# Wanted: Group homomorphism $\operatorname{Aut}(G) \to H$ whose kernel is the set of inner automorphisms

By a straightforward computation, it is not hard to show that the set $\operatorname{Inn}(G)$ of the inner automorphisms of a group $G$ is a normal subgroup of $\operatorname{Aut}(G)$, see for example this question.

Typically, more insight is gained by identifying a normal divisor as the kernel of some homomorphism. For example, $A_n \triangleleft S_n$ is the kernel of the signum $\operatorname{sgn} : S_n \to (\{\pm 1\},\cdot)$, $\operatorname{SL}(n,K) \triangleleft \operatorname{GL}(n,K)$ is the kernel of the determinant $\det : \operatorname{GL}(n,K) \to K^\times$, and the center $Z(G) \triangleleft G$ is the kernel of the homomorphism $G \to \operatorname{Inn}(G)$, $g\mapsto (x\mapsto gxg^{-1})$.

Now I wonder:

For a given group $G$, is there some "natural" group homomorphism $\operatorname{Aut}(G) \to H$ into a group $H$ whose kernel is $\operatorname{Inn}(G)$?

Of course, a solution is given by the canonical projection $\operatorname{Aut}(G) \to \operatorname{Aut}(G)/\operatorname{Inn}(G)$, $\phi\mapsto \phi\operatorname{Inn}(G)$, but this is not what I'm looking for. By "natural", I mean some homomorphism which is of significant importance on its own, like in the above examples the signum or the determinant.

• Let $\Omega$ be set of all conjugacy classes of $G$, then $Aut(G)$ acts on $\Omega$. In that action $Inn(G)$ is contained in the kernel. In some cases, it is exactly equal to $Inn(G)$. – mesel Apr 25 '15 at 18:38
• Well the quotient group ${\rm Aut}(G)/{\rm Inn}(G)$ is known as the outer automorphism group of $G$ and denoted by ${\rm Out}(G)$, but that doesn't really answer the question. I would guess that there is no general anser to the question. – Derek Holt Apr 25 '15 at 18:44
• Another idea is the natural map $Aut(G)\rightarrow Aut(G^\text{ab})$. It certainly contains $Inn(G)$ in its kernel, and at least in cases when the abelianization is large (ie, free groups), the kernel is precisely $inn(G)$. – oxeimon Apr 25 '15 at 19:22