# automorphism group of groups [closed]

Given a group $G$, I would like to calculate $\operatorname{Aut}(G)$. From definition of $\operatorname{Aut}()$ we know:

• $\operatorname{Aut}(G)\le \operatorname{Sym}(G)$

If the group is finitely generated and abelian I should use those properties:

• $\operatorname{Aut}(G \times H) \cong\operatorname{Aut}(G) \times\operatorname{Aut}(H)$

• $\operatorname{Aut}( C_n) \cong \Phi(C_n)$

• $\operatorname{Aut}({\bf Z})\cong C_2$

Where $G,H$ are(not necessarily abelian) groups of coprime order and $C_n$ is cyclic of order n.

In case the group is not abelian, if the group $G$ is complete, then:

• $\operatorname{Aut}(G)=G$

In case of other groups (ex. semi-direct products) how should I proceed?

Edit:

I just want to know if there are other criterions for such calculations. For example: if I have a semi-direct product, what can I do?

## closed as too broad by Martin Brandenburg, Robert Wolfe, Moishe Kohan, RghtHndSd, TrueDefaultSep 14 '14 at 4:02

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• What is a complete group? Also, what are you actually asking about, exactly? – tomasz Sep 13 '14 at 22:53
• As far as I know there isn't a totally general way to compute automorphism groups. You might be interested in: en.wikipedia.org/wiki/SO(8)#Spin.288.29 – user148177 Sep 13 '14 at 23:03
• – user148177 Sep 13 '14 at 23:05
• Complete means centerless (i.e. Z(G)={0}) – user1118686 Sep 13 '14 at 23:13
• Aut(G x H) is not isomorphic to Aut(G) x Aut(H). Perhaps you want the orders of G,H to be coprime. – Martin Brandenburg Sep 14 '14 at 0:15

In either case you still need a way to describe $\operatorname{Aut}(G)$ when $G$ is indecomposable, for which there is no one single method. Automorphisms and central automorphisms of $p$-groups are still not a completely solved problem, for example.