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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!

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    $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Jul 25, 2023 at 8:13

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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.

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