## The centralizer and normalizer of a nonnormal semidirect factor in a normal semidirect factor are equal

Let $H$ and $K$ be groups, let $\varphi : K \rightarrow \mathsf{Aut}(H)$ be a group homomorphism, and identify $H$ and $K$ as subgroups of $G = H \rtimes_\varphi K$.

Prove that $C_H(K) = N_H(K)$.

Recall that $C_H(K) \leq N_H(K)$ always holds.

$(\supseteq)$ Let $(h,1) \in N_H(K)$. Then for every $(1,k) \in K$, $(h,1)^{-1}(1,k)(h,1) \in K$. Carrying out this multiplication, we see that $(h^{-1} \varphi(k)(h),k) \in K$. Thus $h^{-1}\varphi(k)(h) = 1$, and we have $\varphi(k)(h) = h$. Then $(\varphi(k)(h),k) = (h,k)$, and evidently $(1,k) \cdot (h,1) = (h,1) \cdot (1,k)$. Thus $(h,1) \in C_H(K)$.