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).

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