w^{1} = w and so (db) holds.

Let x ∈ X \ Y. Then π(κ_{t}(w)^{x}) = κ_{t}(w)^{x}(1) = κ_{t}(w)(x · 1) = κ_{t}(w)(x) = t. So also (dc) holds.

Suppose now that W is a Y -group. It is readily verified that W^{X} is a group via (fg)(x) = f(x)g(x). Moreover, for f, g ∈ W ⇑_{Y}^{X}, x ∈ X and y ∈ Y we have

(fg)(xy) = f(xy)g(xy) = f(x)^{y}g(x)^{y} = (f(x)g(x))^{y} = (fg)(xy)^{y}

So fg ∈ W ⇑_{Y}. Similarly f^{−1} ∈⇑_{Y} and clearly 1 ∈ W ⇑_{Y}. So W ⇑_{Y}^{X} is a subgroup of W^{X}.

For any J ⊆ X, then restricted map W^{X} → W^{J}, f → f |_{J} is a homomorphism. Thus ρ_{I} and π are homomorphism. W → W, w → w^{y} and W → W, w → 1 are homomorphisms and so also κ_{1} is a homomorphism. beta Lemma 16.3 Let X, Y, W, K be groups with Y ≤ X and Y = W ⋉ K Then there exits a

semidirect product H = V ⋉ X and an embedding β : Y → H such that

(a) V ≅ W^{Y/X} as groups.

(b) V β(y) = V y for all y ∈ Y.

(c) Let y ∈ Y with y ∉ k^{W}, where y = wk, w ∈ W, k ∈ K. Then

(a) V C_{H}(β(y) = V C_{X}(y)

(b) β(y) ∉ y^{V}.

Proof: Let y ∈ Y and y = wk with w ∈ W, k ∈ K. Let ρ be the projection of Y onto K, that is ρ(wk) = k. Define σ : Y → Sym(W) by w^{σ(y)} = w^{ρ}(y). So w^{σ(y)} = w^{k} and W becomes a Y -set W_{σ}. Put V = W_{σ} ⇑_{Y}^{X}. Then by 16.2 X acts on V and we can build the semidirect product, H = V ⋉ X = {(v, x)|v ∈ V, x ∈ X}. We view V and X as subgroups of H. So H = V X and V ∩ X = 1. Let π : V → W and κ = κ_{1} : W → V be as in 16.2. Let v ∈ V.

1. π(v^{y}) = π(v)^{ρ(y)} = π(v)^{k} and π(v^{y^{−1}}) = π(v)^{k^{−1}}. beta-2

By 16.2(c) and keeping in mind that the action of Y on W is given by σ, π(v^{y}) = π(v)^{σ}(y) = π(v)^{k} and so the first part hold. Since ρ(y^{−1}) = ρ(y)^{−1} = k^{−1}, the second statement follows from the first.

2. Define beta-3 β : Y → H, y → (κ(w), y).

Then β is a monomorphism and V β(y) = V · (1, y).

Clearly β is one to one. For i = 1, 2 let y_{i} ∈ Y and y_{i} = w_{i}k_{i} with w_{i} ∈ W, k_{i} ∈ K_{i}. Then

y_{1}y_{2} = w_{1}k_{1}w_{2}k_{2} = w_{1}w_{2}^{k_{1}^{−1}}k_{1}k_{2},

and so

β(y_{1}y_{2}) = (κ(w_{1}w_{2}^{k_{1}^{−1}}), y_{1}y_{2}).

On the otherhand,

β(y_{1})β(y_{2}) = (κ(w_{1}), y_{1})(κ(w_{2}), y_{2}) = (κ(w_{1})κ(w_{2})^{y_{1}^{−1}}, y_{1}y_{2})

As κ is a Y -equivariant homomorphism,

β(y_{1})β(y_{2}) = κ(w_{1}w_{2}^{k_{1}^{−1}}), y_{1}y_{2})

and so indeed β is a homomorphism.

3. Let (v, x) ∈ C_{H}(β(y)). Then κ(w)v^{y^{−1}} = vκ(w)^{y^{−1}}. and xy = yx. beta-4

We compute

β(y)(v, x) = (κ(w), y)(v, x) = (κ(w)v^{y^{−1}}, yx)

and

(v, x)β(y) = (v, x)(κ(w), y) = (vκ(w)^{x^{−1}}, xy)

Thus 3. holds.

beta-5 4. Suppose that V C_{H}(β(y)) ≠ V C_{Y} (y). Then y ∈ k^{W}.

Since β(C_{Y} (y)) ≤ C_{H}(β(y)), V C_{H}(β(y) ≥ V β(C_{Y} (y)) =^{2.} V C_{Y} (y) and so there exists (v, x) ∈ C_{H}(β(y)) with x ∉ Y.

By 16.2(c), π(κ(w)) = w. By 1., π(v^{y^{−1}}) = π(v)^{k^{−1}}. Also since x ∉ Y and κ = κ_{1}, 16.2(dc) implies π(κ(w)^{x^{−1}}) = 1. So applying π to both sides of the equation in 3. we obtain

wπ(v)^{k^{−1}} = π(v).

Put r = π(v). Then wkrk^{−1} = r, wk = rkr^{−1} and y = wk = k^{r^{−1}} ∈ k^{W}.

beta-6 5. If β(y) ∈ y^{V}, then y ∈ k^{W}.

Suppose that β(y) = (1, y)^{(v,1)} for some v ∈ V. Then

(κ(w), y) = (v^{−1}, 1)(1, y)(v, 1) = (v^{−1}, y)(v, 1) = (v^{−1}v^{y^{−1}}, y)

and so κ(w) = v^{−1}v^{y^{−1}}. Applying π to both sides we conclude w = π(v^{−1})π(v)^{k^{−1}}. Put r = π(v) then w = r^{−1}krk^{−1} and y = wk = r^{−1}kr = k^{r}. Thus 5. holds.

We are now in the position to prove the lemma. (a) follows from ?? (b),(e). By 2. (b) holds. By 4. (ca) and by ,5. (cb) holds.

alpha Lemma 16.4 Let G be a finite group and Π a set of primes. Then there exist a finite abelian ℤG-module W and a monomorphism α : G → W ⋉ G such that

(a) W is a Π-group.

(b) W α(g) = W g for all g ∈ G

(c) α(g) ∉ g^{W} for all non-trivial Π-elements g in G.

(d) If G is perfect, then W = [W, G] and W ⋉ G is perfect.

Proof: Let m be the Π-part of |G|. Let B = (ℤ/mℤ)^{G} and H = ℤ/mℤ≀G, where wreathed product is build with respect to regular action of G on G. Then B is the base group of H and H = BG. For f ∈ B put ||f|| = ∑_{g∈G} f(g). Let W = [B, G] = {f ∈ B | ||f|| = 0}. Then W is a Π-group and (a) holds. Also if G = G′, the three subgroup lemma implies [B, G, G] = [B, G] and so W = [W, G] and W G is perfect. Thus (d) holds.

Let b ∈ B be defined by b(1) = 1 and b(g) = 0 for all g ∈ G^{#}. Define α : G → W G, g → g^{b} = [b, g^{−1}]g. Then α is an monomorphism and (b) holds. It remains to prove (c). So let g be a non-trivial Π-element in G and suppose that g^{b} = g^{a} for some a ∈ W. Put n = |g| and c = ba^{−1}. Then c ∈ C_{B}(g). Let I be a left transveral to ⟨g⟩. Then each g in G can be uniquely written as ig^{k} for some i ∈ I and some 0 ≤ k < n. Since c^{g} = c, c(i) = c(ig^{k}). Let s = ∑_{i∈I} c(i). We conclude that ||c|| = ns. Thus

1 = ||b|| = ||ac|| = ||a|| + ||c|| = 0 + ns = ns

in ℤ/mℤ, a contradiction as n divides m.

Lemma 16.5 Let G and F be finite groups and Π a set of primes. Then there exist a beal finite group G^{∗} with G ≤ G^{∗} and normal subgroup V of Gsuch that:

(a) V is an abelian Π-subgroup and G^{∗}/V is simple.

(b) G ∩ V = 1.

(c) Let x be a Π-element in G. Then C_{G^{∗}}(x) has a normal solvable Π-subgroup M(x) with C_{G^{∗}}(x) = M(x)C_{G}(x).

(d) G^{∗} has a subgroup isomorphic to F.

(e) If G is perfect, G^{∗} is perfect.

Let α and Y = A ⋉ G be as in 16.4. Let X be any finite simple group containing Y as a subgroup and such that X has a subgroup isomorphic to F. Let β and V be as in 16.3 applied with K = G. Let G^{∗} = V ⋉ X and M^{∗} = V. Let g be a Π-element in G and put γ = βα. Then by 16.4 α(g) ∉ g^{W}. So by 16.3, C_{G^{∗}}(γ(x)) ≤ V C_{Y} (α(x) Also C_{Y} (α(x)) ≤ W C_{G}(x) and V W γ(C_{G}(x)) = V W C_{G}(x). Thus

C_{G^{∗}}(γ(x)) ≤ V W γ(C_{G}(x))

Put M(x) = C_{V W} (γ(x)). Then C_{G^{∗}}(γ(x)) = M(x)C_{G}(x). Identifying G with its image in G^{∗} under γ we see that all parts of the lemma hold.

Lemma 16.6 Let G be a locally finite group. sobysi

(a) Suppose G is absolutely perfect and (locally solvable) by simple. Then LSol(G) contains all proper normal subgroups of G.

(b) Suppose G is locally “absolutely perfect and (locally solvable) by simple” Then G is absolutely perfect and (locally solvable) by simple.

(c) If G has Kegel-cover K such that for all (H, M) ∈ K, H is perfect and M is solvable, then G is simple.

Proof: Let L = LSol(G).

(a) Let N be a normal subgroup of G with N ≰ L. As G/L is simple, G = NL. Since G is absolutely perfect, G/N is the union of finite perfect subgroups. Now G/N ≅ L/L ∩ N is locally solvable and so all finite subgroups of G/N are solvable. Thus G/N = 1 and G = N.

(b) Let N/L be a non-trivial normal subgroup of G/L. Then N is not locally solvable and so there exists a finite, non-solvable subgroup T of N. Let F be any finite subgroup of G. By assumption there exists an absolutely perfect and (locally solvable) by simple subgroup M of G with ⟨F, T ⟩ ≤ M. Then T ≤ M ∩ N ⊴ M. Thus M ∩ N ≰ LSol(M) and so by (a), M ∩ N = M. Hence F ≤ M ≤ N and N = G.

(c) By (b) it suffices to show that L = 1. Let F be a finite subgroup of L. Then there exists (H, M) ∈ K with F ≤ H and F ∩ M = 1. Since H is perfect, H ≠ Sol(H) and so Sol(H) = M. Thus F ≤ L ∩ H ≤ M and F = 1. Therefore also L = 1.

cc Proposition 16.7 Let G_{0} be a finite, perfect group, Π a non-empty set of primes and for each positive integer n let F_{n} be a finite group. Then there exists a locally, finite simple group G with G_{0} ≤ G and such that:

(a) If x is a Π-element of G, then C_{G}(x) has a locally solvable, normal Π-subgroup M(x) of finite index.

(b) If x is a Π element in G_{0}, then C_{G}(x) = M(x)C_{G_{0}}(x).

(c) F_{n} is isomorphic to a subgroup of G.

Proof: By induction on n ∈ Z^{+} we will produce finite groups G_{n} and a normal subgroup M_{n} of G_{N} such that

cc-1 1.

1. G_{n} is perfect.

2. M_{n} is abelian and G_{n}/M_{n} is simple.

3. G_{n−1} ≤ G_{n} and G_{n−1} ∩ M_{n} = 1.

4. If x is a Π element in G_{n−1}, then there exists a solvable normal Π subgroup M_{n}(x) of C_{G_{n}}(x) with C_{G_{n}}(x) = M_{n}(x)C_{G_{n−1}}(x).

5. G_{n} has a subgroup ismorphic to F_{n} Let n ≥ 0 and suppose we already found G_{1}, … G_{n}. Then G_{n} is perfect and so we can apply 16.5 to G = G_{n} and F = F_{n+1}. Put G_{n+1} = G_{n}^{∗}. Then by 16.5 1. to 5. hold.

Put G = ⋃_{i=1}^{n} G_{n}. Then {(G_{n}, M_{n}) | n ≥ 1} is a Kegel cover for G and by 16.6, G is simple. Let x ∈ G be a Π-element. Then x ∈ G_{n} for some n. Put M^{n}(x) = 1 and inductively, M^{m+1}(x) = M^{m}(x)M_{m+1}(x). It follows from 1.4. and induction that

2. Let m ≥ n. Then M^{m}(x) is a solvable, normal Π-subgroup of C_{G_{m}}(x) and C_{G_{m}}(x) = cc-2 M^{m}(x)C_{G_{n}}(x).

Put M(x) = ⋃_{m=n}^{∞} M^{m}(x). Then by 2., M(x) is a locally solvable, normal Π-subgroup of C_{G}(x) and C_{G}(x) = M(x)C_{G_{n}}(x). Thus the proposition is proved.

Corollary 16.8 Let Π be a non-empty set of primes. Then there exists a non-linear, sobyfi locally finite, simple group G such that

(a) The centralizer of every non-trivial Π-element has a locally solvable Π-subgroup of finite index.

(b) There exists an element whose centralizer is a locally solvable Π-group.

Proof: Let p ∈ Π. Let G_{0} = Alt(2p + 1) and x the product of two commuting p-cycles. Then x ∈ G_{0}, G_{0} is perfect and C_{G_{0}}(x) ≅ C_{p} × C_{p} and is a solvable Π-group. Apply 16.7 to this G_{0} and with F_{n} = Sym(n). The resulting G is not linear and fulfills (a). Moreover,

(b) holds for the element x ∈ G_{0} ≤ G.

Corollary 16.9 msobfi

(a) There exists a non-linear, locally finite, simple group such that the centralizer of every element is (locally solvable) by finite..

(b) Let p be a prime. Then there exists a non-linear, locally finite, simple group with an element whose centralizers is a p-group.

Proof: Apply 16.8(a) with Π the set of all primes and 16.8(b) with Π = {p}.

86CHAPTER 16. NON-LINEAR, LOCALLY FINITE, SIMPLE GROUPS WITH P -GROUPS AS CENTRALISERS

Examples of simple and of characteristically simple groups

Let I be a set. Then the power set P(I) is a algebra over the field F_{2}, where addition is given by symmetric difference and multiplication by intersection. Then empty set is the O and I is the 1 in this algebra. Let A be a subalgebra of P(I) with 1, i.e A is a non-empty set of subsets of I such that

1. If A ∈ A then I \ A ∈ A.

2. If A, B ∈ A, then A ∩ B ∈ A.

Note that then also A ∪ B and A \ B are in A for all A. Let H be a group and let

R = R(H, A) = {f ∈ H^{I} | |f(I)| < ∞, f^{−1}(i) ∈ A for all i ∈ I}.

Let D be the set of finite partions of I into members of A. Then f ∈ H^{I} lies in R(H, A) if and only if Δ_{f} = {f^{−1}(i) | i ∈ I} ∈ D. For A ∈ A let R_{A} consist of whose function f ∈ H^{I} which are constant on A and 1 on I \ A. For A ∈ A let D_{A} be the set of finite partitions of A into members of A. Also for A ∈ A and Δ ∈ D_{A} let R_{Δ} consist of whose function which are constant on each member of Δ and are 1 on I \ A. For f ∈ H^{I} let Supp_{I}(f) = Supp(f) = {f^{−1}(H^{#}). Define a partial order on D by defining Δ ≤ Δ^{∗} if and only if each member of Δ^{∗} is contained in a member of Δ. For Δ, Δ^{∗} ∈ D define Δ ∧ Δ^{∗} = {D ∩ D^{∗} | D ∈ Δ, D^{∗} ∈ Δ^{∗}}. Then Δ ≤ Δ ∧ Δ^{∗} and so D is a directed set.

Lemma 17.1 basicR

(a) Let Δ, Δ^{∗} ∈ D with D ≤ D^{∗}. Then R_{Δ} ≤ R_{Δ^{∗}} ≤ R.

(b) Let A ∈ A. Then R_{A} ≅ H.

(c) Let Δ ∈ D. Then R_{Δ} = ⊕_{D∈Δ}R_{D} ≅ H^{Δ}.

88CHAPTER 17. EXAMPLES OF SIMPLE AND OF CHARACTERISTICALLY SIMPLE GROUPS

(d) Let f ∈ R and Δ ∈ D. Then f ∈ R_{Δ} if and only if Δ_{f} ≤ Δ.

(e) {R_{Δ} | Δ ∈ D is a local system for R. In particular, if H is locally finite then R is locally finite.

Proof: This is readily verified.

cnoid Lemma 17.2 Suppose that H is a non-trivial perfect, simple group. Let B be an ideal in A and put R_{B} = {f ∈ R | Supp(f) ∈ B The the map B⟨N_{B} is a one to one correspodence between the ideals in R and the normal subgrousp of R.

Proof: Note first that B is an ideal in A is equivalent to saying that B is a subset of A with the following two properties:

1. If A ∈ A, B ∈ B with A ⊂ B, then A ∈ B.

2. If B, B^{∗} ∈ A with B ∩ B^{∗}∅, then B ∪ B^{∗} ∈ B.

Let f, g ∈ R_{B} and r ∈ R. Since Supp(f) = Supp(f^{r}) = Supp(f^{−1}, f^{−1}, f^{r} ∈ R_{B}. Let D ∈ Δ_{f} ∧ Δ_{g}. Then f, g and fg are constant on D and D ∈ A. Suppose that fg is not 1 on D, then at lest one of f and g is not one on D. Hence D ⊆ Supp(f) or Supp(g) and so by 1. D ∈ B. As Supp(fg) is the finite disjoint union of such D’s, 2. implies Supp(fg) ∈ B and so R_{B} is a normal subgroup of R.

Next let M be any normal subgroup of R and put E = {Supp(m) | m ∈ M}. We need to show that E is an ideal in A and M = R_{E}. Let m ∈ M and E = Supp(m). Let Δ ∈ D_{E} and put Δ^{∗} = Δ ∧ Δ_{f}. Then m ∈ R_{Δ^{∗}} and and m projects non truvially onto R_{D^{∗}} for all D^{∗} ∈ Δ^{∗}. Hence by 15.4, R_{D^{∗}} ≤ ⟨m^{R_{Δ^{∗}}} ≤ M. It follows that R_{D} ≤ R_{D^{∗}} ≤ M. Hence M contains all elements of R with support E and if A ∈ A with A ≤ E, then A ∈ E. Finally if m_{1}, m_{2} ∈ M withe Supp(m_{1}) ∩ Supp(m_{2}) = 1, then Supp(m_{1}m_{2}) = Supp(m_{1}) ∩ Supp(m_{2}) and also 2. holds. Thus E is indeed an ideal in A and M = R_{E}.

minide Lemma 17.3

(a) Suppose that B is a minimal ideal of A. Then B = {∅, B} for some minimal elements of A.

(b) Suppose that H is a non-trivial perfect, simple group and that M is a minimal normal subgroups of R. Then M = R_{B} for some minimal member of A.

(a) Let E = ⋃ B, e ∈ E and C = {D ∈ B | e ∉ D}. Then C ≠ B. As B is an ideal it easy to see that also C is an ideal. So by minimality of B, C = {∅}. Hence e ∈ B for all ∅ ≠ B ∈ B and all e ∈ E. Thus B = E and (a) holds.

(b) follows from (a) and 17.2.

Rsesi Lemma 17.4 Suppose that H is a non-trivial, perfect simple group. Then R is semisimple if and only if I is the disjoint union of finitely many minimal members if A.

Proof: Suppose first that R = ⨁_{j∈J} R_{j} for some nontrivial simple groups R_{j}, j ∈ J. Then by 17.3 R_{j} = R_{B_{j}} for some minimal member B_{j} of A. Let 1 ≠ f ∈ R_{I}. Then f ∈ R = ⨁_{j∈J} R_{j} and so f = r_{1}r_{2} … r_{n} where r_{i} ∈ R_{j_{i}} for some j_{i} ∈ J. It follows that I = ⋃_{i=1}^{n} B_{j_{i}} and so the ”only if” statement of the lemma is proved.

Conversely, let I be the disjoint union of A_{1}, …, A_{n} where each A_{i} is a minimal member of A. Then each A ∈ A is a union of some of the A_{i}’s and so R = ⊕_{i=1}^{n}R_{A_{i}} ≅ H^{n}.

Definition 17.5 dcaaut

(a) Let I_{1} and I_{2} sets and for i = 1, 2 let A_{i} a subalgebra of P(I_{i}) and ϕ : I_{1} ← I_{2} a map. ϕ is called homomorphism from (A_{1}, I_{1}) to (A_{2}, I_{2}) provided that ϕ(A)_{1} ⊆ A_{2}. ϕ is an isomorphism if ϕ is a bijection and maps A_{1} onto A_{2}. ϕ is an automorphism if I_{1} = I_{2}, A_{1} = A_{2} and ϕ is an ismorphism. Aut(A, I) is the group of automorphism of (A, I).

(b) Let S be a subgroup of Aut(A, I). Then the canonical action of S on R(H, A) is defined by f^{s}(i) = f(i^{s^{−1}}) for all f ∈ R(H, A), s ∈ S. A is called S-simple if S leaves no proper ideal of A invariant.

Lemma 17.6 Suppose that S ≤ Aut(A, I) and A is S-simple. Also suppose that H is non- Rchsi trivial,perfect,simple. Then S normalizes no proper normal subgroup of R. In particular R is characteristically simple and R is a minimal normal subgroup of R⊳<S.

Proof: Clear by 17.2

Lemma 17.7 Let S ≤ Aut(A, I) and for all Δ ∈ D there exists Δ^{∗} ∈ D with Δ ≤ Δ^{∗} essial such that N_{S}(Δ^{∗}) acts transitively on Δ^{∗}. Then A is S-simple.

Proof: Let B be a non-zero ideal S-invariant ideal in A. We need to show that B = A. For this let A ∈ A and ∅ ≠ B ∈ B. Let Δ be the set containg the non-empty members of A \ B, B \ A, A ∩ B, I \ (A ∪ B). Then Δ ∈ D and so by assumption there exists Δ^{∗} ∈ D so that Δ ≤ Δ^{∗} and N_{S}(Δ^{∗}) acts transitively on Δ^{∗}. Since either B \ A or A ∩ B is not empty, Δ contains a memember of B. As B is closed under A-subsets also D^{∗} contains a member of B. As B is S-invarinat and N_{S}(D^{∗}) is transitive on Δ^{∗} all Δ^{∗} ⊆ B. Since D ≤ D^{∗}, we conclude that A \ B and A ∩ B and so also A is a disjoint union of members of B. Thus A ∈ B. As A was arbitrary, B = A.

As an application we now use the above machinery to prove:

Proposition 17.8 There exists a locally finite group G and a minimal normal subgroup nsemn R of G so that R is not semi simple.

Proof: Let K be an infinite set of finite sets. Let I = ∏ K. Let H be any locally finite, non-abelian, simple group. For subset E of K put I_{E} = ∏ E. We will always identify I with I_{E} × I_{K\E}.

E ⊂ ∏ E put A_{E} = E × ∏ K \ E ⊂ I. Let A consists of all the subsets of the form A_{E}, where E runs through the finite subsets of K and E through the subsets of ∏ E. It is easy to see that A is a subalgebra of P(I) with I ∈ A. As K is infinite, A has no minimal elements and by 17.4 R = R(H, A) is not semisimple.

Next we will define a locally finite subgroup S of Aut(A, I) which makes A, S-simple. Then by 17.6 G = R⊳<S fulfills the conclusion of the lemma.

Again let E be a finite subset of K. Define a monomorphsim ϕ_{E} : Sym(I_{E}) ← Sym(I), ϕ_{E}(π) : (a, b) ← ϕ(a), b) for all a ∈ I_{E}, b ∈ I_{K\E} and π ∈ Sym(I_{E}). Put S_{E} = ϕ_{E}(Sym(I_{E}. Then it is easy to see that S_{E} is a subgroup of Aut(A, I). Moreover, if E^{∗} ⊂ E then S_{E^{∗}} ≤ S_{E} and so if we let S be the union of the S_{E}, where E runs through the finite subsets of K, then S is a locally finte subgroup of Aut(A, I).

It remains to show that A is S-simple. For this it suffices to verify the assumptions of 17.7. So let Δ ∈ D. Then Δ = {A_{E_{i}} | 1 leqi ≤ n}, for some finite subsets E_{i} of K and some E_{i} ⊂ I_{E_{i}}. Put E = ⋃_{i=1}^{n} E_{i} and Δ^{∗} = {{a} × I_{I\E} | a ∈ I_{E}. Then Δ ≤ Δ^{∗}, S_{E} normalizes Δ^{∗} and induces Sym(Δ^{∗}0) on Δ^{∗}. In particular, S_{E} and so also N_{S}(Δ^{∗}) acts transitively on Δ^{∗}. So we assumptions of 17.7 are indeed verified and the proposition is proved.

For later use let as denote the group S constructed in the proof of the previous lemma by Diag(∏ K), the diagonal group on ∏ K. Note here that all orbits of S_{E} ≅ Sym(I_{E} on I = ∏ K are isomorphic to I_{E}, so Sym(I_{E} is ”diagonally” embedded into Sym(∏ K). Also the redaer should notice that Diag(∏ K) does depend on K and not only on ∏ K as a set. Also let A_{E} = ϕ_{E}(Alt(I_{E}) and Diag^{∗}(K) be the union of the A_{E}, where as usual E runs through the finite subsets of K. The easy proof of the following propostion is left to the dedicated reader.

simdia Lemma 17.9 Diag^{∗}(K) is a locally finte simple group. | Diag(K)/ Diag^{∗}(K)| ≤ 2 with equality if and only if only if only finitely many of the K ∈ K have even size.

Finitary groups

In this section we will investigate two kinds of finitary groups: locally solvable ones and locally finite, simples one of alternating type.

Definition 18.1 Let X be a group, K a field and V a KX-module. dfinitary

(q) V is called a finitary KX-module if [V, g] is finite dimensional for all x ∈ X.

(b) X is called a finitary group, if there exists a field K and a faithful, finitary KXmodule.

(c) Let x ∈ X. Then deg_{V} (x) = dim_{K}[V, x].

(d) FGL_{K}(V) = {x ∈ GL_{K}(V) | deg_{V} (x) < ∞}.

Proposition 18.2 Let X be a group, K a field and V a faithful KX-module. Then X has LUnip largest normal subgroup LU(X) acting locally unipotent on V.

Proof: Let Y the semidirect product V X and let N be a normal subgroup of X. If Y N is locally nilpotent, it is easy to see that N acts locally unipotently on V. Conversely if N acts locally unipotently on V, then by 15.9 V N is locally nilpotent. Since V ≤ LN(Y), LN(Y) = V (LN(Y) ∩ X). Hence LN(Y) ∩ X is the largest normal subgroup of X acting unipotently on Y.

For the remainder of this section we assume that K is a field, V is a vector space over K, X is a subgroup of FGL_{K}(V) and F is the set of finitely generated subgroups of X.

Lemma 18.3 Let t ∈ X and W a K-subspace of V, finblo

(a) If C_{W} (t) = 0, then dim_{K} W ≤ deg_{V} (t).

(b) If W ∩ W^{t} = 0, then dim_{K} W ≤ deg_{V} (t).

(c) Let k be a positive integer and Δ be a block system for X on W so that dim_{K} W = k for all W ∈ Δ. Let s the number of orbits for t on Δ. Then

\frac{1}{2}| Supp_{Ω}(t) ≤ k · \frac{1}{2}| Supp_{Ω}(t)| ≤ k · (|Δ| − s) ≤ deg_{V} (t).

Proof: (a) W ≅ W/C_{W} (t) ≅ W/W ∩ C_{V} (t) ≅ W + C_{V} (t)/C_{V} (t) ≤ V/C_{V} (t). As dim V/C_{V} (t) = deg_{V} (t), (a) holds.

(b) Since C_{W} (t) ≤ W ∩ W^{t} = 0, (b) follows from (a).

(c) Let W_{1}, … W_{s} be representatives for the orbit fof t on Δ. Let W = ∑_{i=1}^{s} W_{i}. Since each W ∈ Δ is conjugate one of the W_{i}’s and since V = ∑ Δ, V = ⟨W^{⟨t⟩} ≤ W + [V, t]. Thus

k · |Δ| = dim_{K} V =≤ dim_{K} W + dim_{K}[V, t] = s · k + deg_{V} (t).

Thus it remains to show that |Delta| − s ≥ \frac{1}{2} Supp_{Δ}(t). For thus let Δ_{1}, … Δ_{s} be the orbits of t on Δ with Δ_{1}, …, Δ_{r} the orbits with |Δ_{i}| > 1, i.e. Δ_{i} ≤ Supp(t). Then \frac{1}{2}|Δ_{i}| ≤ |Δ_{i}| − 1 for all 1 ≤ i ≤ r and so

|Δ| − s = ∑_{i=1}^{s} |Δ_{i}| − 1 = ∑_{i=1}^{r} |Δ_{i}| − 1 ≥ ∑_{i=1}^{r} \frac{1}{2}|Δ_{i}| = \frac{1}{2}|Supp_{Δ}(t)|.

Ffd Lemma 18.4 Let F ∈ F. Then there exists a finite dimensional KF -submodule U in V and C ≤ C_{V} (F) with V = U ⊕ C. In particular,F is a linear group.

Proof: Let F = ⟨I⟩ for some finite subset I of F. Then C_{V} (F) = ⋂_{i∈I} C_{V} (i) and [V, F] = ∑_{i∈I}[V, i]. Hence as X is finitary, V/C_{V} (F) and [V, F] are finite dimensional. Thus there exists a finite dimensiona K-subspace U of V with [V, F] ≤ U and V = U + C_{V} (F). Since [V, F] ≤ U, U is F -invariant. Let C be a complement to C_{U}(F) in C_{V} (F). Then V = U ⊕ C and the lemma is proved.

lufu Lemma 18.5 Suppose that X acts locally unipotently on V and V is finite dimnsional. Then X acts unipotently on V. In particular, if V ≠ O, then V ≠ [V, X]

Proof: Without loss V ≠ 0. Let F ∈ F. Then F acts unipotently on V and so [V, F] ≠ V. Pick F ∈ F so that [V, F] has maximal K-dimension. Then for all F ≤ F^{∗} ∈ F, [V, F] = [V, F^{∗}] and so [V, X] = [V, F] ≠ V. That X is unipotent on V now follows by induction on dim_{K} V.

unpcen Lemma 18.6 Let D be an KX-composition series on V. Then LU(X) is the largest subgroup of X acting trivially on all of the factors of D.

Proof: Let W be factor of D and N a normal subgroup of X acting locally unipotent on W. Suppose that [W, N] ≠ O. Then as X is irreducible on V, W = [W, N]. Let 0 ≠ w ∈ W. Then w ∈ W = ⟨⟨w^{X}⟩, N] and so there exists F ∈ F with [w ∈ [⟨w^{F} ⟩, F ∩ N]. Put U⟨w^{F} ⟩.

Then w ∈ [U, F ∩ N] ans since F normalizes [U, F ∩ N], U = [U, F ∩ N]. But F ∩ N acts locally unipotent on U and we obtain a contradiction to 18.5. Hence N centralizes W.

Conversely let N be a subgroup of X which centralizes all the factor of D. Let U be a finite dimensional subspace of V and F a finitely generated subgroup of N. Put W = U[V, F]. Then W is finite dimensional and F invariant. Moreover F centralizes all factors of the series D∩W and so F avts unipotently on W. Thus N acts locally unipotently on V

Lemma 18.7 Let H ≤ X so that d = dim_{K}[V, H] is finite and put L = ⟨H^{X}⟩. sbbd

(a) Let C be a series for KX on V. Then at most d factors of L are non centralized by L.

(b) There exists a KX series D on V so that

(a) |D| ≤ 2d + 2.

(b) If W is a factor of D, then either L act trivially on W or W is irreducible as KX-module.

Proof: Let E = [V, H] and let C be any series for KX on V. Then C ∩ E is a finite series

1 = E_{0} < E_{1} < … E_{r} = E

with r ≤ d. For 0 ≤ i ≤ r, put T_{i} = ⋂{C ∈ C | C ∩ E = E_{i} and B_{i+1} = ⋃{C ∈ C | C ∩ E = E_{i+1}. Then

1 = T_{0} ≤ B_{1} < T_{1} ≤ B_{2} … B_{r} < T_{r} … B_{r+1} = V

is an KX-series on V. We claim that for all 1 ≤ i ≤ r, (B_{i}, T_{i}) is a jump of C. For this note first that E ∩ B_{i} = E_{i−1} and E ∩ T_{i} = E_{i}. Let C ∈ C with B_{i} ≤ C ≤ T_{i}. Then E_{i−1} ≤ E ∩ C ≤ E_{i}. As (E_{i−1}, E_{i}) is a jump of E ∩ C we comclude E ∩ C = E_{i−1} or E_{∩}C = E_{i}. In the first case E ≤ B_{i} and in the second T_{i} ≤ C. Thus C = B_{i} or T_{i}.

Thus (B_{i}, T_{i}) is indeed a jump and so T_{i}/B_{i} is a factor of C. Next we show that for all 1 ≤ i ≤ r, L centralizes B_{i}/T_{i−1}. For this note that [B_{i}, H] ≤ B_{i} ∩ E ≤ E_{i−1} ≤ T_{i−1}. Hence H and so also L acts trivally on B_{i}/T_{i−1}. It follows that all factors of C other than the (B_{i}, T_{i}) are centralized by L. Thus (a) holds. Choosing C to be a composition series, we get that T_{i}/B_{i} is irreducible as KX-module and so also (b) is proved.

Corollary 18.8 Let H ≤ X so that d = dim_{K}[V, H] is finite and put L = ⟨H^{X}⟩ and untx LU(L) is unipotent of degree at most 2d + 1 and so also nilpotent of class at most 2d.

Proof: By 18.7 there exists a series of length at most 2d+1 so that each factor of is either irreducible as KX-module or centralized by L. Hence by 18.6 LU(L) centralizes all the factors and so is unipotent of degree at most 2d+1. Thus by 15.9 LU(L) is nilpotent of class at most 2d.

Lemma 18.9 Let x, y ∈ X. Then deg_{V} [x, y] ≤ 2 · x. degcom

Proof: [x, y] = x^{−1}x^{y} and so [V, [x, y]] ≤ [V, x] + [V, x^{y}]

degprp Lemma 18.10 Let H be a primitive of Sym(Ω) with Sol(H) ≠ 1 Then |Ω| ≤ 2| Supp_{Ω}(h) for all 1 ≠ h ∈ H.

Proof: Let 1 ≠ A be an abelian normal subgroup of H. An orbit of A is a block for H, A is transitive. Let ω ∈ Ω. By a Frattini argument, H = H_{ω}A. Also C_{H_{ω}}(A) fixes Ω = ω^{A}, elementwise and so C_{H_{ω}}(A) = 1. In particular, A_{ω} = 1 and so Supp(a) = Ω for all 1 ≠ a ∈ A. So we may assume that h ∉ A. Now C_{H}(A) = C_{H_{ω}}(A)A = A and so ther exists b ∈ A with a = [b, h] ≠ 1. Thus |Omega| = |Supp_{Ω}(a)| ≤ 2| Supp_{Ω}(h).

crsu Lemma 18.11 Let H be a solvable group. Then H has a characteristic subgroup N with C_{G}(N) ≤ N such that N is nilpotent of class at most two.

Proof: By Zorn’s Lemma H has a maximal characteristic, abelian normal subgroup A. Next let N/A be maximal characteristic, abelian normal subgroup of C_{N}(A)/A. Then N is characteristic in H. Suppose that B = C_{H}(N) ≰ N. Since Since A ≤ B ∩ N ≤ Z(B), the maximality of A implies A = B ∩ N. Let E/A be last non-trivial term of the derived series of B/A. Then [NE, NE] = N′[N, E]E′ ≤ A and the maximality of N/A implies E ≤ N and so E ≤ A, a contradiction.

degsol Lemma 18.12 Suppose that X is solvable and acts irreducible on V and that V is finite dimensional.

(a) If X is primitive then dim_{K} V ≤ 4 deg_{V} (t) for all 1 ≠ t ∈ X.

(b) If X is generated by elements of degree less or equal to d, then dim_{K} V ≤ 4d.

Proof: (a) By 18.11 there exists a normal subgroup N of X so that C_{X}(N) ≤ N and N is nilpotent of class at most N. Let 1 ≠ a ∈ Z(N). Since X is primitive, Z(N) is homogenous on V and so V = [V, a]. By ?? a ∈ ⟨t^{x_{i}} ≤ i ≤ 4⟩, where x_{i} ∈ X. Hence deg_{V} (a) ≤ 4 deg_{V} (t) and so (a) is proved.

(b) By (a) we may assume that X acts imprimitively on V. Let Δ be a block system for X on V with |Δ| minimal. Then X acts primitively on Δ. As X is generated by elements of degree less r equal to d, there exists t ∈ X with deg_{V} (t) ≤ d acting non-trivially on Δ. By 18.10 |Δ| ≤ 2| Supp_{Δ}(t). Let k = dim_{K} W, W ∈ Δ By 18.3c, k|Supp_{Ω}(t) ≤ 2 deg_{V} (t) and so

dim_{K} V = k|Δ| ≤ k2 · | Supp_{Δ}(t)| ≤ 4 deg_{V} (t).

XY Lemma 18.13 Suppose that n dim_{K}(V) is finite and that Y ≤ GL_{K}(V) with [X, Y] ≤ K^{#}. Then |X/C_{X}(Y)| ≤ n^{2}.

Proof: Let T be a transversal to C_{X}(Y) in X. View X and Y has subsets of End_{K}(V) and note that End_{K}(V) is a n^{2} dimensinal vector space over K. Hence it suffices to show that T is linear independent over K. Otherwise let M be a minimal linear independent subset of T. Since none of the elements of T are 0, |M| contains at least two elemenst a and b. Since C_{X}(Y)a ≠ C_{Y} (t)b, ba^{−1} ≰ C_{Y} (t), there exists x ∈ X with (ba^{−1})^{x} ≠ ba^{−1} and so [b, x] ≠ [a, x]. For x ∈ X and m ∈ M let λ_{m} = [m, x] ∈ K. Then λ_{m} = m^{−1}m^{x} and

(1)m  m^{x} = λ_{m}

As M is minimal linear dependent,

(2)  0 = ∑_{m∈M} μ_{m}m

for some non-zero μ_{m} ∈ K. Conjuating (2) by x and using (1) we obtain

(3)  0 = ∑_{m∈M} μ_{m}λ_{m}m

Multilying (2) by λ_{a} we obtain

(4)  0 = ∑_{m∈M} μ_{m}λ_{a}m

Subtracting (4) from (3) we conclude

(5).  0 = ∑_{m∈M} μ_{m}(λ_{m} − λ_{a})m

Since μ_{a}μ_{a}(λ_{a} − λ_{a}) = 0, the minimality of M implies μ_{m}(λ_{m} − λ_{a}) = 0 for all m. But as μ_{b} ≠ 0 and λ_{a} ≠ λ_{b}, μ_{b}(λ_{b} − λ_{a}) ≠ 0, a contradiction.

Lemma 18.14 Suppose that X is solvable and n = dim_{K} V. solbf

(a) If K is algebraicly closed and X is primitive, then |X/Z(X)| ≤ n^{2}!.

(b) X has as normal subgroup A so that |X/A| ≤ n^{2}! and A′ acts unipotenly on V.

Proof: (a) Let N a a normal subgroup of X so that C_{X}(N) ≤ N and N has class at most 2. Since X is primitive and K is algebraicly closed, Z(N) ≤ K^{#} and Z(N) ≤ Z(X). Since N has class at most 2, [N, N] ≤ Z(N). Thus by 18.13, N/Z(N)| ≤ n^{2}. Let Y = |C_{X}(N/Z(N)|. Then |X/Y | ≤ (|N/Z(N)| − 1)! ≤ (n^{2} − 1)!. Also [N, Y] ≤ F and so by 18.13, |Y/C_{Y} (N)| ≤ n^{2}. But C_{Y} (N) ≤ Z(N) ≤ Z(X) and so |X/Z(X)| ≤ n^{2}(n^{2}−1)! = n^{2}!.

(b) Without loss K is algebraicly closed.Suppose that W is a proper KX-submodule and put m = dim_{K} W By induction on n there exist normal subgroups A_{1} and A_{2} so that C_{H}(W) ≤ A_{1}, C_{H}(V/W) ≤ A_{2}, A′_{1} act unipotently on W, A′_{2} acts unipotently on V/W, |H/A_{1} ≤ m^{2}! and |H/A_{2}| ≤ (n − m)^{2}!. Put A = A_{1} ∩ A_{2}. Then A′ acts unipotently on W and V/W and so also on V. Futhermore, |H/A| ≤ |H/A_{1}||H/A_{2}| ≤ m^{2}!(n − m)^{2}! ≤ n^{2}!.

Suppose next that X acts irreducible on V and that Δ is a block system for H on V. Let W ∈ Δ and let k = dim_{K} W. Since H has no non-trivial unipotent normal we conclude by induction that again by induction there exists exists normal subgroups B of C_{H}(Δ) so that C_{H}(W) ∩ C_{H}(Δ) ≤ B, B′ centralizes W and |C_{H}(Δ)/B ≤ k^{2}!. Let m = |Δ| = \frac{n}{k} and A = ⋂_{h∈H} B^{h}. Then |C_{H}(Δ)/A| ≤ (k^{2}!)^{m}, |H/C_{H}(Δ)| ≤ m! and |H/A| ≤ (k^{2}!)^{m}m! ≤ (mk)^{2}! = n^{2}.

Finally if X is primitive (b) follows from (a).

bodeso Corollary 18.15 If X is solvable and V is finite dimensional, then the derived length of X is bounded by a function of dim_{K} V.

Proof: Since Unip(X) is nilpotent of class at most dim_{K} V, this follows directly from 18.14

lsolnc Theorem 18.16 Let d be a positive integer and suppose that X is generated be elements of degree at most d. Then LSol(X)/LU(X) is solvable of derived length bounded by a function of d.

Proof: Let L = LSol(X), U = calLU(X) ≤ L and I = {[l, x] | l ∈ l ∈ L, x ∈ X, deg_{V} (x) ≤ d. Then deg_{V} i ≤ 2d for all i ∈ I and H = ⟨I⟩[L, X]. Since L′ ≤ H it suffices to bound the derived length of HU/U in terms of d. By 18.15 there exists an postive integer r so that if F a linear groups of degree at most 8d, then F^{(}r) is unipotent. We will shot that H^{(r)} is locally unipotent.

For this let J be a finite subset of I and put H_{J} = ⟨J⟩. Then [V, X_{J}] is finite dimensional. Let W be any composition factor for H_{J} on W. Then [W, H_{J}] = 0 or W = [W, H_{J}]. Hence in any case W is finite dimensional. Since H_{J} is generated by elements of degree at most 2d on W we conclude from 18.12 that W has dimension at most 8d. So X_{J}^{(r)} is unipotent on W and so on all V. If J^{∗} ⊂ J, then H_{J^{∗}}^{(}r) ≤ H_{J}^{(}r). Since the H_{J}’s form a local system for H, the H_{J}^{(}r) form a local system for H^{(r)}. Thus H^{(r)} is indeed locally unipotent.

bosole Corollary 18.17

(a) Let H ≤ X with [V, H] finite dimensional. Then LSol(⟨H^{X}⟩) is solvable of derived length bounded by a function of dim_{K}[V, H].

(b) LSol(X)/calLU(X) is the union of a countable ascending chain of solvable X-invariant subgroups.

Proof: (a) follows from 18.16 and 18.8. (b) Let X_{i} = ⟨x ∈ X | deg_{V} (x) ≤ i⟨. Then by 18.16, LSol(X_{i}) is unipotent by solvable. Also LSol(X_{i}) = ⋃_{i=1}^{∞} LSol(X_{i}0 and so (b) holds.

Lemma 18.18 Let T, S be a finite groups, p a prime and T_{p} and S_{p} Sylow p-subgroups of sylwr T and S respectively. Let S be acting on the finite set I. Then T_{p} ≀ S_{p} is a Sylow p-subgroup of T ≀ S.

Proof: Clear by an order argument.

Lemma 18.19 Let n be a positive integer, p a prime, and S_{n,p} a Sylow p-subgroups of sylsym Sym(n).

(a) If S is a primitive p-subgroup of Sym(n), then |S| = n = p.

(b) If n = p^{k}, then S(p^{k}, p), is transitive and S_{p^{k},p} ≅ S_{p^{k−1}} ≀ C_{p}.

(c) Let n = ∑_{i=1}^{k} n_{i}p^{i} with 0 ≤ i < p. Then S_{n,p} ≅ ⊕S_{p^{i},p}^{n_{i}}.

(d) S_{n} has derived length ⌊log_{p} n⌋.

(e) Let y ∈ Sym(n) with |y| = p. Then there exists a p-subgroup S of Sym(n) with y ∈ S and der⟨y^{S}⟩ ≥ ⌊log_{p} \frac{Supp(y)}{p}⌋.

Proof: (a) Let A ≤ Z(S) with |A| = p. Then A ⊴ S and as S is primitive, A is transitive. Thus n = p and (a) holds. (b) Since Sym(p^{k}) contains a p^{k}-cycle, S(p^{k}, p) is transitive. Let D be a system of blocks for S(p^{k}, p) with |D| minimal. Then by (a), |D| = p. Let Δ ∈ D. Then S(p^{k}, p is a Sylow p-subgroup of N_{Sym(p^{k})}(D) ≅ Sym(Δ) ≀ Sym(D). So (b) follows from 18.18.

(c) Let Ω_{1}(i) … Ω_{m_{i}}(i) be the orbits of S_{n,p} of length p^{i}. Then S(n, p) ≤ ⊕_{i,j} Sym(Ω_{j}(i) leq Sym(n). Let T_{j}(i) be the projection of S(n, p) onto Sym(Ω_{j}(i) onto Sym(Ω_{j}(i) and let S_{j}(i) be a Sylow p-subgroup of Sym(Ω_{j}(i) with T_{j}(i) ≤ S_{j}(i). Then S(n, p) ≤ ⊕_{i,j}T_{j}(i) and as the latter group is a p-subgroup of Sym(n), S(n, p) = ⊕_{i,j}T_{j}(i). Suppose that m_{i} ≥ p for some i. Then ⊕_{j=1}^{p}T_{j}(i) is a Sylow p-subgroup of Sym(⋃_{j=1}^{p} Ω_{j}(i)) = Sym(p^{i+1}, which is not transitive, a contradcition to (b). Hence m_{i} < p and as ∑_{i} m_{i}p^{i} = n, m_{i} = n_{i}. Thus

(c) holds.

(d) Choose k as in (c) with k minimal. Then n_{k} ≠ 0 and k = ⌊log_{p} n⌋. Then by (c), S_{n,p} and S_{p^{k},p} have the same derived length. By (b) S_{p^{k},p} ≅ S_{p^{k−1,p}} ≀ C_{p}. Also (S_{p^{k−1},n} ≀ C_{p})′ is a subgroup of the base group S_{p^{k−1}}^{p} which projects onto S_{p^{k−1}}. Hence der(S_{p^{k−1},n} ≀ C_{p})′ der S_{p^{k−1},n} = k − 1, by induction. So der S_{p^{k},p} = (k − 1) + 1 = k.

(e) Let Δ_{1}, … Δ_{r} be the non-trivial orbits for y. Then r = \frac{| Supp(y)|}{p}. Pick x_{i} ∈ Δ_{i} and

let Δ = {x_{i} | 1 ≤ i ≤ r}. Let 1 ≤ j < p. Then x_{i} ≠ x_{i}^{y^{j}} ∈ Δ_{i} and so Δ ∩ Δ^{x^{j}} = ∅. Thus Δ is block for y. Let T_{0} be a Sylow p-subgroup of Sym(Δ). Let T_{j} = T_{0}^{x^{j}}. Then T_{j} ∈ Sym(Δ^{x^{j}}) and T = ⟨T_{j} | o ≤ j < p⟩ ≅ T_{0}^{p}. Hence T is a p group and y normalizes T. Hence S = T ⟨y⟩ is a p-subgroup of Sym(n) containg y. Moreover [T, y] ≤ ⟨y^{T} ⟩ ≤ ⟨y^{S}⟩. Let t ∈ T. Then [t^{−1}, y] = tt^{−y} acts as t on Δ and so der⟨y^{S}⟩ ≥ der[T, y] ≥ der T. By (d) der T = log_{p} |Δ| = log_{p} r and (e) is proved.

lfsfa Theorem 18.20 Let G be a locally finite, simple, finitary group which has a Kegel cover all of whose factors are alternating groups. Then G is an alternating group.

Proof: Let p be an odd prime so that G has an element x of order p. By 18.17 there exists a postive integer r so that

lfsfa-1 1. ⟨x^{S}⟩^{(r)} = 1 for all p subgroups S of G with x ∈ S.

Let {(H_{i}, M_{i}) | i ∈ I} be a Kegel cover for G with H̄_{i} = H_{i}/M_{i} ≅ Alt(Ω_{i}). Let J = {i ∈ I | i ∈ H_{i}, x ∉ M_{i}}.

lfsfa-2 2. Let j ∈ J and S̄ a p-subgroup of H̄_{j} with x̄ ∈ S̄. Then der⟨x̄S̄⟩ ≤ r.

For this let R be the inverse image of S̄ in H_{j}. Also let S be a Sylow p-subgroup of R with x ∈ S. As R/M_{j} = S̄ is a p-groups, R = SM_{j}. Hence der⟨x̄S̄⟩ der⟨x^{S}⟩M_{j}/M_{j} ≤ der⟨x^{S}⟩ and so 2. follows from ??.

lfsfa-3 3. Let j ∈ J. Then | Supp_{Ω_{j}} |(x) < p^{r+2}.

Suppose | Supp_{Ω_{j}} |(x) ≥ p^{r+2}. Then by 18.19e there exists a p-subgroup S̄ of H̄_{j} ≅ Alt(Ω_{j}) with x̄ ∈ S̄ and der⟨x̄S̄⟩ ≥ r + 1. But this contradicts 2..

lfsfa-4 4. G is a group of finitary permutations.

Let D be an ultrafilter on I with J ∈ D. Let Ω = ∏_{D} Ω_{i}. Then G acts on Ω. Since J ∈ D, x acts non-trivially on Ω. As G is simple, G acts faithfully on Ω. By 3. and as J ∈ D we conclude that | Supp_{Ω}(x)| < p^{r+2}. Hence x ∈ FSym(Ω). Since G = ⟨x^{G}⟩ and FSym(Ω) is a normal subgroup of Sym(Ω), G ≤ FSym(Ω).

The theorem now is an immediate consequence of 4. and 9.14.

Chapter 19

Elementary equivalence

100 CHAPTER 19. ELEMENTARY EQUIVALENCE

Exercises

exercises

5. Let M be a minimal normal subgroup of G. If M is locally solvable show that M is elsmn elementary abelian (i.e M is abelian and there exists a prime p with m^{p} = 1 for all m ∈ M).

6. Let G be a universal, locally finite group group and H a countable, locally finite group. ehiug Prove that G contains a subgroup isomorphic to H.

7. Let G be a group and N a normal subgroup of G. If N and G/N are locally finite, elfblf prove that G is locally finite.

8. Let G be an LF S-group such that G has a Kegel cover all of whose factors are alter- eac nating groups. Let x ∈ G such that C_{G}(x) is abelian.

(a) If G is finite determine all possibilities for G and x.

(b) Prove that G is finite.

9. Let U be a P. Hall’s universal group. Show that U is simple and explicitly write down a Kegel cover U.

10. Let G be any of non-absolutely simple LFS-groups constructed in section 6. Determine ekn a Kegel-cover for G.

11. Let V be a finite dimensional vector space over the field K and p a prime. Investigate emaxplin the maximal p-subgroups of GL_{K}(V).

12. Let N be a minimal normal subgroup of G. What can you say about the structure eminnor of G?

13. Construct a locally finite simple group with a maximal subgroup which is a p-group. emaxp 14. Let X be a primitive subgroup of Sym(Ω) containing a 3-cycle. Prove that Alt(Ω) ≤ e3cycle X.

15. Let K be an infinite, locally finite field. Determine under all pairs (F, K) where F emssl2 is a finite subfield of K so that the subgroup SL_{2}(F) of SL_{2}(K) is not contained in any maximal subgroup of SL_{2}(K).

eplc 16. Prove that a periodic, linear group with no non-trivial, unipotent, normal subgroups is countable.

euifp 17. Show that there are uncountably many isomorphism classes of transitive, finitary permutations groups.

euup 18. Given a set I, a non-principal ultra-filter D on I and non-empty finite sets A_{i}, i ∈ I. Suppose that for all positive integers n,

{i ∈ I | |A_{i}| ≤ n} ∉ D.

Also suppose that either all A_{i} are finite or that I is countable. Show that ∏_{D} A_{i} is uncountable.

ecwp 19. Let A, B be groups such that B acts transitively on the set I. Let H = A ≀_{I} B. Determine H′.

eilsn 20. Investigate G under the assumption that all infinite subgroups of G are locally subnormal.

ecnas 21. Let G be any of the non-absolutely simple, locally finite simple group constructed in class and

1 = M_{0} ⊴ M_{1} ⊴ M_{2} ⊴ M_{3} … ≤ G

the corresponding series.

(a) Explcitly determine a Kegel coverfor G.

(b) Show that for each i ≥ 0, M_{i+1}/M_{i} is residually finite.

(c) Show that G has a composition series all of whose factors are finite.

emnsc 22. Let N be a minimal normal subgroup of G. Show that N has a sectional cover all of whose factors are characteristically simple.

eqc 23. Let G be a finite group, M ⊴ G and x ∈ G. Prove that |C_{G/M}(x)| ≤ |C_{G}(x)|.

efc 24. Let G be a locally finite simple group with a Kegel cover all of whose factors are alternating groups. Let x ∈ G with C_{G}(x) finite. Prove that G is finite. djckc Definition 20.1 Let K be field, X a group and V an irreducible KX-module. Let D =

End_{KV} (V). Then a JCK-cover for V with respect to X is a set of triples {(H_{i}, V_{i}, M_{i}) | i ∈ I} such that for all i in I:

(a) H_{i} is a finitely generated subgroup of X.

(b) V_{i} is a finitely generated DH_{i}-submodule of V.

(c) M_{i} is a maximal DH_{i}-submodule of V_{i}.

(d) For each finite dimensional D-subspace W of V, there exists j ∈ I so that W ≤ V_{j} and W ∩ M_{j} = 0.

25. Let K,X and V as in the above defintion. Show that there exists a JCK-cover for ejckc V with respect to X. ( Hint: Use the usual argument for Kegel covers and the Jacobson-Chevalley Density Theorem.)

26. Let K be a set of finite sets and J a partion of K into finite subsets. For J ∈ J put ediag T_{J} = ∏ J. Show that Diag(∏ K) ≅ Diag(∏_{J∈J} T_{J}).

27. Let K and J be countable sets of finite sets. Determine necessary and sufficient ediis conditions for Diag(∏ K) and Diag(∏ J) to be isomorphic as abstract groups.

28. Let I be a set, A a subalgebra of P(I) with I ∈ A, H a non-abelian simple group erha and R = R(H, A). Let M be a maximal normal subgroup of R. Show that R = MR_{I} and conclude that R/M ≅ H.

104 CHAPTER 20. EXERCISES

Bibliography

[BBMN] B.L. Birch, R.G. Burns, S.O Macdonald and P.M. Neumann, On the degrees of permutation groups containing separating finite sets, Bull.Austral.Math.Soc. 14(1976) 7-10.

[Ca] P.J. Cameron,Oligomorphic Permutation Groups, LMS Lecture Note Series 152, Cambridge University Press(1990).

[Di1] J. Dixon, Linear Groups

[Di2] J. Dixon, Permutation Groups

[HH] J.I. Hall and B. Hartley, A group theoretical characterization of simple, locally finite, finitary linear groups, Arch.Math 60, 108-114(1993).

[Ha1] J.I. Hall, Infinite Alternating Groups as Finitary Linear Transformation Groups, J. of Alg. 119, 337-359 (1988).

[Ha2] J.I. Hall: Periodic Classical Groups of Finitary Linear Transformationitary Linear Transformations, in preparation.

[He] I.N. Herstein, Topics in Algebra, John Wiley & Sons, New York (1975).

[Hum] S.P. Humphries: Free subgroups of SL(n,Z) generated by transvections, J. Alg. 116(1988)155-162.

[Hup] B. Huppert, Endliche Gruppen I, Springer-Verlag, Berlin (1983).

[KW] O. Kegel and B.A.F. Wehrfritz, Locally Finite Groups, North Holland (1973).

[MPP] U. Meierfrankenfeld, R.E. Phillips and O. Puglisi, Local Groups, J.London Math.Soc.(2) 47, 31-40 (1991).

[Ne2] P. Neumann: The lawlessness of finitary permutation groups, Arch. Math. 26(1975) 561-566.

[Ne2] P. Neumann: The structure of finitary permutation groups, Arch. Math. 27(1976) 3-17.

[Ph1] R.E. Phillips: The Structure of Finitary Linear Transformation, J. Alg. 119 (1988) 400-448.

[Ph2] R.E. Phillips, On Absolutely Simple Locally Finite Groups, Rend. Sem. Mat. Univ. Padova, 79,(1988), 213-220

[PZ] C.E.Praeger and A.E.Zalesskii, Orbit Lengths Of Permutation Groups And Group Rings Of Locally Finite Simple Groups Of Alternating Type, Preprint 1993

[Wi1] H. Wielandt, Unendliche Permutationsgruppen, Lecture notes, Math. Inst. Unversität Tübingen (1960).

[Wi2] H. Wielandt, Finite Permutation Gruppen, Academic Press,New York (1964).

Index

H-isomorphc, 19 E-perfect, 21 E-series, 22 first order language, 45 defect, 25 degree, 39 inverse limit, 43 Jordan-Hölder, 19 linear, 39 locally, 5 primitive, 35 symbols, 45 unitary, 40