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?

  • $\begingroup$ I added the "group-theory" tag to your post. Cheers! $\endgroup$ 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 $\theta\in Aut(G)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.