Ascending subgroups of irreducible finitary linear group

U. Meierfrankenfeld

1 Introduction

Let K a field and V a vector space over K. Let F GL_{K}(V) be the finitary linear group of V over K, namely F GL_{K}(V) = {g ∈ GL_{K}(V)| [V,g] has finite K-dimension }. Subgroups of F GL_{K}(V) are called finitary groups. Recently a good amount of work has been done towards a classification of the locally finite, finitary groups (see [1, 2, 7]). On the otherhand very little is known without the assumption of locally finiteness. This paper is meant as a contribution to the general theory of finitary groups.

\underline{Throughout this paper G is a subgroup of F GL_{K}(V)}. Suppose that G is irreducible and H an ascending subgroup of G, that is there exist a well ordered set I and subgroups H_{i}, i ∈ I, of G including H and G such that H_{i} ◁H_{i+1} and if i is a limit ordinal in I, then H_{i} = ∪_{j<i}H_{j}. Then our main theorem (7.6) asserts that H acts completely reducibly on V. The most important step in the proof of this is (6.2), which provides a component type subgroup in G in the case where G acts irreducibly on V and has a reducible ascending subgroup. As a consequence (7.4) we can prove that ascending subgroups of primitive, infinite dimensional, finitary groups are primitive. In (4.5) we prove a Jordan-Hölder Theorem for finitary modules of finitary groups.

2 Preliminaries

Definition 2.1 Let H be a group and V a KH-module.

1. A series for H on V is a set Γ of submodules of H in V such that

(a) Γ is totally ordered by inclusion,

(b) Γ is complete, i.e. if Λ ⊂ Γ, then ∪Λ ∈ Γ and ∩Λ ∈ Γ

(c) 0 ∈ Γ and V ∈ Γ.

2. Let Γ be a series for H on V. For S in Γ let S_{−} = ∪{T ∈ Γ|T < S}. If S ≠ S_{−} then (S_{−}, S) is called a jump of Γ and \overline{S} = S/S_{−} is called a factor of S

3. A series for H on V is called a composition series if all its factors are irreducible KH-modules.

4. A class F of KH-modules is called closed if every non-zero section of H of an element of F is in F.

5. Let F be a closed class of KH-modules. Then series for H on a KH-module is called a F-series provided that all its factors are in F.

6. A series for H on V is called finite dimensional if all its factors are finite dimensional: i.e if it is a F-series, where F is the class of finite dimensional KH-modules.

Lemma 2.2 Let H ≤ GL_{K}(V) and F a closed class of KH-modules. Then H has a unique minimal submodule m_{V,F}(H) in V such that H has a F-series on V/m_{V} (H).

Proof: LetW be the set of a H-submodules F in V such that H has a F-series on V/F. We need to show that ∩W is in W and for this we may assume that ∩W = O.Thus the goal is to show that V has an F-series for H.

For any series Γ of H on V,define

m(Γ) = \{S ∈ Γ|0≠\overline{S} ∉ Γ\}

and

M(Γ) = {S ∈ Γ|m(γ) ≤ S}

On the set C of H-composition series on V define a partial order ≤ by

Γ ≤ Λ if M(Γ) ⊆ M(Γ)

We will show that

(*) C is linearly ordered.

Suppose for the moment that (*) is true. Then by Zorn’s lemma C has a maximal element Γ. We claim that Γ is a F series. Otherwise m(∥Gamma) ≠ 0. Since by assumption ∩W = 0, there exists F in W with m(Γ) ≰ F. But then m(Γ)/(m(Γ ∩ F ≅ m(Γ) + F)/F has a F-composition series. It is now easy to see that there exists a compostion series Γ^{∗} of H on V with m(Γ^{∗}) ≤ m(Γ) ∩ F and M(Γ) ⊂ M(Γ^{∗}), a contradiction to the maximality of Γ.

It remains to prove (*). We remark that Γ ≤ Λ implies that M(Γ) = {T ∈ Λ|m(Γ) ≤ T }. Indeed, let T ∈ Λ with m(Γ) ≤ T. Since M(Γ) ⊆ M(Λ) ⊆ T, X ≤ T or T ≤ X for any X ∈ M(Γ). Put T_{−} = ∪{X ∈ M(Γ)|X ≤ T } and T_{+} = ∩{X ∈ M(Γ)|T ≤ X}. Then T_{−} ≤ T ≤ T_{+}. Since Γis a composition series, T_{+}/T_{−} is irreducible as H-module. Hence T is equal to T_{+} or T_{−} and so T ∈ M(Γ).

Let D be any chain in C and put

m = ∩{m(d)|d ∈ D}

and

M = ∪{M(d)|d ∈ D}.

Let N be any composition series for H on m and Γ = M ∪ N. We wish to show that Γ is a composition series for H on V. Clearly Γ is totally ordered. To show that Gamma is complete, let Λ be any subset of Γ.

If N ∩ Λ ≠ Ø, then∩Λ ∈ N. If Λ ⊆ M(d) for some d in D, then Λ ∈ M(d). If Λ ⊆ M, but Λ ≰ M(d) for every d in D, then for every d in D there exists T in Λ with T ≰ M(d). Pick e in D with T ∈ M(e). Then d ≤ e. By the above remark, m(d) ≰ T and so T ≤ m(d). Thus m ≤ ∩Λ ≤ ∩{m(d)|d ∈ D} = m. Hence in all cases ∩Λ ∈ Γ.

If Λ ⊆ N, then ∪Λ ∈ N. If Λ ∩ M(d) ≠ Øfor some d in D, then ∪Λ ∈ M(d). Hence also ∪Λ ∈ Γ and Γ is complete. To finish the proof that S is a composition series we have to show that \overline{S} is irreducible for all S in Γ. If S ∈ N this is obvious. If S ∉ N there exists d in D with m(d) < S, since otherwise S ≤ m(d) for all d in D and so S ≤ m and S ∈ N,a contradiction. Thus S and S_{−} are in M(d) and so \overline{S} is a factor of d. In particular, \overline{S} is irreducible and S ∈ F ∪ {0}

We have proven that Γ is a composition series and that S ∈ F ∪ {0} for all S ∈ Γ with S ≰ m. It follows that m(Γ) ≤ m and M ⊆ M(Γ). Thus Gammais an upper bound for D and (*) is proved.

Lemma 2.3 Let Y be a group, R ≤ Y and s in R with [R, R^{s}] = 1.Then R′ ≤ [R, s].

Proof: Let r,t be in R. Then

[r, t] = r^{−1}s^{−1}rt = (r^{−1}t^{−1}(rt)^{s})(rt)^{−1}rt = r^{−1}t^{−1}r^{s}t^{s}[rt, s]^{−1}

=

= r^{−1}r^{s}t^{−1}t^{s}[rt, s]^{−1} = [r, s][t, s][tr, s]^{−1} ∈ [R, s]

Lemma 2.4 Let Y be a group and R and S subgroups of Y. Suppose that the following three conditions hold:

(a) R ∩ R^{s} = 1 and [R, R^{s}] = 1 for all s ∈ S \ N_{S}(R).

(b) S ∩ S^{r} = 1 and [S, S^{r}] = 1 for all r ∈ R \ N_{R}(S).

(c) R does not normalize S and S does not normalize R.

Then one of the following holds:

(i) R′ = S′ = 1, N_{R}(S) = N_{S}(R) = 1 and any two non-trivial elements in R ∪ S have the same order.

(ii) If {T,U} = {R,S} and X = N_{T} (U), then |T/X| = 2, X is abelian, C_{X}(T) = 1 and T inverts X.

Proof: Put A = ∩{N_{R}(S^{r})|r ∈ R} and B = ∩{N_{S}(R^{s})|s ∈ S}. Pick r ∈ R \ N_{R}(S) and s ∈ S\N_{S}(R). Note that [(R∩S)^{s}, R] ≤ [R^{s}, R] = 1 and so (R∩S)^{s} = (R∩S)^{sr} ≤ S∩S^{r} = 1. Hence

(1) R ∩ S = 1.

(2) [R,S] normalizes R and S, R′S′ ≤ [R, S], R′ ≤ A and S′ ≤ B.

For the proof of (2), note that since S is normal in < S^{R} >, [R,S] normalizes S. Furthermore, by (2.3), R′ ≤ [R, S] and so R′ normalizes S and R′ ≤ A.

(3) A′ = B′ = [A, B] = 1.

By (2.3), A′ ≤ [A, S]. But A normalizes S and so A′ ≤ R ∩ S = 1. Now also [A, B] ≤ R ∩ S = 1, and (3) is established.

(4) C_{A}(R) = 1 and C_{B}(S) = 1.

Let t ∈ S . If R = R^{t} then C_{A}(R)^{t} ≤ Z(R)^{t} = Z(R). If R≠R^{t}, then [R, R^{t}] = 1. Thus in any case, [R, C_{A}(R)^{t}] = 1. Hence [C_{A}(R), S] ≤ C_{S}(R) ≤ S ∩ S^{r} = 1 and C_{A}(R) ≤ C_{R}(S) ≤ R ∩ R^{s} = 1.

(5) If R′ ≠ 1, then (ii) holds.

Suppose that R′ ≠ 1. Assume also that R^{S}≠{R, R^{s}} and pick t in S with R^{t} ≰ {R, R^{s}}. Let u, v ∈ R. Then [[t, u], [s, v]] = [u^{−t}u, v^{−s}v] = [u, v]. If u ∈ R′, then by (2) u normalizes S and so [t, u] ∈ S. Now [s,v] normalizes S and so [u, v] ∈ R ∩ S = 1. Thus [R′, R] = 1 and (2) and (4) imply that R′ = 1. Hence R^{S} = {R, R^{s}} and so |S/B| = 2. Note that R ≅ (R × R^{s})/R^{s} = [R, s]R^{s}/R^{s}. It follows that [R,S] is not abelian and so S is not abelian. Therefore |R/A| = 2. By (4) R inverts A, and S inverts B. Thus (ii) holds.

We may assume from now on that R and S are abelian. By (4), A = B =1, and so N_{R}(S) = N_{S}(R) = 1. Moreover, < R^{S} > is abelian and so [S,R,R] = 1 and [R,S,S] =1. Let k be a positive integer. Then [r^{k}, s] = [r, s]^{k} = [r, s^{k}]. So r^{k} ≠ 1 implies [r^{k}, s]≠1 and s^{k} ≠ 1. Hence |r| = |[r, s]| = |s| and (i) holds.

Lemma 2.5 (a) Let Ω be a set with at least 5 elements. Then Alt(Ω) has no proper ascending subgroups.

(b) Let Ω be a finite set, H a primitive subgroup of Sym(Ω) and N a non-trivial subnormal subgroup of H. Then N acts fixed-point freely on Ω.

Proof:(a) This is well known and easy to proof, see for example [8].

(b) We prove (b) by induction on |H|. Suppose 1 ≠ N is subnormal in H and has a fixed-point ω in Ω.

Let N ◁ ◁L ◁ H with L ≠ H. Then L acts transitively and, by induction, imprimitively on Ω. Let R be the largest subgroup of L which acts trivally on all maximal systems of imprimitivity for L on Ω. Then H normalizes R and R acts intransitively on Ω. So R = 1. On the other hand let Π be any maximal system of imprimitivity for L on Ω. Pick P in Π with ω ∈ P. Then N normalizes P. Since L acts primitively on Π we conclude by induction that N acts trivially on Π. So N ≤ R and N = 1; this contradiction completes the proof.

Lemma 2.6 Let Ω be an infinite set, L a transitive subgroup of F Sym(Ω)(= {g ∈ Sym(Ω)| supp(g) is finite}) and H an ascending subgroup of L. Then either H is transitive on Ω or Ω is the disjoint union of H invariant sets of imprimitivity for L on Ω.

Proof: Let L_{i}, i ∈ I, be an ascending normal series from H to L with L_{0} = H. Assume first that H as an orbit of infinite length on Ω, then clearly L_{1} normalizes that orbit and in easy induction proof shows that L normalizes that orbit. Hence H is transitive, a and the Lemma follows in this case.

Assume next that every orbit of H on Ω is finite and that L has a maximal system Π of imprimitivity on Ω. Put D = {d ∈ L|P^{d} = P forallP ∈ Π}. Then by [9], Satz 9.4, L/D = FSym(P) or L/D = Alt(P). Since clearly every orbit for H on P is finite, Alt(P) ≤ HD/D. Thus by 2.5(a), H ≤ D and the lemma holds also in this case.

Assume finally that every orbit of H on Ω is finite and L has no maximal system of imprimitivity. By [5],(2.2), there exists a chain Ω_{1}, Ω_{2}, … of sets of imprimitivity for L on Ω such that ∪_{i}Ω_{i} = Ω and such that no set of imprimitivity of L on Ω lies between Ω_{i} and Ω_{i+1}. Put

M_{0} = {Ω_{i}^{g}|i ≤ 1, g ∈ L, Ω_{i}^{g} is H invariant}

and let M be the set of mimimal elements in M_{0}. We claim that Ω is the disjoint union of the elements in M. Let X, Y ∈ M with X ∩ Y ≠ Ø. Without loss |X| ≤ |Y |. Hence there exists g ∈ L with X ⊆ Y^{g}. Thus Y ∩ Y^{g} ≠ Øand so Y = Y^{g}. By minimality of Y, X = Y.

Suppose ∪M≠Ω, Then there exists an orbit O for H on Ω with O ⊈ ∩M. Since O is finite, there exists i with O ⊆ Ω_{i}. Then H normalizes Ω_{i} and Ω_{i} ∈ M_{0}. Let X ∈ M_{0} be of mimimal order with respect to X ⊈ ∪M. Pick Y ∈ M with Y ⊆ X and let i ≤ 1 maximal with respect to Y ⊆ Ω_{i}^{g} ⊂ X for some g ∈ L. Then H normalizes Ω_{i}^{g}. By choice of g, N_{L}(X) acts primitively on {Ω_{i}^{gk}|k ∈ N_{L}(X)} and so by 2.5(b) H acts trivially on {Ω_{i}^{gk}|k ∈ N_{L}(X)} ⊆ M_{0}. The minimal choice of |X| implies X = ∪{Ω_{i}^{gk}|k ∈ N_{L}(X)} ≤ ∪M, a contradiction.

Hence Ω = ∪M and the Lemma holds also in this last case.

3 The submodules \overline{U}_{V} (G) and \underline{U}_{V} (G)

Definition 3.1 Let H be a group, h ∈ H and W a KH-module.

1. h acts unipotently on W if W (h − 1)^{n} =0 for some non-negative integer n.

2. H acts unipotently on W if each of its elements act unipotently.

3. \overline{U}_{W} (H) =< \{X| X a KH-submodule in W such that H acts unipotently on W/X\}¿.

4. \underline{U}_{W} (H) = ∑\{X| X a KH-submodule in W such that H acts unipotently on X\}.

5. W is a perfect KH-module if W = \overline{U}_{W} (H): i.e if no non-zero factor module of W is a unipotent KH-module.

Lemma 3.2 (a) G acts unipotently on V if and only if V has a series with respect to G all whose factors are central.

(b) The subgroup of G generated by all the ascending unipotent subgroups of G acts unipotently on V.

(c) If G acts irreducibly on V, no non-trivial normal subgroup of G acts unipotently on V.

Proof: see Theorem B in [4].

Lemma 3.3 (a) G acts unipotently on V/\overline{U}_{V} (G).

(b) G acts unipotently on \underline{U}_{V} (G).

(c) Let Y be a KG-submodule in V. Then \overline{U}_{Y} (G) ≤ \overline{U}_{V} (G). Moreover, if \overline{U}_{V} (G) ≤ Y, then \overline{U}_{Y} (G) = \overline{U}_{V} (G).

(d) Suppose that G =< L_{i}|i ∈ I > with L_{i} asc G and L_{i}^{G} ⊆ {L_{j}|j ∈ I}, for all i in I. Then

\overline{U}_{V} (G) = ∑_{i∈I} \overline{U}_{V} (L_{i}) and \underline{U}_{V} (G) = ∩_{i∈I} \underline{U}_{V} (L_{i})

(e) Let S be a set of KG-submodules in V. Then

\overline{U}_{∑_{S}}(G) = ∑\{\overline{U}_{s}(G)|s ∈ S\} and \underline{U}_{∩S}(G) = \underline{U}_{∩\{U_{s}(G)|s∈S\}}(G)

.

Proof: We first prove that

(*) Let x be a unipotent element in F GL_{K}(W). Then W (x − 1)^{deg(x)+1} = 0.

Indeed let d be minimal with W (x − 1)^{d} = 0. Then

W (x − 1) > W (x − 1)^{2} > ... > W (x − 1)^{d−1} > 0.

So deg(x) = dimW (x − 1) ≤ d − 1, and (*) is proved.

Put Z = {X| X a G-submodule such that G acts unipotently on V/X}and let g ∈ G. By (*), V (g − 1)^{deg(g)+1} ≤ X for all X in Z. Thus V (g − 1)^{deg(g)+1} ≤ \overline{U}_{V} (G) and so \overline{U}_{V} (G) ∈ Z. This proves (a)

By (*) \underline{U}_{V} (G)(g − 1)^{deg(g)+1} = 0 for all g in G. So (b) holds. Let Y be a KG-submodule in V. Then G acts unipotently on Y + \overline{U}_{V} (G)/\overline{U}_{V} (G) and so

also on Y/Y ∩ \overline{U}_{V} (G). Thus \overline{U}_{Y} (G) ≤ Y ∩ \overline{U}_{V} (G) ≤ \overline{U}_{V} (G).

If \overline{U}_{V} (G) ≤ Y, then G acts unipotently on V/Y and Y/\overline{U}_{Y} (G). Hence G acts unipotently on V/\overline{U}_{Y} (G) and therefore \overline{U}_{V} (G) ≤ \overline{U}_{Y} (G). Thus (c) holds.

To prove (d), note that G normalizes ∑\{\overline{U}_{V} (L_{i})|i[I\} and that L_{i} acts unipotently on V ∑\{\overline{U}_{V} (L_{i})|i ∈ I\}. By part (b) of (3.2), G acts unipotently on V ∑\{\overline{U}_{V} (L_{i})|i ∈ I\}. Thus \overline{U}_{V} (G) = ∑\{\overline{U}_{V} (L_{i})|i ∈ I\}. A similar argument shows that \underline{U}_{V} (G) = ∩\{\underline{U}_{V} (L_{i})|i ∈ I\}.

(e) This is readily verified and we omit the proof.

We remark that (2.2) provides an alternative proof for part (a) of 3.3. Indeed, let F be the class of one dimensional central KG-modules and X a KG-submodule in V. Then by part (a) of (3.2), G acts unipotently on V/X if and only if G has an F series on V/X.Thus m_{V,F}(G) = \overline{U}_{V} (G).

Proposition 3.4 If G is locally nilpotent and V is a perfect KG-module, then every KGsubmodule of V is perfect.

Proof: Assume first that G is nilpotent and V is finite dimensional. If \overline{U}_{X}(G) ≠ 0, then by induction on the dimension of V, X/\overline{U}_{X}(G) is a perfect submodule of V/\overline{U}_{X}(G). So X is perfect. If \overline{U}_{X}(G) = 0, G acts unipotently on X. Since X is finite dimensional, part (a) of (3.2) implies that C_{X}(G) ≠ 0. Let 1 ≠ z ∈ Z(G)). Since C_{V} (z) ≠ 0,V ≠ [V,z]. Moreover, [V, z] ≅ V/C_{V} (z) and so [V,z] is perfect. It follows by induction that X ∩ [V, z] and X+[V,z]/[V,z] are perfect. Hence X is perfect.

In the general case let F be the set of finitely generated subgroups of G. Note that G acts unipotently on V/ ∪ \{\overline{U}_{V} (F)|F ∈ F\} and so V = ∪\{\overline{U}_{V} (F)|F ∈ F\}. This implies U = ∪\{X ∩ Ū_{V} (F)|F ∈ F\}. Moreover, \overline{U}_{V} (F) ≤ [V, F] and so \overline{U}_{V} (F) is finite dimensional. Since F is nilpotent, the preceeding paragraph implies that X ∩ \overline{U}_{V} (F) is a perfect module for every F ∈ F. Thus X is a perfect KG-module and the lemma is proved.

4 A Jordan-Hölder Theorem for Finitary Modules

In this section we will prove that the non-central factors in a composition-series for G on V are independent from the choice of the composition series. This statement is not true for the central factors. The concept introduced in the following definition is designed to isolate the non-central factors of a series.

Definition 4.1 1. A u-series for G on V with respect to a normal subgroup H of G is a set S of KG-submodules in V such that

(a) S is totally ordered by inclusion,

(b) if T ⊆ S, then ∪T ∈ S and \overline{U}_{∩T} (H) ∈ S,

(c) 0 ∈ S and \overline{U}_{V} (H) ∈ S,

(d) all elements in S are perfect KH-modules.

2. Let S be a u-series for G on V with respect to H. For X in S let

X_{−} = ∪\{T ∈ S|T < S\}, X_{+}/X_{−} = \underline{U}_{X/X_{−}}(H) and \overline{X} = X/X_{+}

If X≠X_{−}, then (X_{−}, X) is called a jump of S and \overline{X} is called a factor of X.

3. A u-composition series for G on V with respect to H is a u-series for G on V with respect to H all of whose factors are irreducible as KG-modules.

4. A u-series (u-composition series) for G on V is a u-series (u-composition series)for G on V with respect to G.

Lemma 4.2 Let H be a normal subgroup of G and for any KH-module X put U(X) = \overline{U}_{X}(H) (a) Let S be a series for G. (a1) Put

T = {U(X)|X ∈ S}

Then T is a u-series for G on V with respect to H.

(a2) Let Y ∈ T and put

X = ∩{Z ∈ S|U(Z) = X}.

Then Y = U(X), Y_{−} = U(X_{−}) and

\overline{Y} ≅ U(\overline{X})/(\underline{U}_{U(\overline{X})}(H)

as a KG-module. In particular, if S is a composition series for G on V, then \overline{(}Y) ≅ ¯(X), T is a u-composition series for G on V with respect to H and the factors of T are the factors of S not centralized by H.

(b) Let Y be a KG-submodule in V and S a u-series for G on V with respect to H.

(b1) Let

T = {U(X ∩ Y)|X ∈ S}.

Then T is a u-series for G on X with respext to H.

(b2) Let Z ∈ T and put

X = U(∩{W ∈ S|U(X ∩ W) = Z}).

Then \overline{Z} is isomorphic to a submodule of \overline{X}. In particular, if S is a u-composition series for G on V with respect to H, then \overline{Z} ≅ \overline{X} and T is a u-composition series for G on X with respect to H.

Proof: (a) By part (c) of (3.3), T is totally ordered and U(W) = W for all W in T. Let R be a subset of T and put Q = {W ∈ S|U(W) ∈ R}. By part (e) of (3.3)

∪R = ∪{U(W)|W ∈ Q} = U(∪Q) ∈ T

and

U(∩R) = U(∩{U(W)|W ∈ Q}) = U(∩Q) ∈ T.

Hence T is a u-series. Pick Y in T and define X as above. Then clearly U(X) = Y. Let W be in S. It follows from the definition of W that W < X if and only if and only if U(W) < Y. Therefore Y_{−} = U(X_{−}) and the remaining assertions in (a2) follow easily.

(b) Clearly T is totally ordered and U(W) = W for all W in T. Let R be a subset of T and put Q = {W ∈ S|U(Y ∩ W) ∈ T }. By part(e) of (3.3),

∪R = ∪{U(Y ∩ W)|W ∈ Q} = U(Y ∩ ∪Q) ∈ T

and

U(∩R) = U(∩{U(Y ∩ W)|W ∈ Q}) = U(Y ∩ ∩Q) = U(Y ∩ U(∩Q)) ∈ .T

Hence T is a u-series for G on Y with respect to H. Pick Z in T and define X as above. Then clearly U(Y ∩ X) = Z. Let W be in S. It follows from the definition of X that W < X if and only if U(Y ∩ W) < Z. Therefore Z_{−} = U(Y ∩ X_{−}). Note that

U(Y ∩ X_{−}) ≤ U(Y ∩ U(X_{+})) ≤ U(Y ∩ X_{−}) = Z_{−}.

So Y ∩ X_{+} ≤ Z_{+}. On the other hand U(Z_{+}) ≤ Z_{−} ≤ X_{−} and Z_{+} ≤ X_{+}. Thus

Z ∩ X_{+} = Y ∩ X_{+} = Z_{+} and \overline{(}Z) = Z/Z_{+} = Z/(ZcapX_{+}) ≅ Z + X_{+}/X_{+}

. Now Z + X_{+}/X_{+} is a submodule of \overline{X} and this finishes the proof of 4.1

Lemma 4.3 Let T and B be KG-submodules of V such that B ≤ T, T/B is irreducible and [T/B, G] ≠ 0. Then there exists a minimal G-supplement to B in T, i.e. a KG -submodule C in T such that T = B + C and every proper KG-submodule in C is contained in B.

Proof: Pick h in G with [T/B, h] ≠ 0 and put H =< h^{G} >. Let U be any G-submodule in V and let S be a composition series for G on U. Since deg(h) is finite, h and H centralize all but finitely many of the factors of S. Let d(S) be the number of factors of S not centralized by H, d(U) = min { d(S) | S a KG composition series on U} and d = min { d(U) | U a KG-submodule of T with T = U+B}. Replacing T by an appropiate supplement,there is no loss in assuming that d(T) = d and that T = \overline{U}_{T} (H). Let S be a composition series for G on T with d(S) = d. Let C be a supplement to B in T.

Suppose that C ≠ T. Put D = \{\overline{U}_{X}(H)|X ∈ S\}. Then |D| = d + 1. For Y ∈ D, let Ỹ = \overline{U}_{Y ∩C}(H). Put E = \{Ỹ |Y ∈ D\}. Then by (4.1), D and E are u-composition series for G with respect to H on T and C, respectively. Thus

d + 1 ≤ d(C) + 1 ≤ |E| ≤ |D| = d + 1,

and so |E| = |D|. Pick Y in D with Y_{−} ≤ K and Y ≰ K. Since |E| = |D|, Ỹ_{−}≠Ỹ, and so Ỹ ≰ Y_{+}. Since Y/Y_{+} is an irreducible KG-module, we conclude that Y = Y_{+} + Ỹ. Hence H acts unipotently on Y/Y + Y_{−}. Since Y is a perfect KH-module, Y = Y + Y_{−} ≤ C, a contradiction to the choice of Y. Thus C = T and every proper submodule of T is contained in B.

Definition 4.4 Let R and S be sets of KG-modules.Then R and S are KG-isomorphic if there exists a bijection α : R → S such that for all MinR, M and α(M) are isomorphic KG-modules.

Theorem 4.5 Let R and S be composition series for G on V. Then the sets of non-central factors of R and S are KG-isomorphic.

Proof: Let T and B be G-submodules in V with B ≤ T, T/B irreducible and [T/B, G] ≠ 0. We claim that T/B is isomorphic to a factor of S. By (4.3) we may assume that every proper G-submodule in T is contained in B. Put Y = ∩{X ∈ S|T ≤ X}. Let R ∈ S with R ¡ Y. Then T ≰ R, T ∩ R is a proper submodule of T and so T ∩ R ≤ B. Thus T ∩ Y_{−} ≤ B. Suppose that B ≰ Y_{−}. Then Y = B + Y_{−} and so T = B + (T ∩ Y_{−}) ≤ B, a contradiction. Thus B ≤ Y_{−} and T ∩ Y_{−} = B. It follows that

T/B = T/T ∩ Y_{−} ≅ T + Y_{−}/Y_{−} = Y/Y_{−}.

Let X be a non-central factor of S. By the preceeding paragraph X isomorphic to a factor of R. Pick g in G with [X, g] ≠ 0. Since g has finite degree, g centralizes all but finitely many factors of R and S. In particular only finitely many factors of R and S are isomorphic to X. Let (B_{i}, T_{i}), 1 ≤ i ≤ n, and (C_{i}, U_{i}), 1 ≤ i ≤ m, be the jumps of R and S, respectively, with factors isomorphic to X.

To complete the proof of the theorem it is enough to show that n = m. We assume without loss that T_{i} ≤ B_{i+1}, 1 ≤ i < n, U_{j} ≤ C_{j+1}, 1 ≤ j < m, and n ≤ m.

Note that S^{∗} = {T_{n} ∩ D|D ∈ S} is a composition series for T_{n} whose set of factors is isomorphic to a subset of the set of factors of S. Suppose that the number of factors of S^{∗} isomorphic to X is smaller than m. Then there exists 1 ≤ i ≤ m with T_{n} ∩ U_{i} = T_{n} ∩ C_{i}. It follows that U_{i} + T_{n}/T_{n} ≅ U_{i}/U_{i} ∩ T_{n} ≅ U_{i}/C_{i} ∩ T_{n} and so U_{i} + T_{n}/T_{n} has a factor-module isomorphic to X. On the other hand {Y/T_{n}|Y ∈ R, T_{n} ≤ R} is a composition series for G on V/T_{n} none of whose factors is isomorphic to X, a contradiction to the first paragraph of the proof.

Thus S^{∗} has m factors isomorphic to X. Replacing S by S^{∗} we therefore may assume that V = T_{n}. Similarly we may assume that V = U_{m}. Put Q = {B_{n} ∩ D|D ∈ S}; then Q is a composition series for B_{n}. By induction on n, we conclude that Q has exactly n-1 factors isomorphic to X. Suppose that m ≥ n; then there exists 1 ≤ i < j ≤ m with B_{n} ∩ U_{i} = B_{n} ∩ C_{i} and B_{n} ∩ U_{j} = B_{n} ∩ C_{j}. Suppose that U_{i} ≤ B_{n}; then U_{i} ≤ C_{i}, a contradiction. Thus V = U_{i} + B_{n} and so U_{j} = U_{i} + (B_{n} ∩ U_{j}) = U_{i} + (B_{n} ∩ C_{j}) ≤ C_{j}, again a contradiction.. Thus n =m and the proof is completed.

Remark 4.6 (a) (4.5) can be rephrased as:

Let R and S be u-composition series for G on V. Then the sets of factors of R and S are KG-isomorphic.

(b) By (4.5), if G has a finite dimensional series on V, then every composition series for G on V is finite dimensional.

Example 4.7 The number of central factors in two distinct composition series for G can be different as the following example shows:

Suppose that K has a element k with k ≠ 0,1. Let V be the vectorspace over K with basis v_{0}, v_{1}, v_{2}, …. Define x_{i}, i ≥ 1, in F GL_{K}(V) by

v_{j}^{x_{i}} = v_{j}, 0≠j≠i, v_{i}^{x_{i}} = kv − i, v_{0}^{x_{i}} = v_{0} + v_{i}

Put V_{i} =< v_{j}|j > i > and U_{i} = V_{i}+ < (1 − k)v_{0} + ∑_{1≤j≤i} v_{j} >. Then it is readily verified that

V > V_{0} > V_{1} > V_{2} > … > 0 mboxandV > U_{1} > U_{2} > … > 0

are composition series for < x_{i}|i ≥ 1 > on V. Moreover, V/V_{0} is a central factor in the first series, while the second one has no central factors.

Using (4.5) we are now able to prove a dual version of (2.2) for finitary modules. We remark that the following lemma is false for non-finitary modules.

Lemma 4.8 Let F be a closed class of KG modules which includes all 1-dimensional central KG-modules. Then there exists a unique KG-submodule M_{V,F}(G) in V maximal with respect to G having a F-series on M_{V,F}(G).

Proof: Let S be the set of all KG-submodules U in V such that G has an F-series on U. Let U be in S. Then by the properties of F, G has a F -composition series on U and by 4.5 every composition series for G on U is an F-series. Order S by inclusion and let R be a chain in S. PutN = ∪R and extend R to a composition series C for G on N. Then, for all U in R, {X ∈ C|X ≤ U} is a composition series for G on U and hence a F-series for G on U. It follows that C is a F-series and so N is in S. By Zorn’s Lemma S has a maximal element M. Let U be in S; then G has a F-series on U + M/M ≅ U/U ∩ M and on M. Thus U + M is in S and U ≤ M.

5 Complete Reducibilty of Normal Subgroups of Irreducible Finitary Groups

Theorem 5.1 If G acts irreducibly on V and H is a normal subgroup of G, then H acts completely reducibly on V.

Proof: Suppose the Theorem is false. If H has an irreducible submodule I in V, then V = ∑_{g∈G} I^{g} and V is completely reducible. So H has no irreducible submodule in V and in particular, H has no non-zero finite dimensional submodule in V. Pick 1≠r ∈ H and put R =< r^{H} >. We will first prove that

(1) [V,R]+Y/Y is finite dimensional for any non-zero H-submodule Y in V.

Since G is irreducible on V, V = ∑_{g∈G} Y^{g} . Also since [V,r] is finite dimensional there exists g_{1}, …, g_{s} in G with [V,r] Z, where Z = ∑_{1≤i≤s} Y^{g_{i}}. Now Z is a H submodule and so [V, r^{h}] = [V, r]^{h} ≤ Z for all h in H. Thus [V, R] ≤ Z. Note that C_{Y} (g_{i}) ≤ Y ∩ Y^{g_{i}} and hence Y/C_{Y} (g_{i}) is finite dimensional for all 1 ≤ i ≤ s. Therefore Y^{g_{i}} + Y/Y, Z + Y/Y and [V, R] + Y/Y are finite dimensional, proving (1).

Since H has no non-zero irreducible submodule in V we can find a descending chain

[V, R] > Y_{1} > Y_{2} > ....Y_{n} > ...

of H submodules Y_{i} in [V,R]. Let Y = ∩_{i≥1}Y_{i}. Then [V,R]/Y is infinite dimensional and (1) implies that Y = 0.(in fact all non-zero KH-submodules of [V,R] have finite codimension in [V,R]). Hence there exists n with [V, r] ∩ Y_{n} = 0. It follows that [Y_{n}, r] = 0. Thus [Y_{n}, R] = 0 and C_{V} (R)≠0. (1) implies that [V, R] + C_{V} (R)/C_{V} (R) is finite dimensional. In particular,

(2) \overline{U}_{V} (R) + \underline{U}_{V} (R)/\underline{U}_{V} (R) is finite dimensional.

For N ≤ G let

U(N) = \overline{U}_{V} (N) ∩ \underline{U}_{V} (N) and d(N) = dim_{K} \overline{U}_{V} (N)/U(N).

Put d = min{d(N)|1≠N ⊴ H} and pick 1≠L^{∗} ⊴ H with d(L^{∗}) = d. Put

L =< N|1≠N ⊴ H, \overline{U}_{V} (N) = \overline{U}_{V} (L^{∗}), U(N)) = U(L^{∗}) > .

By (3.3), \overline{U}_{V} (L) = \overline{U}_{V} (L^{∗}) and U(L)) = U(L^{∗}). We prove next that:

(3) For all g in G, L = L^{g} orL ∩ L^{g} = 1.

Indeed, let g be in G with L∩L^{g}≠1. Then \overline{U}_{V} (L∩L^{g}) ≤ \overline{U}_{V} (L) and U(L)∩\overline{U}_{V} (L∩L^{g}) ≤ U(L ∩ L^{g}). By the minimal choice of d, d(L ∩ L^{g}) ≤ d(L). On the other hand

d(L ∩ L^{g}) = dim_{K} \overline{U}_{V} (L ∩ L^{g})/U(L ∩ L^{g}) ≤ dim_{K} \overline{U}_{V} (L ∩ L^{g})/U(L) ∩ \overline{U}_{V} (L ∩ L^{g}) ≤

≤ dim_{K} \overline{U}_{V} (L ∩ L^{g}) + U(L)/U(L) ≤ dim_{K} \overline{U}_{V} (L)/U(L) = d(L).

Hence equality holds at each place and we conclude that U(L) ∩ \overline{U}_{V} (L ∩ L^{g}) = U(L ∩ L^{g}) and \overline{U}_{V} (L ∩ L^{g}) + U(L) = \overline{U}_{V} (L). In particular, L acts unipotently on \overline{U}_{V} (L)/\overline{U}_{V} (L ∩ L^{g}) and so \overline{U}_{V} (L) = \overline{U}_{V} (L ∩ L^{g}). Thus U(L) = U(L ∩ L^{g}). By symmetry, \overline{U}_{V} (L^{g}) = \overline{U}_{V} (L ∩ L^{g}) and U(L^{g}) = U(L ∩ L^{g}). Therefore \overline{U}_{V} (L) = \overline{U}_{V} (L^{g}) and U(L) = U(L^{g}). By definition of L, L^{g} ≤ L and L ≤ L^{g}. Thus L = L^{g} and (3) is established.

If g is an element of G with L≠L^{g}, then by (3) L ∩ L^{g} = 1. Since L and L^{g} are normal subgroups of H we get [L, L^{g}] = 1. Put L^{⊥} =< L^{g}|g ∈ G \ N_{G}(L) >. Then [L, L^{⊥}] = 1. Put Y = [V, L^{⊥}] ∩ \overline{U}_{V} (L). Since \overline{U}_{V} (L)/U(L) is finite dimensional, there exist l_{1}, …, l_{t} in L^{⊥} with Y ≤ U(L) + Z, where Z = ∑_{1≤i≤t}[V, l_{i}]. In particular, L acts unipotently on Y+Z/Z and so \overline{U}_{V} (L) ≤ Z. Since Z is finite dimensional, \overline{U}_{Y} (L) is a finite dimensional H submodule in V and so \overline{Y}_{V} (L)) = 0. Thus

\overline{U}_{[V,L^{⊥}]}(L) ≤ \overline{U}_{[V,L^{p}erp]∩\overline{U}_{V} (L)}(L) ≤ \overline{U}_{Y} (L) = 0.

It follows that [V, L^{⊥}] ≤ \underline{U}_{V} (L) and in particular, \overline{U}_{V} (L^{)} ≤ \underline{U}_{V} (L). So \overline{U}_{V} (L^{g}) ≤ \underline{U}_{V} (L) for all g ∈ G\N_{G}(L) and \overline{U}_{V} (L) ≤ \underline{U}_{V} (L^{g}) for all g ∈ G\N_{G}(L). Hence \underline{U}_{V} (L) ≤ \underline{U}_{V} (< L^{G} >). Note that G normalizes \underline{U}_{V} (< L^{G} >). Since G acts irreducibly on V, part (c) of (3.2) implies that < L^{G} > does not act unipotently on V. Thus \underline{U}_{V} (< L^{G} >) = 0, \underline{U}_{V} (L) = 0 and \overline{U}_{V} (L) is a non-zero finite dimensional H -submodule in V; this contradiction completes the proof of (5.1).

Lemma 5.2 Let h be any element of G. Then there exists a subnormal subgroup H in G with h ∈ H such that H has a finite u-composition series on V.

Proof: Let S be any composition series for G on V and let T be the set of factors in S which are not centralized by h. Since [V,h] is finite dimensional, T is finite. Let H_{1} =< h^{G} > and inductively H_{i+1} =< h^{H_{i}} >. Pick t in T. If H_{i} acts irreducibly on V on t for all i, let n(t) = 0. Otherwise let n(t) be minimal such that H_{n(t)} acts reducibly on t. Note that in this case by (5.1) H_{n(t)} acts completely reducibly on t. Since H_{n(t)−1} acts irreducibly and finitary on t it is easy to see that all the irreducible H_{n}(t)-submodules are finite dimensional.

Let j ≤ n(t). We claim that H_{j} has a finite u-composition series on t. If n(t) = 0, H_{j} is irreducible. So assume that n(t) > 0. Then H_{j−1} is reducible on t and so t is the direct sum of finite dimensional irreducible H_{j−1}-submodules. Note that h centralizes all but finitely many of these submodules and so [t, H_{j}] is finite dimensional, proving our claim.

Put n = max{n(t)|t ∈ T } + 1 and H = H_{n}. Then H has a finite u-composition series on all elements of T and hence also on V.

6 Component-Type Subgroups for Irreducible Finitary Groups

Definition 6.1 1. d(G) is the sum of the dimensions of the factors of a u-composition series of G in V.

2. d_{G} = min{ d(L)— 1 ≠ L asc G}.

3. Λ = { L asc G — d(L) = d_{G}}.

We remark that d(G) and d_{G} may be infinite and that by 4.5, d(G) does not depend on the choice of the composition series.

Theorem 6.2 . Suppose G is irreducible on V and and d_{G} is finite. Let L ∈ Λ. Then one of the following holds:

(i) There exists L^{∗} ∈ Λ such that L^{∗} =< L^{g}|g ∈ G, L^{g} ≤ L^{∗} > and [L^{∗}, L^{∗g}] = 1 for all g in G with L^{∗}≠L^{∗g}.

(ii) L is of order p for some prime p and < L^{G} > is a solvable and locally finite p-group. (iii) L has a normal subgroup B such that |L/B| = 2, B ∈ Λ, B is isomorphic to a subgroup

of the multiplicative group of a finite field extension of K, L inverts B and C_{B}(L) = 1.

Proof: Let G_{i}, i ∈ I be an ascending normal series from L to G. For X, Y ≤ G write X ∼ Y if d(< X, Y >) = d(X) = d(Y). Note that

(1) If X ∼ Y and A, B ≤< X, Y > with d(A) = d(B) = d(X), then A ∼ B.

Next we prove that

(2) Let X and Y be in Λ such that X normalizes Y and Y normalizes X. Then X ∼ Y or X ∩ Y = 1.

Suppose X ∩ Y ≠1 and let t be a composition factor for XY on V. If [t, X ∩ Y] = 0, then d = d(X ∩ Y) = d(X) = d(Y) implies that 0 = [t, X ∩ Y] = [t, X] = [t, Y] = [t, XY]. If [t, X ∩ Y]≠0, then since X ∩ Y is normal in XY, t is the direct sum of irreducible X ∩ Y - submodules. It follows that X ∩ Y has no central composition factor on t. Hence d(XY) = d(X ∩ Y) = d and X ∼ Y.

(3) Let X ∈ Λ and Σ ⊆ Λ such that for every S in Σ , S normalizes X and S ∼ X, then d(< X, < Σ >>) = d and S ∼ R for any S and R in Σ.

Put T =< X, < Σ > and let t be a composition-factor for T on V. If X centralizes t, then since d(X) = d(XS) for all S in Σ, S acts unipotently on t. By part (b) of (3.2), T acts unipotently on t. If X does not centralize t, then since X is normal in T, (5.1) implies that X has no central composition factor on t. Hence d = d(X) = d(T) = d(< S, T >) and (3) is proved.

(4) Suppose L is a group of prime order. Then (ii) holds.

Since L is an ascending, locally finite and locally solvable p-subgroup of G, an easy induction argument shows that < L^{G} > is a locally finite and locally solvable p-subgroup of G. Proposition 1 in [4] implies that < L^{G} > is solvable and this completes the proof of (4).

Let 0 be the minimal element in I. Put L_{0} = L and inductively define L_{i}, (i ∈ I) by L_{i} = ∪{L_{i}|j < i}, if i is an limit ordinal and L_{i+1} =< L_{i}^{g}|L_{i} ∼ L_{i}^{g}, g ∈ G_{i+1} >.

(5) If k ∈ I and s ∈ G_{k} with d(L_{k}) = d and L_{k} ∼ L_{k}^{s}, then L_{k} = L_{k}^{s}.

Let r be minimal in I with s ∈ G_{r}. Put J = {j ∈ I|j < k and r ≤ j + 1}. Let j in J and h in G_{j+1} with L_{j} ∼ L_{j}^{h}. Then L_{j}^{h} ≤ L_{j+1} ≤ L_{k}. By assumption d(< L_{k}, L_{k}^{s} >) = dand so by (1) L_{j} ∼ L_{j}^{hs}. Since hs ∈ G_{j+1}, L_{j}^{hs} ≤ L_{j+1} ≤ L_{k}. This holds for all such h and so L_{j+1}^{s} ≤ L_{k}.

If k is not a limit ordinal, k-1 is in J and hence L_{k}^{s} ≤ L_{k}. If k is a limit ordinal then r < k and L_{k} = ∪{L_{j}|j ∈ J} and again L_{k}^{s} ≤ L_{k}. By symmetrie, L_{k}^{s^{−1}} ≤ L_{k} and L_{k} = L_{k}^{s}.

(6) Suppose that neither (ii) nor (iii) hold. Then each of the following holds for all i in I:

(a) For each g in G_{i}, L_{i} = L_{i}^{g} or [L_{i}, L_{i}^{g}] = 1.

(b) If i is not a limit ordinal, L_{i−1} is normal in L_{i}.

(c) L_{i} ∈ Λ.

Assume (6) is false and let i be minimal in I such that (6) fails. Put J = {j ∈ I|j < i}. Suppose that (a) and (b) hold. Then by (a), L_{i} is normal in < L_{i}^{G_{i}} > and so L_{i} is an

ascending subgroup of G. If i is not limit ordinal, then by (b), (2) and the definition of L_{i}, d(L_{i}) = d. Thus (c) hold in this case, a contradiction. Hence i is a limit ordinal. Let t be a non-central composition factor for L_{i} on V. Since d = d(L_{i}) for all j in J and since L_{j} is normal in L_{j+1}, we conclude that \underline{U}_{t}(L_{j}) = \underline{U}_{t}(L_{j+1}). An easy induction argument shows that \underline{U}_{t}(L_{j}) = \underline{U}_{t}(L_{i}) = 0. Since d is finite, t is finite dimensional and so every L_{j} normalizes a non-central irreducible submodule in t. Let m = max { dim X | there exists j in J such that X is a L_{j}- submodule with no central L_{j} -composition factor}. Note that 0 < m ≤ d. Choose j and X as above with dim X = m. Then L_{j} as no central composition factor in < X^{L_{j+1}} > and so L_{j+1} normalizes X and L_{j+1} has no central composition factor in X. By induction, L_{i} normalizes X and so X = t, and L_{j} has no central composition factor in t. This implies that d(L_{i}) = d and L_{i} ∈ Λ.

Hence (a) or (b) fails. Suppose first that i is a limit ordinal. Then (a) must fail. Pick g ∈ G_{i} with L_{i}^{g} ≰ L_{i} and [L_{i}, L_{i}^{g}]≠1.Then there exist j,k,l,m in J such that L_{j}^{g} ≤ L_{i}, [L_{k}, L_{l}^{g}]≠1 and g ∈ G_{m}. Put q =max{j,k,l,m}. Then L_{q}^{g} ≰ L_{q}, [L_{q}, L_{q}^{g}]≠1 and g ∈ G_{q}, a contradiction to the minimal choice of i. Hence i = k +1 for some k in I. Set R = L_{k}. We will first prove that

(*) R is normal in < R^{G_{i}} >.

Suppose not and pick h in G_{i} so that S = R^{h} does not normalize R. Pick s in S with R^{s}≠R. Since s ∈ G_{k}, [R, R^{s}] = 1. By (4) d(RR^{s})≠d. Hence there exists a composition factor t for < R, S > in V such that R has a central and a non-central composition factor on t. Note that R is normal in < R^{G_{k}} > and so also S is normal in < S^{G_{k}} >. In particular, R∩S is subnormal in < R, S >. Suppose that R∩S≠1. Since R, S and R∩Sact completely reducibly on t and d = d(R ∩ S) = d(R) = d(S) we conclude that [t, R] = [t, R ∩ S] = [t, S] = t and so R has no central composition factor on t, a contradiction. We conclude that R ∩ S = 1. Similarly, since R ∩ R^{s} is centralized by RR^{s} and d(RR^{s})≠d, R ∩ R^{s} = 1.

Suppose that R is abelian and let C be the set of factors of some u-composition series of < R^{<s>} > on V. Let 1≠r ∈ R. Then < r > is normal in R and so ascending in G. Thus d(< r >) = d and it follows that \overline{U}_{V} (R) ≤ [V, r] . Hence \overline{U}_{V} (< R, s >) is finite dimensional and so C is finite. Let E be a subset of C and let R_{1} and R_{2} be in R^{<s>} with [e, R_{i}]≠0 and [f, R_{i}] = 0 for all e in E, f ∈ C \ E and i ∈ {1, 2}. Then since R_{i} is normal in R^{<s>} we conclude that [e, R_{i}] = e for all e in E and so d(R_{1}R_{2}) = d. Thus by (5) R_{1} = R_{2}. Since C has only finitely many subsets we conclude that |R^{<s>}| is finite. So s^{k} normalizes R for some positive integer k. Since < R^{S} > and < S^{R} > are abelian, [R, S] ≤ Z(< R, S >) and so [r^{k}, s] = [r, s^{k}] ∈ R. Hence (r^{k})^{s} ∈ R ∩ R^{s} = 1 and R and S are of exponent k. Choosing s such that s^{p} ∈ N_{S}(R) for some prime p we see that R and S are of exponent p. It follows that |R/C_{R}(u)| = p for every non-central composition factor u of R on V. Then C_{R}(u) is an ascending subgroup of G with d(C_{R}(u)) < d, C_{R}(u) = 1 and |R| = p, a contradiction to (4).

Thus R′≠1. Suppose that R normalizes S. By (2.3), R′ ≤ [R, S] and so R′ ≤ R ∩ S = 1, a contradiction. So R does not normalize S and we can apply (2.4). Let A = N_{R}(S), then A is a abelian, R inverts A and C_{A}(R) = 1. Moreover, since d(A) = d and A asc G, A acts faithful on each of its u-composition factors on V. Thus by Schur’s Lemma, A is isomorphic to a subgroup of the multiplicative group of a finite field extension of K. Suppose L ≤ A and pick n in I minimal with L_{n} ≰ A. Then clearly n = j+1 for some j ∈ I. Pick g ∈ G_{r} with L_{j} ∼ L_{j}^{g} and L_{j}^{g} ≰ A. Since L_{j} ≤ A, L_{j}^{g} is abelian and so A∩L_{j}^{g} ≤ C_{A}(R) = 1. Thus |L_{j}^{g}| = 2 and since R inverts T, L_{j} ≤ C_{T} (R) = 1, a contradiction. Thus L ≰ T. Put B = L ∩ A. Then (iii) holds.This contradiction to the assumptions establishes (*).

By (*), L_{k} is normal in L_{i} and so (a) must fail. Let g ∈ G_{i}. If L_{i} ∼ L_{i}^{g}, then by (5) L_{i} = L_{i}^{g}. If L_{i} ≁ L_{i}^{g} then (2) implies [L_{i}, L_{i}^{g}] = 1. This completes the proof of (6).

By (6) either (ii) or (iii) holds or (i) holds with L^{∗} = L_{a}, where a is in I with G = G_{a}.

Remark 6.3 Retain the assumption of (6.2). Replacing L by B in case (iii) it follows that there always exists L in Λ which fulfills (i) or (ii).

7 Ascending Subgroups of Finitary Groups

Throughout this section H is a ascending subgroup of G, G_{i}, i ∈ I be an ascending normal series from L to G,0 is the minimal element in I and a is the maximal element. Let F be the class of finite dimensional KH-modules and put m_{V} (H) = m_{V,F}(H) and M_{V} (H) = M_{V,F}(H), where m_{V,F}(H) and M_{V,F}(H) are defined in (2.2) and (4.8), respectively.

Lemma 7.1 (a) Let U be a KH-submodule in V and M_{U}^{V} (H) the inverse image of M_{V/U}(H) in V. Then G normalizes m_{U}(H) and M_{U}^{V} (H).

(b) If G is irreducible and H reducible on V, then every composition series of H in V is finite dimensional. In particular, d(< h^{H} >) is finite for all h in H.

Proof: Suppose (a) is not true and choose i mimimal in I so that G_{i} does not normalize both of m_{U}(H) and M_{U}^{V} (H). Then clearly i is not a limit ordinal and so G_{i−1} is normal in G_{i} and normalizes m_{U}(H) and M_{U}^{V} (H). Let g ∈ G_{i}. Then m_{U}(H) ∩ m_{U}(H)^{g} is of finite codimension in m_{U}(H). Moreover, G_{i−1} and therefore also H normalizes m_{U}(H) ∩ m_{U}(H)^{g}. It follows immediately from the definition of m_{U}(H) that m_{U}(H) has no proper H-submodule of finite codimension. Hence m_{U}(H) ∩ m_{U}(H)^{g} = m_{U}(H) = m_{U}(H)^{g}. Thus G_{i} normalizes m_{U}(H). Similarly, M_{U}^{V} (H)^{g} + M_{U}^{V} (H)/M_{U}^{V} (H) is a finite dimensional KH-submodule of V/M_{U}^{V} (H) and G_{i} normalizes M_{U}^{V} (H), a contradiction which establishes (a).

(b) Let U be a proper H-submodulein V. Since G acts irreducibly on V, (a) implies that M_{U}^{V} (H) = V and m_{U}(H) = 0. Thus H has finite dimensional composition series on U and on V/U and therefore also on V. By (4.5) every composition series for H on V is finite dimensional. Note that < h^{H} > centralizes all but finitely many of those composition factors and (b) is verified.

Note that part (b) of (7.1) implies:

Corollary 7.2 Suppose G is irreducible on V and has a reducible ascending subgroup. Then d_{G} is finite.

Definition 7.3 1. A set Π of proper K-subspaces in V such that V = ⊕Π and P^{g} ∈ Π, for all P ∈ Π, g ∈ G , is called a system of imprimitivity for G on V. A K-subspace P of V is called an subspace of imprimitivity for G on V if P^{G} is a system of imprimitivity.

2. H ≤ GL_{K}(V) is called imprimitive, if there exists a system of imprimitivity for G on V. Otherwise G is called primitive.

Proposition 7.4 If G is primitive and V infinite dimensional, then every non-trivial ascendent subgroup of G acts irreducibly and primitively on V.

Proof: Suppose that H is reducible. Using (5.1) the Wedderburn components of a nontrivial reducible normal subgroup of G form a system of imprimitivity. So G has no nontrivial reducible normal subgroup. In particular, G has no non-trivial abelian or solvable normal subgroup. By (7.2) d_{G} is finite and this permits us to apply (6.2). Note that case (ii) cannot occur. In case (iii) we replace L by B. Thus (i) holds for some L ∈ Λ. Without loss L^{∗} = L . Then L is subnormal in G and so by (5.1) V is completely reducible for L. Since d(L) = d_{G} is finite we conclude that [V,L] is finite dimensional. Therefore < L^{G} > is reducible, a contradiction.

So H is irreducible. Suppose that Π is a system of imprimitivity for H on V. By the previous paragraph, H has no reducible normal subgroup and so H acts faithfully on Π as a transitive group of finitary transformations. Suppose that H has no maximal system of imprimitivity on Π. Then any element in H acts trivally on some system of imprimitivity, a contradiction. Thus we may assume that H acts primitively on Π. By Satz 9.4 in [9], H = Alt(Π) or F Sym(Π). Let P ∈ Π. Suppose that [P, h] ≠ 0. for some h ∈ N_{H}(P). Then |P^{C_{H}(h)}| is infinite and so [V,h] is infinite dimensional, a contradiction. Hence [P, N_{H}(P)] = 0 and V is the natural permutation module for H, but then the even permutation module is a proper H-submodule in V, a contradiction.

Proposition 7.5 (a) If Π is a infinite system of imprimitvity for G on V, and H is an ascending subgroup of G acting transitively on Π, then G′ ≤ H.

(b) Suppose that G is irreducible and imprimitive on V and V that is infinite dimensional; then G′ is the unique mimimal irreducible ascending subgroup of G.

Proof: (a) Suppose G′ ≰ H and pick j in I minimal with G′_{j} ≰ H. Then clearly j is not a limit ordinal and so j = t+1 for some t in I. Let L = G′_{t}. Then L is a normal subgroup of G_{j} and L ≤ H. Moreover, since G_{k} is transitive on Π, and Π is infinite, L is transitive on Π. Let x, y ∈ G_{j} and A =< x, y >. Since [V,A] is finite dimensional, there exists a finite subset Σ of Π with [P, A] = 0 for all P ∈ Π \ Σ. By (2.3) in [6], there exists l in L with Σ ∩ Σ^{l} = Ø. Hence [A, A^{l}] = 1. Thus by (2.3) A′ ≤ [A, l] ≤ [G_{j}, L] ≤ L and so G′_{j} ≤ L ≤ H, a contradiction.

(b) By (a) it remains show that G′ acts irreducibly on V. Otherwise the abelian group G/G′ acts transitively and finitary on the infinite set of Wedderburn components of G′ in V. This is impossible.

Theorem 7.6 If G is irreducible on V, then every ascending subgroup of G acts completely reducibly on V.

Proof: Suppose H is not completely reducible on V. Let i in I be minimal such that G_{i} is completely reducible. If i = k+1 for some k in I, then G_{k} is normal in G and by (5.1), G_{k} is completely reducible, a contradiction. So G = ∪{G_{k}|k < i}. Suppose first that V is finite dimensional. For k in I let A_{k} be the K-subalgebra of End_{K}(V) generated by the elements of G_{k}. Under the elements less than i, choose k such that dim_{K}A_{k} is maximal. Then A_{k} = A_{j} for all k ≤ j < i. Thus A_{i} = ∪{A_{j}|j < i} = A_{k}. It follows that G_{k} and G_{i} have the same submodules in V, a contradiction to the fact that G_{i} is completely reducible on V and G_{k} is not.

Suppose next that V is infinite dimensional. Since H is reducible, (7.4) implies that G is imprimitive. Let Π be a system of imprimitivity for G om V.

Assume first that H does not act transitively on Π. Then by (2.6), V is the direct sum of H-invariant subspaces of imprimitivity for G on V. Let U be any H-invariant subspaces of imprimitivity for G on V. Then U is finite dimensional, N_{G}(U) acts irreducibly on U and H is an ascending subgroup of N_{G}(U). By the finite dimensional case, H is completely reducible on U. Since this holds for all such U, H is completely reducible on V.

Assume next that H is transitive on Π. Then by 7.5, G′ ≤ H, and H acts irreducibly on V, a contradiction, which completes the proof of (7.6).

We are now able to prove an improved version of (5.2)

Lemma 7.7 Let h be in G, then < h^{<h^{G}>>} has a finite u-composition series on V.

Proof: Retain the notation established in the proof of (5.2). By that proof it is enough to show that n(t) ≤ 1 for all t in T. Suppose first that G acts primitively on t. Then by (7.4), n(t) = 0. Suppose next that G has a system of imprimitivity Π on t and let D be the subgroup of G which is maximal with respect to acting trivial on Π. Now assume that h is in D. Then H_{1} ≤ D and n(t) = 1. So we may assume that h is not in D and that G acts primitively on Π. But then by Satz 9.2 im [9], G/D = Alt(Π) or FSym(Π) and so G′D ≤ H_{1}D. It follows by induction that G′D ≤ H_{i}D for all i . Thus by (7.5), G′ ≤ H_{i}C_{G}(t) and H_{i} acts irreducibly on t. Thus n(t) =0, and (7.7) is proved.

8 Free Groups as Infinite Dimensional Finitary Groups

Example 8.1 Let I be a set with at least two elements, F a free group with generators f_{i}, i in I, and M a free abelian group with generators v_{i}, i in I. For i in I let z_{i} be an integer with |z_{i}| ≤ 6 . Then the representation of F on M defined by

v_{i}^{f_{i}} = v_{i} and v_{j}^{f_{i}} = v_{j} + z_{i}v_{i}

for all i,j in I with i ≠ j, is faithful. Moreover, if K is a field with z_{i}≠0 (in K) for all i in I, then K ⊗M is an irreducible KF-module. If char K = 0, then K⊗ M is a faithful F -module.

Proof: For i in I let M(i) = {∑ a_{k}z_{k} ∈ M||a_{i}| > 2(∑_{k≠i} |a_{k}|)}. We will first prove

(*) Let i,j in I with i ≠ j and t a nonzero integer. Then M(j)^{(f_{i}^{t})} ⊆ M(i).

Let m = ∑ a_{k}v_{k} ∈ M(j) and m^{(f_{i}^{t})} = ∑ b_{k}v_{k}. Since

v_{i}^{(f_{i}^{t})} = v_{i} and v_{j}^{(f_{i}^{t})} = v_{j} + tz_{i}v_{i}

we conclude that b_{k} = a_{k} for k ≠ i and b_{i} = tz_{i}(∑_{k≠i} a_{k}) + a_{i}. Thus

∑_{k≠i} |b_{k}| = ∑_{k≠i} |a_{k}| = ∑_{k≠i,j} |a_{k}| + |a_{j}| < |a_{j}|/2 + |a_{j}| = 3/2(|a_{j}|).

On the other hand,

|b_{i}| = |tz_{i}(∑_{ki} a_{k}) + a_{i}| ≤ |tz_{i}||a_{j}| − |tz_{i}|( ∑_{k≠i,j} |a_{k}|) − |a_{k}| ≤ |tz_{i}||a_{j}| − |tz_{i}|(∑_{k≠i} |a_{k}|) =

= |tz_{i}|(|a_{j}| − ∑_{k≠i} |a_{k}|) ≤ 6(|a_{j}|/2) = 3|a_{j}|.

Hence |b_{i}| > 2(∑_{k≠i} |b_{k}|), m^{(f_{i}^{t})} ∈ M(i) and (*) is established.

Let f = f_{i(1)}^{t(1)} · . . . · f_{i(s)}^{t(s)} ∈ F with i(r) ≠ i(r+1) and t(r) ≠ 0. Let j ∈ I with j≠i(1). We

claim that M(j)^{f} ⊆ M(i(s)). Indeed, put x = f_{i(s)}^{t(s)} and h = fx^{−1}. Then by induction on s, M(j)^{h} ⊆ M(i(s − 1)) and so by (*),

M(j)^{f} = (M(j)^{h})^{x} ⊆ M(i(s − 1))^{x} ⊆ M(t(s))

.

We are now able to prove that f acts non-trivially on M. If i(1) = i(s) pick j ∈ I \ i(1) and if i(1) ≠ i(s) let j = i(1). Put y = f_{i(1)}^{t(1)}. Note that v_{j} ∈ M(j) and so in any case v_{j}^{y} ∈ M(i(1)). Since i(1) ≠ i(2) we conclude that v_{j}^{f} ∈ M(t(s)). Since j ≠ t(s), v_{j} ∉ M(t(s)). Thus v_{j}≠v_{j}^{f} and F acts faithfully on M.

Let K be a field with z_{i}≠0 for all i in I. Let V = K ⊗ M and U a nonzero KF-submodule in V. Clearly C_{V} (F) = 0. Hence [U, f_{i}]≠0 for some i in I and so v_{i} ∈ [U, f_{i}] ≤ U. It follows that for j ≠ i, v_{j} ∈ [Kv_{i}, f_{j}] ≤ U. Thus U = V and V is irreducible as a KF-module.

Remark 8.2 (a) The first assertion of (8.1) is also an easy consequence of the main theorem in [3]. Indeed Humphrey proves a much stronger version of (8.1) for finite I’s. (8.1), F has an irreducible, infinite dimensional, and finitary representation over K. It is easy to see that any such representation must be primitive. It follows from (7.4) that every non-trivial ascending subgroup of F acts primitively. In particular, the intersection of the normal and irreducible subgroups of F is trivial and (7.5) does not hold in general for a finitary primitive groups.

References

[1] J.I. Hall, Infinite Alternating Groups as Finitary Linear Transformation Groups, J. Alg. 119 (1988), pp. 337-359.

[2] J.I. Hall: Periodic Classical Groups of Finitary Linear Transformations, in preparation.

[3] S.P. Humphries: Free subgroups of SL(n,Z) generated by transvections, J. Alg. 116(1988)155-162.

[4] U. Meierfrankenfeld, R.E. Phillips and O. Puglisi, Locally Solvable Finitary Linear Groups, Preprint(1991).

[5] P. Neumann: The lawlessness of finitary permutation groups, Arch. Math. 26(1975) 561-566.

[6] P. Neumann: The structure of finitary permutation groups, Arch. Math. 27(1976) 3-17.

[7] R.E. Phillips: The Structure of Finitary Linear Transformation, J. Alg 119 (1988) 400-448.

[8] R.E. Phillips, On Absolutely Simple Locally Finite Groups, Rend. Sem. Mat. Univ. Padova, Vol. 79 (1988), pp. 213-220.

[9] H. Wieland: Unendliche Permutationsgruppen, Lecture notes, Math. Inst. Unversitt Tbingen (1960).

Department of Mathematics Michigan State University East Lansing Michigan 48824