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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 sum 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 ${\rm Aut}(E) \cong {\rm Aut}(S) \wr S_{n}.$

4 typo; added 11 characters in body

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 sum 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 ${\rm Aut}(E) \cong {\rm Aut}(S) \wr S_{n}.$

3 typos, grammar

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 $S$ (n in fact, are is one of the obvious simple direct factorsof factors of $E$) and the automorphism group of $S$ permutes the minimal normal subgroups of $E$.

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