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This problem asks to show that if $A$ acts on $G$ via automorphisms, where either $A$ or $G$ is soluble and both $A$ and $G$ are finite groups and $G$ is nontrivial, then $G$ possesses an $A$-invariant $p$-subgroup.

My original attempt at a solution was this:

If $G$ is soluble, then $F(G)>1$, in which case any nontrivial Sylow subgroup of $F(G)$, being characteristic all the way up to $G$, will do the trick.

If $A$ is soluble, choose as above some Sylow $p$-subgroup of $F(A)$, say $P$. Certainly $P$ is normal in $A$ and my claim here is that there exists some $P$-invariant Sylow $q$-subgroup of G, regardless of the condition that $p$ divides $|G|$ or not. If $p$ does not divide $|G|$, then a direct appeal to the extended version of the Sylow theorems for coprime actions proves the claim. If $p$ divides $|G|$, then consider the action of $P$ on $Ω = Syl_p(G)$, which is nonempty. Then $|Ω| =$ sum of $P$-orbits, each a $p$-power, while $|Ω| = 1 (mod p)$. Some $P$-orbit must therefore be a singleton set, meaning that $P$ fixes this particular Sylow $p$-subgroup of $G$. Thus, in all cases the claim holds.

Now suppose that the $P$-invariant Sylow subgroup of $G$ is $Q$, where it's possible that $p$ divides $|Q|$. Then for any $u \in P$ and $a \in A$, $aua^{-1} \in P$. Hence $aua^{-1}(Q)=Q$, implying that $a(Q)Qa^{-1}(Q)=Q$. Thus $a(Q)$ is a subgroup of $N_G(Q)$ and also a Sylow $q$-subgroup of $G$, because $a$ (being an automorphism) maps Sylow subgroups to Sylow subgroups. Therefore $a(Q)$ is a member of $Syl_q(N_G(Q))$, the sole member of which is $Q$. Finally $a(Q)=Q$ for all $a \in A$ and we are done.

The last paragraph contains an error, but I'm not sure what that is. I haven't used anywhere that $P>1$. But if $P$ is trivial, then $Q$ can be an arbitrary Sylow subgroup of $G$, and of course, $Q$ is not necessarily fixed by $A$.

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    Is $aua^{-1}(Q) = a(u(a^{-1}(Q)))$ rather than $a(Q)u(Q)a^{-1}(Q)$? That would not imply $a(Q) \leq N_G(Q)$, I believe, and explain the problem. In particular, $a(1(a^{-1}(Q)))= a(a^{-1}(Q))=Q$ for all $a$, regardless of whether $a(Q)$ normalizes anything.2012-01-23
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    Right. Thanks for this. Can you suggest a way to deal with the case where $A$ is soluble?2012-01-23
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    Today is a little crazy, but I think a full proof is in Kurzweil-Stellmacher, which I believed used similar language and lemmas. In fact it is so similar in my fuzzy memory, I worry that you are already using that book and somehow it is an exercise. I'll start writing up a solution, but I suspect it'll be tomorrow before I'm done.2012-01-23
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    And by invariant p-subgroup, we mean non-identity invariant p-subgroup of course.2012-01-23

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