On the rank of a finite group of odd order with an involutory automorphism

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if and only if G −φ = {1} (cf Lemma 1(i) in the next section).The following theorem was proved in [10,Theorem B].

Theorem 1 Let G a finite group of odd order admitting an involutory automorphism φ such that the rank of G φ is at most r . Then the rank of [G, φ] is r -bounded.
Recall that a rank of a finite group G is the least number r such that each subgroup of G can be generated by at most r elements. Throughout this manuscript we use the term "(a, b, c . . . )-bounded" to mean "bounded from above by some function depending only on the parameters a, b, c . . . ".
Since in a finite group of odd order with an involutory automorphism φ there is a kind of (very vague) duality between G φ and G −φ , in this paper we address the question whether a rank condition imposed on the set G −φ has an impact on the structure of G. We emphasize that G −φ in general is not a subgroup of G and therefore the usual concept of rank does not apply to G −φ . Instead we impose the condition that each subgroup of G generated by a subset of G −φ can be generated by at most r elements. Our main result is as follows.

Theorem 2 Let G be a group of odd order admitting an involutory automorphism φ and suppose that any subgroup generated by a subset of G −φ can be generated by r elements. Then [G, φ] has r -bounded rank.
It is noteworthy that in the literature there are several papers dealing with finite groups admitting a (not necessarily involutory) automorphism whose fixed-point subgroup is of rank r (see for example [5,6]). In particular, [5] contains a result similar to the above Theorem 1. Thus, it seems plausible that some analogues of Theorem 2 are valid for the case where the order of φ is bigger than two.

Nilpotent groups with involutory automorphisms
We start with a collection of well-known facts about involutory automorphisms of groups of odd order (see for example [3, Lemma 4.1, Chap. 10]). Lemma 1 Let G be a finite group of odd order admitting an involutory automorphism φ. The following conditions hold: It is well known that a maximal abelian normal subgroup of a nilpotent group coincides with its centralizer. We will require the following related result. Lemma 2 Let G be a nilpotent group of odd order, φ an involutory automorphism of G, and A a maximal φ-invariant abelian normal subgroup of G. Then A = C G (A).
Proof Let C = C G (A) and assume that the result is false, that is, In any case we can choose u ∈ U \ A satisfying either u φ = u or u φ = u −1 . Take H = A u and note that A < H . Furthermore, H is a φ-invariant abelian normal subgroup of G. This yields a contradiction.
Note that the previous lemma fails if φ is allowed to be a coprime automorphism of arbitrary order. For example, the quaternion group of order 8 admits an automorphism α of order 3 and the maximal α-invariant abelian normal subgroup is central. Proof If n = 0, then the result follows from Lemma 1(iv), so assume that n ≥ 1 and use induction on n. Let for any i ≥ 2 and so, obviously, M ≤ Z (G). This concludes the proof. We now fix some notation and hypotheses that will be used throughout this section.

Hypothesis 1
Let p be an odd prime, r a positive integer and G a finite p-group with an involutory automorphism φ such that G = [G, φ]. Assume that any subgroup generated by a subset of G −φ can be generated by r elements. Lemma 4 Assume Hypothesis 1 and suppose that G is of exponent p. There exists a number l = l(r ), depending on r only, such that the rank r (G) of G is at most l.
Proof Let A be a maximal φ-invariant abelian normal subgroup of G. The subgroup A −φ is an r -generated abelian subgroup of exponent p and so that is, the quotient group G/A is nilpotent of class 2r . We deduce that G has r -bounded nilpotency class as well. Since G = G −φ is r -generated by hypothesis, it follows that the rank r (G) of G is r -bounded, as desired.
The following result from [10, Lemma 2.2] is also useful.

Lemma 5
Let G be a group of prime exponent p and rank r 0 . Then there exists a number s = s(r 0 ), depending only on r 0 , such that |G| ≤ p s .

Lemma 6
Let G be a group satisfying Hypothesis 1. There exists a number λ = λ(r ), depending only on r , such that γ 2λ+1 (G) is powerful.
Proof Let s(r 0 ) be as in Lemma 5 and let l(r ) be as in Lemma 4 where λ = s(l(r )). In order to show that N ≤ N p , we assume that N is of exponent p and prove that N is abelian.
Note that the subgroup N −φ is of exponent p. By Lemma 4 the rank of N −φ is at most l(r ). It follows from Lemma 5 that |N −φ | ≤ p s(l(r )) = p λ . Now Lemma 3 yields N ≤ Z 2λ+1 (G) . By using the well-known fact that [γ i (G), Z i (G)] = 1, for any positive integer i and any group G, we conclude that N is abelian, as required.
Lemma 7 Assume Hypothesis 1. For any i ≥ 1, there exists a number m i = m i (i, r ), depending only on i and r , such that γ i (G) is an m i -generated group.
In view of the Burnside Basis Theorem [9, 5.3.2], we can pass to the quotient G/Φ(N ) and assume that N is elementary abelian. Now N −φ is an elementary abelian r -generated group, so | N −φ | ≤ p r . Thus, by Lemma 3, we have N ≤ Z 2r +1 (G) and deduce that G has nilpotency class bounded only in terms of i and Therefore N is m i -generated for some (i, r )-bounded number m i . This concludes the proof.

Proposition 1 Under Hypothesis 1 the rank of G is r -bounded.
Proof Let s(r 0 ) be as in Lemma 5 and l(r ) as in Lemma 4 Lemma 7 tells us that d is an r -bounded integer and N is powerful by Lemma 6. It follows from [1, Theorem 2.9] that r (N ) ≤ d, and so the rank of N is r -bounded. Since the nilpotency class of G/N is r -bounded (recall that λ depends only on r ) and G = G −φ is r -generated, we conclude that r (G/N ) is r -bounded as well. Now r (G) ≤ r (G/N ) + r (N ) and the result follows.

Main results
Throughout this section the Feit-Thompson Theorem [2] is used without explicit references and p stands for a fixed odd prime. Given a finite soluble group G, we denote by r p (G) and l p (G) the rank of a Sylow p-subgroup and the p-length of G, respectively. Recall that l p (G) is by definition the number of p-factors (that is, factors that are p-groups) of the lower p-series of G given by: We aim to establish the following generalisation of Proposition 1.

Theorem 3 Let G be a group of odd order admitting an involutory automorphism
Let r be a positive integer and assume that any subgroup generated by a subset of G −φ can be generated by r elements, then r p (G) is r -bounded.
We start with an extension of Lemma 3.

Lemma 8 Let G be a group of odd order admitting an involutory automorphism φ such that G = [G, φ]. Let M be a φ-invariant normal subgroup of G and assume that
|M −φ | ≤ p n , for some nonnegative integer n. Then M ≤ Z 2n+1 (O p (G)).
Proof The proof can be reproduced word-by-word following that of Lemma 3. We argue by induction on n, being Lemma 1(iv) the case n = 0. Let n ≥ 1. If M Z (O p (G)), then by Lemma 1(iv) we have N −φ = 1, where N = M ∩ Z 2 (O p (G)). This implies that |(M/N ) −φ | < |M −φ |. Thus we can pass to the quotient G/N and use the inductive hypothesis. The result follows.
For the sake of simplicity we fix the following hypothesis that we will use in the next arguments.
Hypothesis 2 Let r be a positive integer and G a group of odd order admitting an involutory automorphism φ such that G = [G, φ]. Assume that any subgroup generated by a subset of G −φ can be generated by r elements.
As usual, we denote by F(G) the Fitting subgroup of a group G.
One key step forward to the proof of Theorem 3 consists in showing that there exists an r -bounded number f such that the f th term of the derived series of G is nilpotent. For our purpose we will require the following result which is an immediate corollary of Hartley-Isaacs Theorem B in [4]. In the proof of the next proposition we will use the well-known theorem of Zassenhaus (see [11,Satz 7] or [8, Theorem 3.23]) stating that for any n ≥ 1 there exists a number j = j(n), depending only on n, such that, whenever k is a field, the derived length of any soluble subgroup of G L(n, k) is at most j.

Proposition 3 Assume Hypothesis 2. There exists a number f = f (r ), depending only on r , such that the f th term G ( f ) of the derived series of G is nilpotent.
Proof Let δ = δ(r ) be as in Proposition 2 and f = j(δ) the number given by the Zassenhaus theorem.
Suppose that the proposition is false and let G be a group of minimal possible order for which Hypothesis 2 holds while G ( f ) is not nilpotent. Then G has a unique minimal φ-invariant normal subgroup M. Indeed, suppose that G has two minimal φinvariant normal subgroups, say M 1 and M 2 . Then M 1 ∩M 2 = 1, being both elementary abelian p-groups for some prime p. Since |G/M 1 | < |G|, the minimality of G implies that (G/M 1 ) ( f ) is nilpotent. For a symmetric argument (G/M 2 ) ( f ) is nilpotent too. This yields a contradiction since G ( f ) can be embedded into a subgroup of G/M 1 × G/M 2 which is nilpotent, being isomorphic to the direct product of (G/M 1 ) ( f ) and We claim that M = C G (M). Since M is a p-subgroup, for some prime p and it is unique, the Fitting subgroup F = F(G) is a p-subgroup too. If Φ(F) is nontrivial, then we immediately get a contradiction because F(G/Φ(F)) = F/Φ(F) and, again by the minimality of G, we know that  (δ(r )). Then G ( f ) ≤ F, which concludes the proof.
As a by-product of the previous result we obtain a bound for the p-length of G.

Corollary 1 Assume Hypothesis 2. Then l p (G) is r -bounded, for any p ∈ π(G).
Proof By Proposition 3 we know that G ( f ) is nilpotent for some r -bounded number f . This implies that the Fitting height h(G) ≤ f . The result easily follows since it can be shown, by induction on the Fitting height h(K ), that l p (K ) ≤ h(K ) for any finite soluble group K and for any prime p ∈ π(K ).
The next result will be useful for a reduction argument inside the proof of Theorem 3.

Lemma 9 Let G be a group of odd order admitting an involutory automorphism φ.
Assume that G = P B, where P is a φ-invariant normal elementary abelian psubgroup and B is a cyclic subgroup such that B = B −φ . If r(P −φ ) = r , then the rank of [P, B] is at most 2r.
Then it follows from Lemma 1(i) that and so x commutes with b 2 . Since b has odd order, it follows that C 0 ≤ C G (b), as claimed. Thus C 0 ≤ Z (G). Choose now a 1 , . . . , a 2r elements that generate P modulo C 0 . By using linearity in P and the fact that C 0 is central in G, we deduce that [P, b] is generated by [a 1 , b], . . . , [a 2r , b]. Hence the result.
We are ready to embark on the proof of Theorem 3.

Proof of Theorem 3
Recall that G is a group satisfying Hypothesis 2 and we want to show that r p (G) is r -bounded for any fixed prime p ∈ π(G).
First, we show that G is generated by r -boundedly many elements from G −φ . If G is a p-group, then the claim follows from the Burnside Basis Theorem since . . , P s } are the Sylow subgroups of G, so the result easily follows from the case of p-groups. Assume now that G is not nilpotent. Let h = h(G) ≥ 2. Since we know from the proof of Corollary 1 that h is rbounded, it is sufficient to show that G is generated by (h, r )-boundedly many elements from G −φ . We argue by induction on h. Let F = F(G). By induction there are boundedly many elements a 1 , . . . , a d ∈ G −φ such that G = F a 1 , . . . , a d . Let D = F −φ , a 1 , . . . , a d . Note that D has an r -bounded number of generators from Lemma 1(v). Thus N ≤ D. Recall that by Lemma 1(i) we have F = F φ F −φ . Hence the image of F in G/N is contained in (G/N ) φ and, therefore, it is central by Lemma 1(iv). Since G = F D, it follows that D/N becomes normal in G/N and, therefore, D is normal in G (because N ≤ D). Now φ acts trivially on the quotient G/D, that is [G, φ] ≤ D. Since G = [G, φ], we have G = D. This concludes the proof that G can be generated by r -boundedly many elements from If G is a p-group, then the theorem follows immediately from Proposition 1. Assume that G is not a p-group and use induction on l = l p (G) that is r -bounded by Corollary 1. So it is sufficient to show that r p (G) is (l, r )-bounded. By induction assume that there exists r 1 , depending only on l and r , such that r p (K ) ≤ r 1 for any φ-invariant quotient K of G having l p (K ) at most l − 1.
Since l = l p (G/O p (G)), we can assume that O p (G) = 1. Take P = O p (G). Note that r p (G) ≤ r (P) + r p (G/P).
Since l p (G/[P, G]) ≤ l − 1, by induction the rank r p (G/[P, G]) ≤ r 1 . Then it is sufficient to bound the rank of P.
Let us show first that P has an r -bounded number of generators. Passing to the quotient G/Φ(P), we can assume that P is elementary abelian. As showed above, we know that G can be generated by t = t(r ) elements from G −φ , say d 1 , . . . , d t . Note that [P, G] = [P, In view of Lemma 9 each [P, d i ] has rank at most 2r . Therefore the rank of the image of P in G/Φ(P) is at most 2rt and by induction on l, r p (G/P) is r -bounded, so P has an r -bounded number of generators, as claimed.
Next, we claim that for any i ≥ 2 there exists a number m i = m i (i, r ), depending only on i and r , such that V = γ i (P) has m i -bounded number of generators. We can pass to the quotient G/Φ(V ) and assume that V is elementary abelian. Now V −φ is an elementary abelian r -generated group, so | V −φ | ≤ p r . Thus, by Lemma 8, we have V ≤ Z 2r +1 (P) and deduce that the nilpotency class of P/Φ(V ) is bounded only in terms of i and r . Since P has an r -bounded number of generators, we conclude that r (P/Φ(V )) is (i, r )-bounded as well. Therefore V is m i -generated for some (i, r )-bounded number m i , as claimed.
Let s(r 0 ) be as in Lemma 5 and let l(r ) be as in Lemma 4. Take M = γ 2λ+1 (P), where λ = s(l(r )). We want to prove that M is powerful. In order to show that M ≤ M p , we assume that M is of exponent p and prove that M is abelian. Note that the subgroup M −φ is of exponent p. By Lemma 4 the rank of M −φ is at most l(r ). It follows from Lemma 5 that |M −φ | ≤ p s(l(r )) = p λ . Now Lemma 8 yields that M ≤ Z 2λ+1 (P) . Since [γ i (P), Z i (P)] = 1, for any positive integer i, we conclude that M is abelian, as required.
Let now d 0 be the minimal number such that M is d 0 -generated. It was shown above that d 0 is an r -bounded integer. Since M is powerful, it follows from [1, Theorem 2.9] that r (M) ≤ d 0 , and so the rank of M is r -bounded. Since the nilpotency class of P/M is r -bounded and P has an r -bounded number of generators, we conclude that r (P/M) is r -bounded as well. Now r (P) ≤ r (P/M) + r (M) and the result follows.
It is now easy to give the proof of our main result, Theorem 2, which states that if G is a group satisfying Hypothesis 2, then the rank of G is r -bounded.

Proof of of Theorem 2
Without loss of generality we can assume that G = [G, φ]. By a result of Kovács [7] for any soluble group H we have r (H ) ≤ max{r p (H ) | p ∈ π(H )} + 1. Therefore it is enough to check that r p (G) is bounded in terms of r only for any p ∈ π(G). This is immediate from Theorem 3.