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Let $G$ be a simple group of order $p^{2}-1$ where $p$ is Mersenne prime.

I am looking for a contradiction when $p\mid |Aut(G)|$. Does anyone knows how we can get a contradiction? Thanks.

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Johannas (mathoverflow.net/users/22991) asks (in a now-deleted answer): Contradiction to what? That $G$ is simple? $$ $$ Rashed answers: Yes Contradiction to $G$ is simple. –  Anton Geraschenko Jul 28 '12 at 20:13
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Yes, if you mean contradiction to the simplicity of $G.$ Let $P$ be a Sylow $p$-subgroup of ${\rm Aut}(G).$ Let $Q$ be a $P$-invariant Sylow $q$-subgroup of $G$ where $q$ is an odd prime divisor of $p-1$ (there is one such as the number of Sylow $q$-subroups of $G$ is certainly not divisible by $p,$ while $P$ permutes them, so fixes at least one). $P$ permutes the non-identity elements of $Q$ by conjugation (thinking of $P$ as a subgroup of the semidirect product $GP).$ Since there are certainly fewer than $p$ such elements, each orbit must be of length $1$. In other words, $[P,Q] =1$ in the semidirect product. Let $R$ be a $P$-invariant Sylow $2$-subgroup of $G.$ Since $p+1$ is a power of $2$ and $q$ was an arbitrary odd prime divisor of $p-1,$ we have $G = RC_{G}(P).$ Hence $[G,P] = [R,P] \leq R. $ But $[G,P] \lhd G.$ If $P \neq 1,$ then we have $[G,P] \neq 1.$ Since $G$ is simple, $[G,P] = G \leq R.$ But then $G$ is a $2$-group and $|G| = 2,$ which is not the case.

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