MATEMATIQKI VESNIK

58 (2006), 31–37

UDK 512.552

originalni nauqni rad research paper

FINITE GROUPS ADMITTING SOME COPRIME OPERATOR GROUPS

Enrico Jabara

Abstract. Let G be a finite group, with a finite operator group A, satisfying the following conditions: (1) (|G|, |A|) = 1; (2) there exists a natural number m such that for any α, β ∈ A^{♯} we have: [ C_{G}(α), \underbrace{C_{G}(β), …, C_{G}(β)}_{m}] = \{1\}; (3) A is not cyclic. We prove the following: (1) If

the exponent n of A is square-free, then G is nilpotent and its class is bounded by a function depending only on m and λ(n) (= n). (2) If Z(A) = {1} and A has exponent n, then G is nilpotent and its class is bounded by a function depending only on m and λ(n).

1. Introduction

Let G be a finite group and A a group acting on G. Since we are not supposing that the action is faithful, A will be referred to as an operator group on G. Giving some conditions on A and on G, deep information about the structure of G can be obtained. The most famous result is probably due to Thompson, who proved that if |A| is a prime number and C_{G}(A) = {1} then the finite group G is nilpotent (see Theorem 10.2.1 of [5]).

The aim of this paper is to extend some results obtained in [11] and [12]. We consider the following hypotheses:

(⋆) Let G be a finite group, with a finite operator group A, satisfying the following conditions:

(1) (|G|, |A|) = 1;

(2) there exists a natural number m such that for any α, β ∈ A^{♯} we have:

[ C_{G}(α), \underbrace{C_{G}(β), …, C_{G}(β)}_{m}] = \{1\};

(3) A is not cyclic.

Here, if X is a group, X^{♯} denotes the set X \ {1}.

AMS Subject Classification: 20 D 45

In [11] it is proved that, under the hypotheses (⋆), if |A| = p^{2} (p a prime number) then G is nilpotent with nilpotency class bounded by a function depending only on m and p. In [12] the case in which A is cyclic has been studied, proving that if A = ⟨α⟩ has square-free exponent n, C_{G}(A) = {1} and the hypotheses (⋆_{1}) and (⋆_{2}) are satisfied, then G is nilpotent with class bounded by a function depending only by m and n. In the cases considered, the exponent n of A is square free, that is n = λ(n) where λ(n) is the square free part of n defined as follows: given a natural number n and its factorization, n = p_{1}^{h_{1}} · p_{2}^{h_{2}} · . . . · p_{r}^{h_{r}} (p_{i} ≠ p_{j} if i ≠ j) we put λ(n) = p_{1} · p_{2} · . . . · p_{r}.

Using Lie methods, we prove:

Theorem 1. Let G and A be two groups satisfying the hypotheses (⋆). If the exponent n of A is square-free, then G is nilpotent and its class is bounded by a function depending only on m and λ(n) (= n).

Theorem 2. Let G and A be two groups satisfying the hypotheses (⋆). If Z(A) = {1} and A has exponent n, then G is nilpotent and its class is bounded by a function depending only on m and λ(n).

Theorems 1 and 2 make us conjecture that if G and A satisfy the hypotheses (⋆) then only exceptionally G is not nilpotent. In fact

Theorem 3. If G and A satisfy the hypotheses (⋆_{1}) and (⋆_{2}) and if G is not nilpotent, then A has the structure of the complement of some finite Frobenius group.

Using the classification of finite simple groups, we can prove that any group satisfying (⋆_{1}) and (⋆_{2}) is soluble (these hypotheses can be further weakened, see for example [13] and [3]; if C_{G}(A) = {1} a nice proof can be found in [10]).

Using a result of Kurzweil ([8]), it can be proved in general that if G and A satisfy the hypotheses (⋆_{1}) and (⋆_{2}) then G has Fitting length at most 3 and this bound is reached. There exists in fact a finite group G with an automorphism σ of order a prime number p with (|G|, p) = 1, [G, ⟨σ⟩] = G and such that C_{G}(σ) is nilpotent and G has Fitting length 3 (see for example [2] and [6]).

2. Notations and preliminary results

All the groups considered in this paper are finite, even if the results of Theorems 1, 2 and 3 holds also under the weaker hypothesis that G is locally finite. We denote by G and A a pair of groups satisfying the hypothesis (⋆_{1}) and by p, q, r and s prime numbers; the function λ has already been defined. If Γ is a set of parameters we say that a number is Γ-bounded if it is bounded by a function depending only on elements of Γ.

We use standard notation for groups and Lie rings (see [5], [9], [4], [7], [1]). In particular if G is a nilpotent group and A is an operator group of G with (|G|, |A|) = 1, then we can associate to G a nilpotent Lie ring L of the same class of G on which A acts (essentially) as on G.

If X and Y are subgroups of a group H (or subrings of a Lie ring L) the subgroup (the subring) [X, _{m}Y] is defined inductively by [X, _{1}Y] = [X, Y] and [X, _{m}Y] = [[X, Y], _{m−1}Y].

The following result will be used without further reference in many proofs by induction on |G|. Its proof is immediate.

Lemma 1. The hypotheses (⋆) are inductive on G, that is they are inherited by A-invariant subgroups and A-invariant quotients of G.

Lemma 2. Let V be a vector space of finite dimension over a field of characteristic p > 0. Let B be a non abelian group of order rs with r > s distinct prime numbers and (p, rs) = 1. If B acts on V then:

V = ∑_{β∈B^{♯}} C_{V} (β).

Proof. Let 1 ≠ ⟨α⟩ be the normal subgroup of order r of B; then C_{V} (α) is a B-invariant subspace of V. By Maske Theorem (Theorem 3.3.1 of [5]) V admits a B-invariant complement W of C_{V} (α). We put X = B\⟨α⟩ and W_{0} = ∑_{κ∈X} C_{W} (κ). If we suppose W_{0} ≠ W, then α and any κ ∈ X act on \overline{W} = W/W_{0} fixed points free. In this case Theorem 5.3.14.iii of [5] shows that the group ⟨α⟩⟨κ⟩ is cyclic, against the hypothesis.

Lemma 3. Let X be a group and assume that either

(a) X is non abelian of square-free exponent of X or

(b) Z(X) = {1}.

If X does not contain elementary abelian subgroups of order p^{2} for any p ∈ π(X), then X is metacyclic and it contains a non abelian subgroup of order rs with r > s distinct prime numbers.

Proof. By hypothesis all the Sylow p-subgroups of X, p odd, are cyclic, while the Sylow 2-subgroups are cyclic or generalized quaternion (Theorem 5.4.10.ii of [5]).

The Sylow 2-subgroups of X cannot be generalized quaternion. This is clear in case (a); in case (b) either

(i) O_{2′}(X) = {1}, then for Brauer-Suzuki Theorem (see chapter 12 of [5]) Z(X) contains an involution: a contradiction;

or (ii) O_{2′}(X) ≠ {1}; since all the Sylow subgroups of O_{2′}(X) are cyclic, O_{2′}(X) is (cyclic or) metacyclic (Theorem 10.1.10 of [9]). A generalized quaternion group cannot act faithfully on a (cyclic or) metacyclic group and therefore we should have Z(X) ≠ {1} against the hypothesis.

Therefore X is metacyclic (Theorem 10.1.10 of [9]) and X = ⟨x⟩⟨y⟩ with ⟨x⟩ normal in X. If Z(X) = {1} we have ⟨x⟩ ∩ ⟨y⟩ = {1} because ⟨x⟩ ∩ ⟨y⟩ ≤ Z(X) and therefore (|⟨x⟩|, |⟨y⟩|) = 1 by hypothesis. In any case there exists an element x_{0} of ⟨x⟩ of prime order r. Since C_{X}(x_{0}) ≠ X there exists y_{0} ∈ ⟨y⟩ of order a prime number s ≠ r that does not centralize x_{0}. Then ⟨x_{0}⟩⟨y_{0}⟩ is the subgroup we are looking for.

Lemma 4. If G and A are groups satisfying the hypothesis (⋆_{1}) then:

(a) if N is a normal A-invariant subgroup of G, then

C_{G/N}(A) = C_{G}(A)N/N ;

(b) for any p ∈ π(G), there exists an A-invariant Sylow p-subgroup of G;

(c) if A is not abelian and its exponent is square-free or if Z(A) = {1} we have

G = ⟨ C_{G}(α) | α ∈ A^{♯} ⟩.

Proof. (a) and (b) are a direct consequence of Schur-Zassenhaus Theorem (see Theorem 6.2.2 of [5]).

If A contains an elementary abelian subgroup of order p^{2} (p a prime number) then the conclusion is given by Theorem 6.2.4 of [5]. Hence, by Lemma 3, A contains a non abelian subgroup of order rs with r and s distinct prime numbers. The conclusion can be obtained using Lemma 2 and point (a).

Lemma 5. Let B be a group of operators of the Lie ring L. If B has exponent n and if nL = L we have:

(a) if I is a B-invariant ideal of L then C_{L/I}(B) = (C_{L}(B) + I)/I;

(b) if B is not cyclic and n is square-free or if Z(B) = {1} then

L = ∑_{β∈B^{♯}} C_{L}(β).

Proof. (a) is a simple extension of Lemma 2.2.1 of [11]. If C ≤ B is elementary abelian of order p^{2} (p a prime number) then clearly pL = L, Lemma 2.2.2 of [11] gives

L = ∑_{γ∈C^{♯}} C_{L}(γ)

and the conclusion. If B do not contains elementary abelian subgroups of order p^{2} then by, Lemma 3, B contains a non abelian subgroup of order rs. The conclusion follows from Lemma 2 and point (a).

Lemma 6. Let L be a metabelian Lie ring and B an operator group of L of exponent n. Suppose that nL = L and that [C_{L}(α), _{m}C_{L}(β)] = {0} for any α, β ∈ B^{♯}. If B is not cyclic and n is square-free or if Z(B) = {1} then L is nilpotent of class {m, λ(n)}-bounded.

Proof. Let L′ = [L, L]; by Lemma 5 (b) we have:

L = ∑_{κ∈K^{♯}} C_{L}(κ) , L′ = ∑_{κ∈K^{♯}} C_{L′}(κ).

Since L is metabelian, for any x ∈ L′ and any a, b ∈ L we have [x, a, b] = [x, b, a]. If t = m(rs − 1) + 1, by the bilinearity of [ , ] we get [L′, _{t}L] = {0} and therefore L is nilpotent of class at most t + 1. It is now enough to observe that rs ≤ n = λ(n).

Lemma 7. Let L be a soluble Lie ring and B be an operator group of L of exponent n. Suppose that nL = L and that [C_{L}(α), _{m}C_{L}(β)] = {0} for any α, β ∈ B^{♯}. If one of the following conditions is satisfied:

(a) B is not cyclic and n is square-free,

(b) Z(B) = {1}, then L is nilpotent of class {m, λ(n)}-bounded.

Proof. Let d be the derived length of L.

It follows from Lemma 6, using Proposition 7.1.1 of [1] (or Theorem 2.4 of [11]), that L is nilpotent of class {m, λ(n), d}-bounded. Hence the statement is proved if we show that d is {m, λ(n)}-bounded. Let β ∈ B^{♯}, by Lemma 5 (b) we have:

[L, _{m}C_{L}(β)] = [ ∑_{κ∈B^{♯}} C_{L}(κ), _{m}C_{L}(β)] = ∑_{κ∈B^{♯}}[C_{L}(κ), _{m}C_{L}(β)] = 0.

Theorem 2.3 of [11] gives the conclusion.

3. Proofs of the theorems

To prove Theorems 1 and 2 it is sufficient to show that if the group G satisfies their hypotheses, then G is nilpotent; the bound of the nilpotency class is then a consequence of Lemma 7 and of the correspondence between (finite) nilpotent groups and nilpotent Lie rings (described in [7] or in chapter VIII of [4]).

We suppose, by contradiction, that there exist counterexamples to Theorem 1 or Theorem 2. We choose among these counterexample, one such that |G| + |A| is minimal. Let p ∈ π(G) be an odd prime number and P an A-invariant Sylow p-subgroup of G. Let N = N_{G}(Z(J(P))). If N ≠ P then |N| + |A| < |G| + |A| and N is nilpotent. By Glauberman-Thompson Theorem (Theorem 8.3.1 of [5]) we have that G has a normal p-complement, which is A-invariant. If N = G then {1} ≠ Z(J(P)) is a normal, nilpotent and A-invariant subgroup of G. In both cases we have F (G) ≠ {1} and since |G/F (G)| + |A| < |G| + |A|, G/F (G) is nilpotent. With a similar argument it can be proved that G/F (G) is minimal A-invariant and therefore there exists q ∈ π(G) such that G/F (G) is an elementary abelian q-group (on which A acts irreducibly).

We suppose that G admits two minimal normal A-invariant subgroups N_{1} and N_{2}, then G/N_{1} and G/N_{2} are nilpotent. Since N_{1} ∩ N_{2} = {1}, the group G should be nilpotent: a contradiction.

We only have to examine the case in which G has a unique minimal normal A-invariant subgroup N; in this case N, and therefore F (G), is a p-subgroup for some p ∈ π(G), p ≠ q. If Φ(F (G)) ≠ {1} we get a contradiction, when we consider G/Φ(F (G)). We can therefore suppose F (G) = P, an elementary abelian p-subgroup of G and G = P Q with P and Q elementary abelian Sylow subgroups. As in the proof of Lemma 3 we obtain that A contains a non abelian subgroup B of order rs with r and s distinct prime numbers. Since the hypotheses (⋆) are still valid for the non-cyclic subgroups of A, if B ≠ A we have |G| + |B| < |G| + |A| and G should be nilpotent. Therefore A = B and |A| = rs. Lemma 4 (c) gives P = ⟨ C_{P} (α) | α ∈ A^{♯} ⟩ and Q = ⟨ C_{Q}(α) | α ∈ A^{♯} ⟩.

Let α be an element of A with C_{Q}(α) ≠ {1} and let y be a non trivial element in C_{Q}(α). Take x_{1}, x_{2} ∈ P, and recall that [x_{1}, y], [x_{2}, y] ∈ P. From the fact that P is abelian, we get [x_{1}x_{2}, y] = [x_{1}, y]^{x_{2}}[x_{2}, y] = [x_{1}, y][x_{2}, y].

By hypothesis we have [C_{P} (β), _{m}⟨y⟩] = {1} for all β ∈ A^{♯} and therefore [C_{P} (β), ⟨y⟩] = {1} because (|P |, |⟨y⟩|) = 1. Since P = ⟨C_{P} (β) | β ∈ A^{♯}⟩ we obtain [P, ⟨y⟩] = {1}. Then 1 ≠ y ∈ C_{G}(P) ≤ P is the contradiction that concludes the proof.

The proof of Theorem 3 holds even without the hypothesis that G is soluble and it is therefore independent from the classification of finite simple groups.

For any p ∈ π(G), by Lemma 4 (b) there exists an A-invariant Sylow p-subgroup of G. We can consider two cases. ✓ For any odd prime number p ∈ π(G) the group G has a normal p-complement.

If P is a Sylow 2-subgroup of G, then P is normal (and therefore A-invariant) and we can choose q ∈ π(G) and a Sylow q-subgroup Q of G such that P Q is not nilpotent.

✓ There exists an odd prime number p ∈ π(G) such that G has not a normal p-complement. If T is an A-invariant Sylow p-subgroup of G then the subgroup N = N_{G}(Z(J(T))) is not nilpotent. If we consider the A-invariant subgroup P = O_{p}(N), we can conclude that there exists q ∈ π(N) and an A-invariant Sylow q-subgroup Q of N such that P Q is not nilpotent. In any case G admits a non-nilpotent A-invariant {p, q}-subgroup P Q with

P normal in P Q. It is possible to choose a non-nilpotent A-invariant section \overline{S} in P Q of minimal order. It is easy to prove that \overline{S} = \overline{P} \overline{Q} where \overline{P} and \overline{Q} are respectively an A-invariant Sylow p-subgroup and q-subgroup of \overline{S} and F (\overline{S}) = \overline{P}. In particular, by Theorem 6.1.3 of [5], C\overline{S}(\overline{P}) = \overline{P}. We can consider two cases. ✓ Any element of A^{♯} acts fixed point free on \overline{P}. Then the semidirect product

\overline{P} A is a Frobenius group with complement A and the statement is proved. ✓ There exists α ∈ A^{♯} with C\overline{P} (α) ≠ \{1\}. Let x be a non trivial element of

C\overline{P} (α). By the minimality of |\overline{S}| we have \overline{P} = ⟨x⟩^{\overline{Q}A}. If there exists β ∈ A^{♯} with C\overline{Q}(β) ≠ \{1\} then taken y ∈ C\overline{Q}(β), y ≠ 1, by hypothesis we have [x, _{m}y] = 1 and therefore, since \overline{P} = ⟨x⟩^{\overline{Q}A}, [\overline{P} , _{m}y] = \{1\} and (|\overline{P} |, |\overline{Q}|) = 1, [\overline{P} , y] = \{1\}, against the fact that C\overline{S}(\overline{P}) = \overline{P}. Therefore any element of A^{♯} acts fixed points free on \overline{Q}. Then the semidirect product \overline{Q}A is a Frobenius group with complement A and the statement is proved.

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(received 24.08.2004, in revised form 30.05.2006)

Dipartimento di Matematica Applicata, Universitá di Ca’ Foscari, Dorsoduro 3825/e, I-30122 Venezia - Italy

E-mail: jabara@unive.it