# Holomorph of a group $G$, then the automorphism of $G$ are inner automorphisms

In my course notes of algebra it says:

Let $G$ be a group. Then $\mathrm{Aut}(G)$ acts on $G$ in a natural way through automorphisms. This allows us to consider $A:= G \rtimes \mathrm{Aut}(G)$. In this group every automorphism of $G$ is an inner automorphism. This group is called the holomorph of $G$.

I don't understand the statement concerning the inner automorphisms.

The first part means $G\rtimes \mathrm{Aut}(G)$ being a group with group operation $$(g,\theta) \cdot (h,\psi) = (gh^{\theta^{-1}}, \theta \psi)$$ right?

And an automorphism $\phi$ on a group $G$ is an inner automorphism if $g^\phi = h^{-1}gh$ for a certain $h\in G$.

I would think the last statement means something like: For any $\phi\in \mathrm{Aut}(G)$ $$(g,\theta)^\phi = (h,\psi)^{-1} (g,\theta) (h,\psi) \qquad \text{for a certain } (h,\psi) \in A$$

But this doesn't make any sense, since $(g,\theta)^\phi$ is not even defined. Can someone help me make sense of this last statement?

• I added the "group-theory" tag to your post. Cheers! – Robert Lewis Mar 4 '18 at 0:09

The precise wording would be that 'every automorphism of $G$ extends to an inner automorphism of $A$'.
Namely, conjugation by $(1,\theta)$ restricted to $G\subseteq A$ will give back just $\theta\in Aut(G)$.