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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. –  Jack Schmidt Jan 23 '12 at 18:42
    
Right. Thanks for this. Can you suggest a way to deal with the case where $A$ is soluble? –  the_fox Jan 23 '12 at 19:27
    
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. –  Jack Schmidt Jan 23 '12 at 21:19
    
And by invariant p-subgroup, we mean non-identity invariant p-subgroup of course. –  Jack Schmidt Jan 23 '12 at 21:21
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2 Answers

You have covered the case that $G$ is solvable.

If $A$ is solvable, let $P\lhd A$ be a nontrivial $p$-subgroup of $A$, and consider $C_G(P)$, the fixed points of $P$ acting on $G$. Since for all $x\in P$ and $a\in A$, we have $x^a\in P$, it's not hard to see $C_G(P)$ is $A$-invariant. Applying induction on $|G|$ then reduces to the case of $C_G(P)=\lbrace1\rbrace$.

Since $|G|\equiv |C_G(P)|\equiv 1\pmod{p}$ in this case, we have that $|G|$ and $|P|$ are coprime, and thus for every prime $q$ dividing $|G|$, there is a unique Sylow $q$-subgroup $Q$ of $G$ which is $P$-invariant (this is theorem 3.23 in FGT). Again, because for all $x\in P$ and $a\in A$, we have $x^a\in P$, it follows that $Q$ is actually $A$-invariant.

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I don't see much difference between your argument and mine. Can you elaborate on your last sentence? –  the_fox Jan 24 '12 at 3:01
    
The difference is saying $(Q)x^a=Q$ is not the same as saying $(Q)a^{-1}Q(Q)a=Q$. Let $T=(Q)a^{-1}$. Note that if $T\neq Q$, then $(T)x\neq T$ by uniqueness, and thus $(T)xa\neq Q$, contradicting the fact $x^a\in P$. –  user641 Jan 24 '12 at 3:11
    
Also the case $C_G(P)=\lbrace1\rbrace$ eliminates the case $P$ is trivial, which was a problem with your argument. –  user641 Jan 24 '12 at 3:13
    
Since $C_G(P)$ is $A$-invariant, an inductive argument works if $C_G(P)$ is a strict subgroup of $G$, taking into account the special treatment required when $C_G(P)$ is trivial. What if $P$ acts trivially on $G$? –  the_fox Jan 24 '12 at 6:48
    
How many automorphisms do you know that act trivially on their group? :) –  user641 Jan 24 '12 at 9:06
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A much stronger claim is false (actually obviously so; I'll not delete in hopes it helps point out why the actual claim should be seen as a generalization of something true):

Let G be the alternating group of degree 5, and let A be the group of inner automorphisms induced by the alternating group of degree 4. Let p be 3. Then a non-identity p-subgroup P of G has order 3, and is generated by a 3-cycle. If A normalizes the 3-cycle, then the 3-cycle cannot move the point 5, since P acts regularly, and so A must send the points moved by P to points moved by p. Hence P is in fact a normal Sylow 3-subgroup of A, which does not exist.

In particular, A is a finite soluble group of automorphisms acting on the finite group G, p is a prime, but G has no non-identity A-invariant p-subgroups.

Coprime action would be enough to fix this, and is Kurzweil–Stellmacher's Proposition 8.2.3 on page 185.

Without coprime action every soluble, non-nilpotent group A = G provides a counterexample, simply because each such group has a non-normal Sylow p-subgroup for at least one prime p.

However, as Derek Holt points out, surely the problem is just suggesting that there is at least one p, not that all p work. This is very similar to Fitting's theorem in the A = G soluble case: every finite soluble group contains a non-identity normal p-subgroup for at least one prime p.

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I've got to run. I think this is right, and asked GAP to verify it, but again today is very rushed, so my apologies in advance if something went wrong. –  Jack Schmidt Jan 23 '12 at 21:45
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I assumed the problem was to show that there is a (presumably nontrivial) $A$-invariant $p$-subgroup for some prime $p$, not necessarily for all primes $p$ dividing $|G|$. –  Derek Holt Jan 23 '12 at 22:30
    
That sounds much better. Otherwise take g=a solvable but not nil potent. –  Jack Schmidt Jan 23 '12 at 22:42
    
I think I might have missed the point of this answer. Surely, the exercise asks for just one prime, not all. Also, it's an exercise from Issac's book, FGT, not Kurzweil-Stellmacher. In particular, it's exercise 3E.1 –  the_fox Jan 24 '12 at 1:42
    
You should include the hint then! –  user641 Jan 24 '12 at 2:00
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