## Extending automorphism groups to inner automorphism groups.

I wanted to make clear just a quick technical thing. If G is a group, N is a normal subgroup, and $\phi_g \in \text{Inn}(G), \phi_g(h) = g h g^{-1}$ then $\phi_g$ is an automorphism of N, right? However, is it the case that we cannot say that $\phi_g$ is an inner automorphism, since we are not guaranteed that $g \in N$? I think this is the case, but I just want to be clear.

Furthermore, is it then possible to extend any group to a larger group such that all automorphisms are given by inner automorphisms from the larger group? More precisely, if G is a group, is there always an injective homomorphism $G \to H$ such that $\text{Aut}(G) \cong \text{Inn}(H)$ ?

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Hi Kreizhn!

 Quote by Kreizhn I wanted to make clear just a quick technical thing. If G is a group, N is a normal subgroup, and $\phi_g \in \text{Inn}(G), \phi_g(h) = g h g^{-1}$ then $\phi_g$ is an automorphism of N, right? However, is it the case that we cannot say that $\phi_g$ is an inner automorphism, since we are not guaranteed that $g \in N$? I think this is the case, but I just want to be clear.
That is correct.

 Furthermore, is it then possible to extend any group to a larger group such that all automorphisms are given by inner automorphisms from the larger group? More precisely, if G is a group, is there always an injective homomorphism $G \to H$ such that $\text{Aut}(G) \cong \text{Inn}(H)$ ?
Indeed, let G be our group we wish to extend. Let T be a group such that there exists an epimorphism

$$\phi:T\rightarrow Aut(G)$$

(for example, take $T=Aut(G)$), then the semidirect product $G\rtimes_{\varphi} T$ is an extension you're looking for. Indeed, automorphism of G has the form $\phi(h):=\phi_h$. And by construction of the semidirect product, we have that for each g in G

$$\phi_h(g)=hgh^{-1}$$

So the automorphisms of G are inner automorphisms of the semidirect product...

 I shall have to go back and look at my notes on semi-direct products a bit more to understand this fully, but thanks.