Sylow subgroups invariant under an automorphism

Let $G$ be a finite group and $\sigma$ an automorphism of $G$. Suppose $p$ is a prime and $\sigma$ has prime order $q \neq p$. It's easy to see that $\sigma$ fixes a Sylow $p$-subgroup of $G$ if $\sigma$ is fixed-point free or if $q$ does not divide the order of $G$.

Suppose that $q$ does divide the order of $G$. What reasonable assumptions on $G$ or on $H$, the fixed point subgroup of $\sigma$, would we have to make to guarantee that $\sigma$ fixes a Sylow $p$-subgroup of $G$? Any ideas? Thanks.

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If asking for $\sigma$ to fix a Sylow $p$-subgroup is too much to hope for, are there any conditions which guarantee that $\sigma$ will even fix a non-trivial $p$-group? – AJB May 28 '11 at 18:42
The normalizer of a Sylow $p$-subgroup of $G$ containing a Sylow $q$-subgroup of $G$ (or, equivalently, the number of Sylow $p$-subgroups being coprime to $q$) would be sufficient. – Derek Holt May 28 '11 at 19:43
Thanks. This would work. If it happens that $q$ does not divide the number of Sylow $p$-subgroups then $\sigma$ will fix a Sylow $p$-subgroup. What if $q$ does divide the number of Sylow $p$-subgroups? Are there any conditions we can impose to guarantee that $\sigma$ fixes any non-trivial $p$-subgroup of $G$? – AJB May 28 '11 at 20:21

This is a very general question, perhaps too general for a definitive answer, and as Jack Schmidt pointed out (in a now deleted comment), it is already a delicate question when $\sigma$ is an inner automorphism. There are bad examples (a little different from Jack's) for all symmetric groups of prime degree greater than $3$. If $p$ is a prime, and $G$ is the symmetric group $S_{p}$, then a Sylow $p$-subgroup $P$ of $G$ is self centralizing of order $p$, so $N_{G}(P)$ has order dividing $p(p-1)$. In fact the order is $p(p-1)$. Hence the only elements of order prime to $p$ which normalize a Sylow $p$-subgroup are powers of a $p-1$-cycle. Such elements (apart from the identity) have a unique fixed point, and all other cycles of equal length dividing $p-1$. There are many elements of prime order $q \neq p$ in $S_p$ which are not of this form, for example any element $\sigma$ of prime order $q$ dividing $p-2$.
Another type of example is provided by ${\rm GL}(n,p)$. If we take a prime $q$ such that $q$ divides $p^{n}-1$ but does not divide $p^{m}-1$ for any $m <n$, then ${\rm GL}(n,p)$ contains an element $\sigma$ of order $q$ which must act irreducibly on the natural module. Hence $\sigma$ can not normalize any non-trivial $p$-subgroup $P$ of ${\rm GL}(n,p)$, for if it did, it would stabilize the space of fixed points of $P$, which is proper and non-zero. Note that $P$ can be made as large as desired by making $n$ large enough (though the choice of $q$ will need to vary).
A weak form of the TTT states that if a Sylow $q$-subgroup $Q$ of $G$ contains a maximal Abelian normal subgroup $A$ with 3 or more generators, and such that $C_G(a)$is solvable for each non-identity element $a$ of $A$, then all maximal $A$-invariant $p$-subgroups of $G$ are conjugate via an element of $O_{q'}(C_{G}(A))$ ( $p$ a prime different from $q$). In particular, the number of such subgroups is prime to $q$. Hence, for example, if we assume $Q$ is $\sigma$-invariant (which we may, possibly after replacing $\sigma$ by an $H$-conjugate, where $H$ is the semi-direct product $G\langle \sigma \rangle$), then $Q$ will permute the maximal $A$-invariant $p$-subgroups by conjugation. The number of these is prime to $q$, so the number fixed by $Q$ (under conjugation) is prime to $q$. Those fixed by $Q$ are the maximal $Q$-invariant $p$-subgroups of $G$. Since $\sigma$ normalizes $Q$, these are in turn permuted by $\sigma$ under conjugation. Since their number is prime to $q$, one of them must be fixed by $\sigma$. Hence if $A$ normalizes a non-trivial $p$-subgroup of $G$, so does $\sigma$.