## In a semidirect product, the centralizer of the normal factor in the nonnormal factor is the kernel of the defining homomorphism

Let $H$ and $K$ be groups, let $\varphi : K \rightarrow \mathsf{Aut}(H)$ be a homomorphism, and let $G = H \rtimes_\varphi K$. Identify $H = H \times 1$ and $K = 1 \times K$ as subgroups of $G$.

Prove that $C_K(H) = \mathsf{ker}\ \varphi$. (Recall that $C_K(H) = C_G(H) \cap K$.)

$(\subseteq)$ Let $(1,k) \in C_K(H)$. By definition, if $(h,1) \in H \times 1$, then $(h,1) \cdot (1,k) = (1,k) \cdot (h,1)$. By the definition of multiplication in a semidirect product, we have $(h \cdot \varphi(1)(1), k) = (1 \cdot \varphi(k)(h), k)$, so that $(h,k) = (\varphi(k)(h),k)$. Comparing first coordinates, $h = \varphi(k)(h)$. Since $h \in H$ is arbitrary, we have $\varphi(k) = \mathsf{id}_H$, so that $k \in \mathsf{ker}\ \varphi$.

$(\supseteq)$ Suppose $(1,k) \in \mathsf{ker}\ \varphi \leq K$. Then $\varphi(k) = \mathsf{id}_H$. Now let $(h,1) \in H \times 1$ be arbitrary. Then $(1,k) \cdot (h,1) = (\varphi(k)(h), k)$ $= (h,k)$ $= (h \cdot \varphi(1)(1), k)$ $= (h,1) \cdot (1,k)$. Thus $(1,k) \in C_K(H)$.

If we identify $H$ and $K$ with their isomorphic copies in $G$, it will be much simpler: $k \in C_K(H)$ iff $khk^{-1}=\varphi(k)(h)=h, \forall h \in H$ iff $k \in \mathsf{ker} \varphi$.