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Suppose that $G$ is a non-abelian simple group and $r$ is a prime number such that $r$ not divide $|G|$. It is a question for me that whether $r$ can divide |Aut($G$)|?

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Do you want $G$ to be non-abelian? –  j.p. Dec 24 '12 at 7:52
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Take a look at the Suzuki groups, according to wikipedia they can have automorphisms of order $r = 3$. –  j.p. Dec 24 '12 at 8:08
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For a list of simple groups together with the order of the outer morphism group (recall $Out(G)=Aut(G)/Inn(G)$, where $Inn(G)$ are the conjugation automorphisms) see wikipedia: en.wikipedia.org/wiki/List_of_finite_simple_groups –  Lior B-S Dec 24 '12 at 8:34
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${\rm PSL}_2(32)$ with $r=5$ is an example. –  Derek Holt Dec 24 '12 at 15:17
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Some more examples: $\text{PSL}_2(128)$ for $r=7$, $\text{PSL}_2(243)$ for $r=5$, $\text{PSL}_3(32)$ for $r=5$. –  Alexander Gruber Dec 25 '12 at 6:30

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