Proof of $_()∩=1$,$$ is a holomorph, $$ automorphism group

Question

I was told that this was a simple proof, by definition of the centraliser and by definition of the intersection with the automorphism group, but I just don't see it.

• $A = \mathrm{Aut}(G)$
• $H = G \rtimes A$
• $C_H(G) = \{h \in H| hg = gh, \forall g \in G\}$

Does $C_H(G) \cap A = 1$ just come from the fact that $H$ doesn't contain automorphisms? I would love some clarity on this, I think maybe I might be confused on what kind of elements are in the holomorph.

Answer 1

Expanding on @Derek's comment, the best course of action is probably to just check what it means for two elements $(1,a)\in A$ and $(g,1)\in G$ to commute (here I identify the groups with their canonical antiprojections in the cartesian product $G\times A$): $(1,a)(g,1)=(a(g),a)$, whereas $(g,1)(1,a)=(g,a)$, which means that the only $a$ satisfying the condition is the identity.