Perfect Frobenius Complements

U. Meierfrankenfeld

May 17, 2002

Abstract

Let H be a finite Frobenius group with a perfect Frobenius complement G. Two new proofs that G is isomorphic to SL_{2}(5) are given.

1 Introduction

A Frobenius group is a transitive but not regular permutation group on a set Ω such that every non-trivial element has at most one fixed-point. Let H be a finite Frobenius group with kernel N and complement G. That is, N consists of the identity and all the fixed-point free elements; and G is the stabilizer of some element in Ω. In [Fr, V] Frobenius proved that N is a normal subgroup of H. In [Za1, Satz 16] ( and in revised form in [Za2]) Zassenhaus showed that if G is perfect then G ≅ SL_{2}(5). Both of these proofs involve character theory. A further proof of Zassenhaus’ theorem based on the elementary theory of exceptional characters can be found in [Be]. In this note we will give two new proofs of Zassenhaus’ theorem without using character theory (except that we assume Frobenius’ theorem).

Theorem A Let H be a finite Frobenius group with complement G. If G is perfect, then G ≅ SL_{2}(5).

The first proof is based purely on standard group theory text book material. The second proof is slightly shorter but relies on a couple of simple facts about modular representations of finite groups ( see the end of the introduction for the details).

Note that the action of G on Ω and the action of G on N by conjugation are isomorphic. Hence C_{N}(g) = 1 for all g ∈ G^{#}. Let p be a prime dividing the order of N and put V = Ω_{1}(Z(O_{p}(N)). By [Th], N is nilpotent and so V ≠ 1. ( For perfect G a more elementary argument is possible: Let q the smallest prime dividing the order of G and S a Sylow q-subgroup of G. By [Go, 10.3.10], S is cylic or q = 2 and S is generalized quaternion. As G is perfect, Burnside’s p-complement theorem [Go, 7.4.5] implies that S is not cyclic. Thus q = 2 and G contains an involution t. But then t inverts N and so N is abelian and again V ≠ 1.) Now V is a GF (p)G-module and all non-trivial elements of G act fixed-point freely on V. Hence Theorem A follows at once from ( and is in fact equivalent to) Theorem B, which is also a theorem of Zassenhaus:

Theorem B Let G be a non-trivial, finite perfect group, K a field and V a faithful KG-module so that all non-trivial elements of G act fixed-point freely on V. Then G ≅ SL_{2}(5).

In section 3 we establish some basic facts about G which will be used in both of our proofs of Theorem B. Section 4 contains the first proof of Theorem B, while in section 5 we prove:

Theorem C Let G be a non-trivial finite perfect group with cyclic or dihedral Sylow 2-subgroups, K a field of characteristic 2 and W a faithful KG-module so that all non-trivial elements of odd order of G act fixed-point freely on W. Then G ≅ Alt(5).

We finish the introduction by showing how Theorem C implies Theorem B, and thus obtain our second proof of Theorem B.

The starting point are the following simple facts from the theory of modular representations of finite groups ( which can be extracted for example from [CR]):

1. Let p be a prime and G a finite group so that p does not divide the order of G. Then every GF (p)G module is isomorphic to the reduction modulo p of some module for G in characteristic 0.

2. Let G be a finite group, p a prime, V a G-module in characteristic 0 and W a reduction of V modulo p. Then a p′-element in G acts fixed-point freely on V if and only if it acts fixed-point freely on W.

Let G and V be as in Theorem B. Without loss K is a ground field. Since |G| is co-prime to the characteristic of K, V is the reduction of some G-module X in characteristic zero. Then all non-trivial elements of G act fixed-point freely on X. Let W be a reduction of X modulo 2. Then all non-trivial 2′-elements of G still act fixed-point freely on W.

Assume t is an involution in G and let v ∈ V. Then v + v^{t} is fixed by t, v^{t} = −v and so t is the unique involution in G. Hence t ∈ Z(G). Moreover, w = w^{t} for all w ∈ W. By [Go, 10.3.1] the Sylow 2-subgroups of S are cyclic or quaternion and so the Sylow 2-subgroups of G/⟨t⟩ are cyclic or dihedral. By Lemma 3.6 below, O_{2}(G) = ⟨t⟩ and so G/⟨t⟩ acts faithfully on W.

Hence we can apply Theorem C, to G if |G| is odd, and to G/⟨t⟩ if |G| is even. We conclude that |G| is even and G/⟨t⟩ ≅ Alt(5). Thus by [Hu, V25.7], G ≅ SL_{2}(5).

Acknowledgement: I would like to thank the Mathematisches Seminar der Universität Kiel for its hospitality. I also would like to thank B. Doerr, P. Hewitt, H. Schnabel, B. Stellmacher and most of all H. Bender for useful comments on an earlier draft of this paper.

2 SL_{2}(3)

In this section we proof a lemma on SL_{2}(3) with fixed-point freely acting elements of order three. This lemma is at the heart of both of our proofs of theorem B.

Proposition 2.1 Let H ≅ SL_{2}(3), A = O_{2}(H), K a field and V a KHmodule. Let d be an element of order three in H and a ∈ A \ Z(H) so that also ad has order three. Let b = a^{d^{2}} and 1 ≠ z ∈ Z(H). Suppose that

(i) The elements of order three in H act fixed-point freely on V.

(ii) vz = −v for all v ∈ V.

Then

(a) The following relations hold in the ring End_{K}(V):

(1 + a)d = b − 1 and − 2d = (a − 1)(b − 1).

(b) If charK = 2, then C_{V} (a) = C_{V} (b) = C_{V} (A).

(c) If charK ≠ 2, then any A-invariant subspace of V is also H-invariant.

Proof: Since d has order three, (1 + d + d^{2})(d − 1) = 0. As d is fixed-point free we conclude

(1)  1 + d + d^{2} = 0.

As ad also as order three, 1+ad+adad = 0. Multiplying this equation with d from the right we get d+ad^{2}+adad^{2} = 0. Hence by (1), d−a−ad+adad^{2} = 0. Furthermore, dad^{2} = a^{d^{2}} = b and so d − a − ad + ab = 0. Thus d − ad = a − ab and (1 − a)d = a(1 − b). As a^{2} = z = −1 we can multiply the last equation with a from the left to obtain

(2)  (a + 1)d = b − 1.

Since (a − 1)(a + 1) = a^{2} − 1 = −2 we get

(3)  −2d = (a − 1)(b − 1).

In particular, (a) holds. Suppose now that charK = 2 and let v ∈ C_{V} (a). Then v = va, v + va = 0 and so v(1 + a)d = 0. Hence by (2) v(b − 1) = 0 and v ∈ C_{V} (b). As A = ⟨a, b⟩, (b) holds.

So suppose that charK ≠ 2. Then by (3), d = −\frac{1}{2}(a−1)(b−1) and so every subspace invariant under A is also invariant under H = A⟨d⟩.

3 Alt(4)

In this section we assume that G is a non-trivial finite group with the following three properties:

(i) G is perfect.

(ii) The Sylow 2-subgroups of G are dihedral or cylic.

(iii) Every subgroup of G of order pq, p and q odd primes, is cyclic.

The main result of this section is Propostion 3.9 which establishes a subgroup isomorphic to Alt(4) in G.

Let p be an odd prime dividing the order of G, T a Sylow p-subgroup and S a Sylow 2-subgroup of G.

Lemma 3.1 All p-subgroups of G are cylic.

Proof: By (iii) applied to the case p = q, all abelian p-subgroups of G are cyclic. Hence the lemma follows from [Go, 5.4.10i].

The following observations will be useful later on.

Lemma 3.2 (a) If A is a p′-group acting on a cyclic p-group B, then either [B, A] = 1 or C_{B}(A) = 1.

(b) If α is an automorphism of order 2 of the cyclic p-group B, then α inverts B.

Lemma 3.3 N_{G}(T)/C_{G}(T) is a 2-group and each p-subgroup of G is inverted by some element in G.

Proof: By induction on p. Suppose first that q is an odd prime dividing the order of N_{G}(T)/C_{G}(T). Then q ≠ p and as T is cyclic, q divides p − 1 and so q < p. Let R be a Sylow q-subgroup of N_{G}(T) and E = C_{R}(T). If E = 1 then Ω_{1}(R)Ω_{1}(T) is not cylic, a contradiction to (iii). Thus E ≠ 1. By induction there exists y ∈ N_{G}(E) which inverts E. Note that T is a Sylow p -subgroup of C_{G}(E). Thus by the Frattini argument [Go, 1.3.7] we may assume that y normalizes T. Now R is a Sylow q-subgroup of N_{G}(T) ∩ N_{G}(E) and so by another application of the Frattini argument we may assume that y also normalizes R. Since y does not centralize E, it does not centralize R. Thus by 3.2 y inverts R and so R = [R, y]. As the autmorphism group of T is abelian we conclude R = [R, y] ≤ C_{G}(T), a contradiction.

Thus N_{G}(T)/C_{G}(T) is a 2-group. By Burnside’s p-complement theorem [Go, 7.4.5], N_{G}(T) ≠ C_{G}(T). Hence T is inverted by some element of G and as any p-subgroup is conjugate to a subgroup of T, 3.3 is proved.

Lemma 3.4 C_{G}(S) ≤ S.

Proof: Suppose C_{G}(S) contains an element x of order p. Then S is a Sylow 2-subgroup of N_{G}(⟨x⟩) and centralizes x. But this contradicts 3.3.

Lemma 3.5 All involutions in G are conjugate.

Proof: Since S is dihedral or cylic, S has a cyclic subgroup of index two. Since G has no subgroup of index two, Thompson transfer [Su, 5.1.8] implies that all the involutions in G are conjugate to the unique involution in this cyclic subgroup.

Lemma 3.6 S is dihedral of order at least four and Z(G) = O_{2}(G) = 1.

Proof: By 3.3, G has even order and no element of odd order is in the center of G. Thus Z(G) ≤ Z(S). By Burnside’s p-complement theorem, S is not cylic and so is S is dihedral of order at least four. Hence G has more then one involution and so by 3.5 S ∩ Z(G) = 1 and so Z(G) = 1. Since O_{2}(G) is cyclic or dihedral, Aut(O_{2}(G)) is solvable and as G is perfect, O_{2}(G) ≤ Z(G).

Lemma 3.7 Let A be a fours group in S. Then 3 divides |N_{G}(A)/C_{G}(A)|.

Proof: It suffices to show that N_{G}(A) acts transitively on A^{#}. If A = S this follows from 3.5 and a theorem of Burnside [Go, 7.1.1]. So suppose A ≠ S and let a, b be any two distinct involutions in A. Let c be the third involution in A. By 3.5, c ∈ Z(S)^{g} for some g ∈ G. Then S^{g} is Sylow 2-subgroup of C_{G}(c) and so we may assume that A ≤ S^{g}. As S^{g} is dihedral, A = C_{S^{g}}(A) < N_{S^{g}}(A). Since N_{S^{g}}(A) fixes c it must permute a and b.

Let F ≤ Z(S) with |F | = 2.

Lemma 3.8 C_{G}(F) = O(C_{G}(F))S.

Proof: Put R = O^{2}(C_{G}(F)). If R has odd order we are done. So suppose that R has even order. Since R has no subgroup of index two we get as in 3.5 and 3.6 that all involutions in R are conjugate and R ∩ S is dihedral of order at least four. But then F ≤ R ∩ S and we get a contradiction as F is normal in R.

Proposition 3.9 G has a subgroup isomorphic to Alt(4).

Proof: By [Go, 6.2.2i] there exists an S-invariant Sylow 3-subgroup L of O(C_{G}(F)). Let A be a fours group in S. We consider the cases that C_{G}(A) is a 3′-group and that 3 divides |C_{G}(A)| seperately.

3.9.1 If C_{G}(A) is a 3′-group, then A is contained in a subgroup of G isomorphic to Alt(4).

Indeed, let D be a Sylow 3-subgroup of N_{G}(A). By assumption C_{D}(A) = 1 and by 3.7, D does not centralizes A. Thus D ≅ C_{3} and DA ≅ Alt(4).

3.9.2 If C_{G}(A) is not a 3′-group, then 1 ≠ L is a Sylow 3-subgroup of C_{G}(A), S ≠ A and if B is a fours group in S not conjugate to A in S, then B inverts L.

Indeed by 3.4 we first conclude that S ≠ A. Let L^{∗} ∈ Syl_{3}(C_{G}(A)). As S is dihedral, F ≤ A and so L^{∗} ≤ C_{G}(F). By 3.8 L^{∗} ≤ O(C_{G}(F)). Since L^{∗} is A-invariant we conclude from [Go, 6.2.2ii,iii] that some conjugate of L^{∗} under C_{G}(A) is contained in L. Hence we may assume without loss that L^{∗} ≤ L. Thus by 3.2, A centralizes L and so L = L^{∗}. Hence ⟨A^{S}⟩ centralizes L. Note that S is a Sylow 2-subgroup of N_{G}(L) and so by 3.3 S inverts L. As S is dihedral, S = ⟨A^{S}⟩B and so B inverts L.

We are now able to prove 3.9. In case 3.9.1 we are done. So assume 3.9.2 holds. Then B does not centralize L. If C_{G}(B) is not a 3′-group, then 3.9.2 applied to B gives the contradiction L ≤ C_{G}(B). Thus C_{G}(B) is a 3′-group and by 3.9.1 B is contained in a subgroup isomorphic to Alt(4).

The next lemma is well known. For completeness we provide a simple ( and also well known) counting argument.

Lemma 3.10 If the centralizer of some involution in G has order four, then G ≅ Alt(5).

Proof: Recall that by 3.5 G has a unique conjugacy class of involutions. Moreover, C_{G}(F) = S and all elements in G have order either odd or two.

3.10.1 Let M and M^{∗} be a maximal abelian subgroup of G of odd order with M ≠ M^{∗}. Then |N_{G}(M)/M| = 2 and M ∩ M^{∗} = 1.

Let b be an element of prime order in M and C = C_{G}(b). Then C has odd order and by 3.4 there exists an involution z in G which inverts b. Then C_{C}(z) = 1 and so C is abelian and C = M. In particular, b ∉ M^{∗} and M ∩ M^{∗} = 1. As any involution normalizing M has to invert M, M can not be normalized by a fours group. Thus N_{G}(M) ∩ C_{G}(z) = ⟨z⟩ and by a Frattini argument applied to M⟨z⟩ ⊴ N_{G}(M),

N_{G}(M) = M(N_{G}(M) ∩ C_{G}(z)) = M⟨z⟩.

Thus 3.10.1 holds.

Let M_{1}, M_{2}, …, M_{k} be representatives for the conjugacy classes of maximal abelian subgroups of odd order in G, n = |G| and m_{i} = |M_{i}|. By 3.10.1 each non-trivial element of odd order in G lies in exactly one conjugate of the M_{i}’s. Moreover, there are \frac{n}{4} involutions and so

n = 1 + \frac{n}{4} + ∑_{i=1}^{k} \frac{n}{2m_{i}} · (m_{i} − 1).

Multipliying by \frac{2}{n} we obtain

\frac{3}{2} > ∑_{i=1}^{k} \frac{m_{i} − 1}{m_{i}}.

Since \frac{2}{3} + \frac{6}{7} = \frac{32}{21} > \frac{3}{2}, we conclude k = 2, m_{1} = 3, and m_{2} = 5. Hence n = 60. In particular, the subgroup of G isomorphic to Alt(4) has index five in G and so G ≅ Alt(5).

4 The first proof of Theorem B

Let G and V be as in Theorem B. Morever, we assume without loss that K is algebraicly closed. Let S be a Sylow 2-subgroup of G. By [Go, 10.3.1] we have

Lemma 4.1 (a) S is cylic or generalized quaternion.

(b) Every subgroup of G of order pq, p and q primes, is cyclic.

If G has odd order then G fullfils the assumptions but not the conclusion of section 3. Thus G contains an involution t. Then t inverts V, t is unique and t ∈ Z(G). Put Ḡ = G/⟨t⟩. Then S̄ is cyclic or dihedral and so we can apply the results of section 3 to Ḡ. In particular, there exists H ≤ G with \overline{H} ≅ Alt(4). Let A = O_{2}(H) and D ∈ Syl_{3}(H). Then A ≅ Q_{8} and H ≅ SL_{2}(3). Without loss A ≤ S. Let F̄ be a subgroup of order two of Z(S̄). Then F ≅ C_{4} and F ≤ A. By 2.1c we have

Lemma 4.2 All A-invariant subspaces of V are also invariant under H.

Lemma 4.3 Let H ≤ R ≤ G so that R normalizes a 2-dimensional subspace of V. Put E = ⟨H^{R}⟩. Then E = H or E ≅ SL_{2}(5). Moreover, C_{R}(E) = Z(R) = O(Z(R))Z(H) and R/Z(R) ≅ Alt(4), Sym(4) or Alt(5).

Proof: Let W be a 2-dimensional subspace of V normalized by R. By the fixed-point free action R acts faithfully on W and we may view R as a subgroup of GL_{K}(W). Let M be a maximal abelian subgroup of R. As K is algebraicly closed, W is the direct sum of two 1-dimensional M-submodules. As M is maximal, M ≰ Z(R) and so these submodules are non-isomorphic and uniquely determined by each m ∈ M \Z(R). Hence M ∩M^{∗} = Z(R) for any two distinct maximal abelian subgroups M and M^{∗} of R. Moreover |N_{R}(M)/M| ≤ 2. Let M_{1}, M_{2}, … M_{k} representatives for the classes of maximal abelian subgroups of R, m_{i} = |M_{i}/Z(R)|, n = |R/Z(R)| and ϵ_{i} = |N_{N_{R}(M_{i})}/M_{i}|. Then

(1)  n = 1 + ∑_{i=1}^{k} \frac{n}{ϵ_{i}m_{i}} (m_{i} − 1).

If k = 1 we conclude that R = M_{1} is abelian, a contradiction to H ≤ R. Hence we may assume from now on that k ≥ 2. By (1)

(2)  1 = \frac{1}{n} + ∑_{i=1}^{k} \frac{m_{i} − 1}{ϵ_{i}m_{i}}.

Since \frac{m_{i}−1}{ϵ_{i}m_{i}} ≥ \frac{1}{4} we get k ≤ 3.

Suppose first that ϵ_{1} = 1. Then \frac{m_{1}−1}{m_{1}} < 1 − \frac{1}{4}(k − 1) and so k = 2 and m_{1} ≤ 3. If m_{1} = 2 we compute from (2) that n = 2m_{2} and so M_{2} is of index two in R. Then as H has no subgroup of index two, H ≤ M_{2}, a contradiction. If m_{1} = 3 we get n = \frac{6m_{2}}{3−m_{2}} . Thus m_{2} = 2, n = 12 and R = HZ(R).

Suppose next that ϵ_{i} = 2 for all i. Then by (1), k > 2. Thus k = 3 and so by (2) \frac{1}{m_{1}} + \frac{1}{m_{2}} + \frac{1}{m_{3}} = \frac{2}{n} + 1. In particular, at least one of the m′_{i}s has to be 2. Say m_{1} = 2. Then \frac{1}{m_{2}} + \frac{1}{m_{3}} = \frac{2}{n} + \frac{1}{2}. Then at least one of m_{2} and m_{3} has to be at most 3. Say m_{2} ≤ 3 and m_{2} ≤ m_{3}.

If m_{2} = 2, then n = 2m_{3}, a contradiction as above.

If m_{2} = 3, then n = \frac{12m_{3}}{6−m_{3}} . Thus m_{3} is 3, 4, or 5 and n = 12, 24 or 60 respectively. Hence HZ(R) has index 1, 2 or 5 in R. As m_{2} = 3 and e_{2} = 2 elements of order three in H are inverted by some element in R. So the case of index 1 is impossible while in the remaining two cases it is easy to see that R/Z(R) ≅ Sym(4) and Alt(5), respectively.

Furthermore, as S is generalized quaternion, O_{2}(Z(R)) = Z(H) and Z(R) = O(Z(R))Z(H).

Lemma 4.4 A ≤ O_{2}(N_{G}(F)). In particular, O^{2}(N_{G}(F)) ≤ C_{G}(A).

Proof: Let g ∈ N_{G}(F), a ∈ A \ F, E = ⟨aa^{g}⟩ and D = ⟨a, a^{g}⟩. Then D̄ is dihedral, and EF as index at most 2 in DF. Since E centralizes F, EF is abelian. Since K is algebraically closed, EF normalizes a 1-dimensional subspace in V. Hence DF normalizes a 2-dimensional subspace W in V. Since A = ⟨a⟩F ≤ DF, we conclude from 4.2 that also H normalizes W. Let R = ⟨D, H⟩ = ⟨a^{g}, H⟩ and E = ⟨H^{R}⟩. Then R = ⟨a^{g}⟩E and we conclude from 4.3 that R̄ ≅ Alt(4), Sym(4) or Alt(5). Hence DF ≤ N_{R}(F) ≅ Q_{8} or Q_{16}. In particular, A and A^{g} commute modulo F. Thus ⟨A^{N_{G}(F)}⟩ is a 2-group and so A ≤ Q := O_{2}(N_{G}(F)). Clearly each element of odd order in N_{G}(F) centralizes F and, as Q is quaternion, also Q.

Lemma 4.5 S = A.

Suppose S ≠ A and let B be a quaternion group of order eight in S not conjugate to A in S.

Suppose that B is contained in a subgroup H^{∗} ≅ SL_{2}(3). Put R = ⟨H, H^{∗}, S⟩. As S has a cyclic subgroup of index two, there exists a 2-dimensional KS-submodule W in V. Then by 4.2 applied to H and H^{∗}, R normalizes W. As |S| ≥ 2^{4} we conclude from 4.3 that R/Z(R) ≅ Sym(4). But then A = O_{2}(H) = O_{2}(R) = O_{2}(H^{∗}) = B, a contradiction.

Thus B is not contained in an SL_{2}(3). From 3.9.1 applied to B in place of A, we conclude that 3 divides |C_{G}(B)|. As F ≤ B, 3 divides |C_{G}(F)| and so by 4.4, 3 also divides |C_{G}(A)|. As the Sylow 3-subgroups of N_{G}(A) are cyclic this implies that all elements of order three in N_{G}(A) are already in C_{G}(A), a contradiction to H ≤ N_{G}(A).

We are now able to complete our first proof of Theorem B. Since A = S, 3.4 implies that CḠ(Ā) = Ā is a 2-group. Hence by 4.4 also CḠ(F̄) is a 2-group and so CḠ(F̄) = Ā. Thus by 3.10, Ḡ ≅ Alt(5) and by [Hu, V25.7], G ≅ SL_{2}(5).

5 Theorem C

Let G and W be as in Theorem C. As in [Go, 10.3.1] we have that subgroups of order pq, p and q odd primes, are cyclic. Thus we can apply the results of section 3. In particular by 3.9 there exists H ≤ G with H ≅ Alt(4). Put A = O_{2}(H) and let S be a Sylow 2-subgroup of G containing A. Let 1 ≠ a ∈ Z(S) ≤ A.

Lemma 5.1 A = C_{G}(a)∩C_{G}(C_{W} (a)) and in particular, A is normal in C_{G}(a).

Let B = C_{G}(a) ∩ C_{G}(C_{W} (a)). By definition, B centralizes C_{W} (a). Since [W, a] ≤ C_{W} (a) and [W, a] is isomorphic to W/C_{W} (a) as C_{G}(a)-module, B also centralizes W/C_{W} (a). It follows that [W, B, B] = 0. Hence B is elementary abelian. By 2.1b, A ≤ B. Since S is a dihedral group, S has no elementary abelian subgroup of order larger than four, and so B = A.

Lemma 5.2 S = A

Suppose that S ≠ A and let B be a fours group in S distinct fom A. If B is not contained in an Alt(4) then by 3.9.2 (with the roles of A and B interchanged), A inverts an element of order three in C_{G}(B) ≤ C_{G}(a), a contradiction since by 5.1, A is normal in C_{G}(a). Thus B is contained in an Alt(4) and hence 2.1b (applied to B in place of A) yields B ≤ C_{G}(a) ∩ C_{G}(C_{W} (a)). Thus by 5.1 B ≤ A, a contradiction.

Lemma 5.3 C_{G}(a) = S.

By 5.2 S = A is a fours group. By 5.1, C_{G}(a) normalizes A and so stabilizes the series 1 ≤ ⟨a⟩ ≤ A. Thus O^{2}(C_{G}(a)) centralizes A and so by 3.4 C_{G}(a) is a 2-group. Thus C_{G}(a) = S.

Theorem C now follows from 3.10.

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Ulrich Meierfrankenfeld

Department of Mathematics

Michigan State University

East Lansing, Mi 48824

meier@math.msu.edu

http://www.math.msu.edu/^{∼}meier/