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Let $G=PSL(2,q)$ where $q$ is prime power. What is Aut$(G\times G)$ and Aut$(G\times G\times G)$? Also if $G=A_{n}$ where $A_{n}$ is the alternating group of degree $n$, then what is Aut$(G\times G)$?

Thanks in advance

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In general, if $S$ is a finite non-Abelian simple group, and $E$ is a direct product of $n$ copies of $S$, then ${\rm Out}(E) = {\rm Aut}(E)/E$ is isomorphic to ${\rm Out}(S) \wr S_{n}.$ This is because every minimal normal subgroup of $E$ is isomorphic to $S$ (in fact, is one of the obvious simple direct factors of $E$) and the automorphism group of $S$ permutes the minimal normal subgroups of $E$. To provide more detail in order to make up for the lack of a reference: The $n$ "obvious" simple direct factors of $E$ are called the components of $E.$ The direct product of $n$ copies of ${\rm Aut}(S)$ obviously sits inside ${\rm Aut}(E).$ Furthermore, the assumed isomorphisms between the $n$ components may be included to show that ${\rm Aut}(S) \wr S_{n}$ embeds in ${\rm Aut}(E).$ On the other hand, the permutation action of ${\rm Aut}(E)$ on the components of $E$ gives a homomorphism from ${\rm Aut}(E)$ to $S_{n}.$ The kernel of this homomorphism is the intersection $K$ of the normalizers of the individual components. Since $E$ contains its centralizer in ${\rm Aut}(E)$, the group $K/E$ is isomorphic to a subgroup of a direct product of $n$ copies of ${\rm Out}(S).$ Hence this establishes that $|{\rm Aut}(E)| \leq | {\rm Aut}(S) \wr S_{n}|.$ But we have the inequality the other way round, so ${\rm Aut}(E) \cong {\rm Aut}(S) \wr S_{n}.$

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See this wikipedia article. The result mentioned there implies that the automorphism group is the wreath product power of the automorphism group of $G.$

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@Igor: I think your answer is basically the same as Geoff's. Though he doesn't provide a handy reference, I'm always reluctant to rely on the (usually anonymous) Wikipedia articles and their somewhat random reference lists. Too often those articles contain small errors, serious omissions, or distortions, though this one probably is OK. –  Jim Humphreys Jul 16 '12 at 18:10
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