TITLE: Prime divisors of the order of an automorphism group
QUESTION [7 upvotes]: Let $G$ be an unsolvable group of order $(p+1)t$, where $t\mid (p-1)/2$. Is it true that $p$ does not divide $|Aut(G)|$?

REPLY [5 votes]: I believe the answer is yes. Let $\alpha$ be an automorphism of $G$ of order $p$. By the Frattini argument, for any prime $q$ dividing $|G|$, a Sylow $q$-subgroup of $G$ is normalized by an element of order $p$ in the semidirect $S$ product of $G$ with $\langle \alpha \rangle$. Since all elements of order $p$ are conjugate in $S$, it follows that $\alpha$ normalizes some Sylow $q$-subgroup $Q$.
But, since the only prime that could divide both $p+1$ and $t$ is 2, we have $p > |Q|$ whenever $q>2$ and hence $\alpha$ centralizes $Q$. We also have $p > |Q|$ for $q=2$, and hence $\alpha=1$, except in the case when $p$ is a Mersenne prime.
So we have reduced to the case when $p$ is a Mersenne prime, and hence $t$ is odd. Now $\alpha$ centralizes a subgroup $T$ of $G$ of order $t$ and acts nontrivially on a Sylow 2-subgroup $Q$, which has order $p+1$ (and must be elementary abelian). In the semidirect product $S$, we have $[\langle \alpha \rangle, G] = [\langle \alpha \rangle, TQ] = [\langle \alpha \rangle, Q] = Q$, so $Q$ is normal in $G$. Hence, since the odd order subgroup $T$ is solvable, $G$ is solvable, contradiction.