I'm dealing with the following problem in Isaacs Finite Group Theory [4D.3], I would appreciate if you could help:

Problem: Let $A$ act via automorphisms on $G$ , where $(\vert G \vert, \vert A \vert ) = 1.$ Suppose that $A$ acts trivially on every $A$-invariant proper subgroup of $G$, but that the action of $A$ on $G$ is nontrivial. Show that $G$ is $p$-group for some prime $p.$

My attempt: For a contradiction, assume that $G$ is not a $p$-group. Using Glauberman Lemma, we can see that there exists an $A$-invariant Sylow $p$-subgroup of $G$ for each prime. Due to hypothesis, for any $A$-invariant Sylow $p$-subgroup $P$, it must be $[P,A]=1$. That is each such a Sylow $p$-subgroup is contained in $C_G(A)$. From the hypothesis again, we know that $C_G(A) \neq G.$ How can I get the desired result from this point?


  • 1
    $\begingroup$ From the solution-verification tag wiki: "A question with this tag should include an explanation for why the argument presented is not convincing enough." $\endgroup$
    – Shaun
    Sep 18 at 18:14
  • $\begingroup$ Thanks, question draft corrected. $\endgroup$
    – Yılmaz
    Sep 18 at 18:30
  • 1
    $\begingroup$ You are still misusing the solution-verification tag: you don't present a solution, and you don't ask about a given solution. It's easy here: since you don't have a solution, how can you be asking for a solution verifcation? $\endgroup$ Sep 18 at 18:41
  • 2
    $\begingroup$ For each prime $p$ dividing $|G|$, we know that $A$ centralizes some Sylow $p$-subgroup $G_p$ of $G$, and $G$ is generated by the subgroups $G_p$, so $A$ centralizes $G$, contradiction. $\endgroup$
    – Derek Holt
    Sep 18 at 21:54
  • $\begingroup$ I got it, thanks. $\endgroup$
    – Yılmaz
    Sep 19 at 7:30


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