automorphism groups

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?

• It is not always true that ${\rm Aut}(G)\cong {\rm Inn}(G) \rtimes {\rm Out}(G)$. See math.stackexchange.com/questions/456073 – Derek Holt Aug 7 '13 at 7:35
• @DerekHolt: Is there a name for group $G$, where $Aut(G)\cong Inn(G)\rtimes Out(G)$ holds. – Mebat Aug 7 '13 at 8:27
• I've never come across such a name! I do not think that this property has been extensively studied. – Derek Holt Aug 7 '13 at 12:59

$\text{Aut}(\cdot)$ 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 $(x,y)\mapsto(y,x)$.