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now I'm searching for general results to computations to automorphism groups.

Let G be a group. Is there any nice way of computing $\operatorname{Aut}(G)$ without using $\operatorname{Aut}(G)\cong \operatorname{Inn}(G)\rtimes \operatorname{Out}(G)$? I think: no!

Does $\operatorname{Aut}(\cdot)$ behave nicely in the sense, that it commutes with (semi)direct products?

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    $\begingroup$ It is not always true that ${\rm Aut}(G)\cong {\rm Inn}(G) \rtimes {\rm Out}(G)$. See math.stackexchange.com/questions/456073 $\endgroup$
    – Derek Holt
    Commented Aug 7, 2013 at 7:35
  • $\begingroup$ @DerekHolt: Is there a name for group $G$, where $Aut(G)\cong Inn(G)\rtimes Out(G)$ holds. $\endgroup$
    – Mebat
    Commented Aug 7, 2013 at 8:27
  • $\begingroup$ I've never come across such a name! I do not think that this property has been extensively studied. $\endgroup$
    – Derek Holt
    Commented Aug 7, 2013 at 12:59

1 Answer 1

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$\text{Aut}(\cdot)$ doesn't commute with direct products. For example, if $G$ is any nontrivial group, then $\text{Aut}(G\times G)$ includes not only automorphisms of the form $(x,y)\mapsto(\alpha(x),\beta(y))$ with $\alpha,\beta\in\text{Aut}(G)$ (which form a copy of $\text{Aut}(G)\times\text{Aut}(G)$), but also $(x,y)\mapsto(y,x)$.

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