The rank of the automorphism group of a ﬁnite group

Let G be a ﬁnite group whose order is divisible by e primes (counting repetitions). Then the automorphism group of G has Prüfer rank at most e 2 , meaning that each subgroup of the automorphism group of G can be generated by at most e 2 elements.


Introduction
If G is a finite group, then d(G) denotes its minimal number of generators, e(G) the number of prime divisors (counting repetitions) of its order |G| and rk(G) its rank; that is, the maximum of the d(X ) as X ranges over the subgroups of G. Clearly d(G) ≤ rk(G) ≤ e(G). Also [r] denotes the largest integer not exceeding the real number r . Theorem 1 If G is a finite group, then always r k(Aut(G)) ≤ e(G) + [e(G) 2 /2] ≤ e(G) 2 .
I mention and use rk(Aut(G)) ≤ e(G) 2 in [12] as if it is well-known. It seems this is not the case and I have failed to find any proof or indeed any mention of it in the literature. Thus I feel now the need to publish a proof. My proof is in the main elementary. The only really non-elementary facts I use are that if S is a finite perfect simple group, then d(S) = 2, rk(Out(S)) ≤ 5, |S| is divisible by 4 and e(S) ≥ 4 (even 5 unless S = Alt (5)). I also use results from [5,8], see below for details, although these can be avoided at the expense of substantially lengthening the proof by using Theorem 2 below to reduce to the soluble case. Clearly rk(G) ≤ e(G) and it is easy to prove that |Aut(G)| ≤ |G| d(G) ≤ |G| e(G) , but these easy facts seem to be of no help in proving the theorem. Note also that rk(Aut(G)) need not be bounded by (rk(G)) 2 , though in significant special cases (e.g. G elementary abelian) it is. As an obvious counter example, the cyclic group G of order 8 has rk(G) = 1 and rk(Aut(G)) = 2. More generally it is easy to see that if G is the direct product of r cyclic groups of order 2 f for f ≥ 3 and r ≥ 1, then Aut(G) contains an elementary abelian subgroup of rank r 2 + r . Hence here rk(G) = r while rk(Aut(G)) ≥ r 2 + r > r 2 . (Also e(G) 2 Our second theorem below we originally used in the proof of Theorem 1 essentially to reduce to the soluble case, but thanks to [5,8], which only very recently were brought to my attention, we no longer need to do this (At the end of this paper we sketch this alternative approach to Theorem 1). However, I think Theorem 2 is of independent interest.

Theorem 2 Let G be a finite subgroup of G L(n, F), where n is a positive integer and F is a field of positive characteristic p. Suppose G has a soluble normal subgroup S such that G/S is a direct product of r perfect simple groups. Then r ≤ [n/2].
Clearly this bound is attained for all n and all F with |F| > 3; recall G L(2, 2) and G L(2, 3) are soluble.

Proof of the results
By a theorem of Guralnick and Lucchini (independently), see [5,8], the rank of a finite group is bounded by one more than the maximum of the ranks of its Sylow subgroups. The following lemma is the analogue of Theorems 1 and 2 for the symmetric groups. Note that Guralnick and Lucchini's theorem together with bounds in [10] yield that in Theorem 2 we have r ≤ 3n/4 if p > 2 and r ≤ 2n if p = 2. These are weaker than the claims of Theorem 2; also they do not seem to help with its proof.

Proof of Theorem 2
If n = 1 we have r = 0. Suppose n = 2. If G is metabelian then r = 0. If not then G is absolutely irreducible and hence is isomorphic to a subgroup of G L(2, p f ) for some f (e.g. [2] 29.21). Thus from Dickson's list of the subgroups of P SL(2, p f ), see [6] II.8.27, we have r ≤ 1. Let G be a counter example to the theorem with n minimal. Then n ≥ 3 and r ≥ 2. Hence G = G 1 G 2 . . . G r , where the G i are normal subgroups of G, the G i /S i are perfect simple for S i = G i ∩ S and G/S is the direct product of the G i S/S. We break the proof into a number of sublemmas.
(a) If N is a normal subgroup of G, then G/N and N have the same structure as G.
Specifically N S/N is soluble and G/N S is the direct product of the G i N S/N S (each being isomorphic to G i S/S or 1 ). Also N ∩ S is soluble, N /(N ∩ S) N S/S and the latter is the direct product of the (N S ∩ G i S)/S (again each of which is isomorphic to G i S/S or 1 ).
Thus we may replace G by H . Equivalently we may assume that The first claim is obvious. Let V = F (n) taken as a right G L(n, F)-module in the obvious way and let W be an then G acts as a soluble group on both W and V /W and hence acts as a soluble group on V . But G i is not soluble. Thus each G i /S i is avoided by at least one of C G (W ) and C G (V /W ) and then by the minimal choice of n we have This contradiction shows that G is irreducible.  (N ∩ S). Now the G i are normal in G and G permutes the V i transitively. Thus G i S ∩ N is not contained in C N (V 1 )(N ∩ S) for each i in I and therefore r 1 = |I | ≤ (dim F (V 1 ))/2. Now t ≥ 2 and n = t.dim F V 1 , so This final contradiction yields that G is primitive. Alt (8). Neither involves the direct product of 2 perfect simple groups (|Alt(8)| = 2.60.168 and is not divisible by 5 2 ) and trivially 1 ≤ [n/2]. Then n = 2 or 4 and so i 2.e(i) < n and r ≤ i e(i) < n/2. This final contradiction yields that E(G) = 1 . Again set E = E(G) and define K ≤ G by S ≤ K and G/S = K /S × E S/S. By (e) and (f) we have Neither E nor C is abelian. Thus if U is an irreducible F E-submodule of V , then V is a direct sum of copies of U by (d), so 1 < m = dim F (U ) < n and n = mk for some integer k < n. Further, by (c), (d) and [9] 1.15 the group G is isomorphic to a subgroup of (G L(m, F) × G L(k, F))/A, where A is a central subgroup of this direct product. Since m < n and k < n, by the choice of n we have r ≤ m/2 + k/2 ≤ n/2. This final contradiction completes the proof of Theorem 2.

The theorem follows.
We conclude with a sketch of a proof of Lemma 2 and hence of Theorem 1 that uses Theorem 2 but avoids using [5,8]. Let G be a subgroup of G L(n, p). As in the first sentence of the proof of Lemma 2 above we may assume n ≥ 3. By results of Roggenkamp, see [3] 7.8, 7.9(i) and 7.20, there exists a Sylow subgroup S of G with d(G) ≤ pr (G) + 1 + rk(S), where pr (G) denotes the presentation rank of G. Thus if pr (G) = 0 (e.g. if G is soluble, see [3] 6.9), then d(G) is bounded using the bounds for rk(S) in [10].
Let N denote the soluble radical of G. If pr (G) = 0, then d(G/N ) = d(G) by Theorem B(ii) of [4]. Clearly H = G/N has trivial soluble radical. Then there exists a normal subgroup E(H ) = S 1 × S 2 × . . . S r = E say of H with C H (E) = 1 , where the S i are perfect simple groups and 1 ≤ r ≤ n/2 by Theorem 2. Now H permutes the S i by conjugation. Applying the arguments from the final two paragraphs of the above proof of Theorem 1, but now to each orbit of H in this action, and using the weak bound [n 2 /4] in Lemma 1, whose proof does not use [5,8], we can again bound d(G). Specifically we obtain d(G) ≤ 2t + 5r + [r 2 /4], where t is the number of such orbits.
Using 1 ≤ t ≤ r ≤ [n/2] we obtain d(G) ≤ n + [n 2 /2] for all n ≥ 3 except when n = 4 = 2r . If n = 4 = 2r we only obtain in this way that d(G) ≤ 15 < n 2 . To obtain the better bound when n = 4 = 2r we need a more delicate study of the simple groups S 1 and S 2 . But in this elementary way we have at least bound rk(G L(n, p)) independently of [5,8] by n 2 always and by n + [n 2 /2] for all n = 4.