# Abelian subgroups of the group of automorphisms of a finite group

This is a follow-up question from my post here, which has been moved according to a comment. For context, here is the setup.

Let $$G$$ be a nontrivial finite group. In his book "Finite Group Theory", M. Isaacs proves the following two results:

Corollary 3.3: Let $$\sigma \in \operatorname{Aut}(G)$$. Then, $$o(\sigma) < |G|$$.

Corollary 3.4: Let $$P$$ be an abelian $$p$$-subgroup of $$\operatorname{Aut}(G)$$ and suppose $$p \nmid |G|$$ . Then, $$|P| < |G|$$.

When we consider both of the results together, I'm led to the following:

Question: If $$A \leq \operatorname{Aut}(G)$$ is an abelian group of order $$n = p_1^{k_1} \cdots p_m^{k_m}$$ such that $$p_1, p_2, ..., p_m \nmid |G|$$, is it true that $$|A| < |G|$$?

We can, of course, decompose $$A = P_1 \times P_2 \times ... \times P_m$$, where $$P_j$$ is a $$p_j$$-group. From 3.4, we know $$|P_j| < |G|$$, so $$|A| < |G|^m$$. This suggests that this question is false, since we could, in principle, find "big" $$p_j$$-subgroups of $$\operatorname{Aut}(G)$$ and take their direct product, but I don't have any examples to back it up.

Any help is appreciated! Thanks in advance!

• I would guess that the answer to this is yes, but I haven't thought much about how to prove it. It is true when $G$ is elementary abelian, when the largest such subgroup has order $|G|-1$, and elementary abelian group generally have the largest automorphism groups. Jul 25, 2023 at 8:13

The answers is yes. As in the proof of 3.4, it suffices to show that $$A$$ has a regular orbit on $$G$$. Using theorem by Hartley-Turull (3.31), we may assume that $$G=G_1\times \ldots\times G_s$$ is a direct product of elementary abelian groups $$G_i$$ (as in the proof of 3.34). If $$A$$ has a regular orbit on each $$G_i$$, so it has on $$G$$ (choose $$x_i\in G_i$$ and consier $$x=x_1\ldots x_s$$). Hence, we may assume that $$G$$ is an $$\mathbb{F}_pA$$-module for some prime $$p\nmid|A|$$. By Maschke's theorem $$G$$ is semisimple. By the same argument as before, we may assume that $$G$$ is simple. For every $$a\in A$$, $$C_G(a)$$ is an $$A$$-invariant submodule since $$A$$ is abelian. Since $$A$$ acts faithfully, we must have $$C_G(a)=1$$ for all $$a\ne 1$$. Equivalently, $$C_A(g)=1$$ for alle $$g\in G\setminus\{1\}$$. So in fact, every non-trivial $$A$$-orbit is regular.