D+F 5.5.2

Let $H,K$ be groups, $\varphi: K \longrightarrow Aut(H)$ and $H,K \leq H \rtimes_\varphi K$.
Prove that $C_H(K)=N_H(K)$
Note: we may also assume that $H \unlhd G$ and $H \cap K = \{e\}$ by the construction of the semi-direct product $H \rtimes_\varphi K$.

$\textit{Proof}$ :
"$\subseteq$" If $h\in{C_H(K)}$ then $hk=kh \quad \forall k\in{K}$. Rearranging, $hkh^{-1}=h \Longrightarrow hKh^{-1}=H \Longrightarrow h\in{N_H(K)}$

"$\supseteq$" Let $h\in{N_H(K)}$. Look at the commutator $[h,k]=hkh^{-1}k^{-1}$ :

$h(kh^{-1}k^{-1}) \in{H}$ since $H \unlhd G=H \rtimes_\varphi K$
$(hkh^{-1})k^{-1} \in{K}$ since $h \in{N_H(K)}$. We also have that $H \unlhd G$ and $H \cap K = \{e\}$. Thus, $$hkh^{-1}k^{-1}=e \Longrightarrow hk=kh \Longrightarrow h\in{C_H(K)} \quad \blacksquare$$