Problem 5.5.1 (If $G = H \rtimes_\varphi K$, then $C_K(H) = \ker(\varphi)$)

Question

Let $H$ and $K$ be groups, let $\varphi$ be a homomorphism from $K$ into $\mathrm{Aut}(H)$ and, as usual, identify $H$ and $K$ as subgroups of $G = H \rtimes_\varphi K$.

Prove that $C_K(H) = \ker(\varphi)$ (recall that $C_K(H) = C_G(H) \cap K$).

Richard Ganaye · Accepted Answer

\begin{proof} 
For clarity, we don't identify $H$ and $K$ as subgroups of $G = H times_\varphi K$, but we write
$$	ilde{H} = \{(h,1) \mid h \in H \}  = H 	imes\{1\} \simeq H,\qquad 	ilde{K} =  \{(1,k) \mid k \in K \} = \{1\} 	imes K \simeq K.$$
Here
$$\varphi
\left\{
\begin{array}{ccl}
K & 	o & \mathrm{Aut}(H),\
k & \mapsto & \varphi_k
    \left\{
    \begin{array}{ccl}
    H & 	o & H\
    h &\mapsto &k\cdot h
    \end{array}
    ight. 
\end{array}
ight.
$$
 For every $k \in K$,
\begin{align*}
(1,k) \in C_{	ilde{K}}(	ilde{H}) &\iff \forall h \in H,\ (h,1)(1,k) = (1,k)(h,1)\
&\iff \forall h \in H,\ (h,k) = (\varphi_k(h), k)\
&\iff \forall h \in H,\ h = \varphi_k(h)\
&\iff \varphi_k = \mathrm{id}_H\
&\iff k \in \ker(\varphi),
\end{align*}
In conclusion, for all $k \in K$, $(1,k) \in C_{	ilde{K}}(	ilde{H}) \iff k \in \ker(\varphi),$
so that
$$C_{	ilde{K}}(	ilde{H}) = \{1\} 	imes \ker(\varphi).$$
If we identify $K$ and $	ilde{K}$, $	ilde{H}$ and $H$, this shows
$$C_K(H)= \ker(\varphi).$$
\end{proof}

{\bf Note:} If $H$ is abelian, for all  $(h,k) \in G$ and all $x \in H$, 
\begin{align*}
(h,k) (x,1)(h,k)^{-1} &=(h\varphi_k(x)h^{-1}, 1)\
&= (\varphi_k(x), 1).
\end{align*}Hence, for all $(h,k) \in G$,
\begin{align*}
(h,k) \in C_{G}(	ilde{H}) &\iff \forall x \in H,\ (\varphi_k(x), 1) = (x,1)\
&\iff \forall x \in H,\ \varphi_k(x) = x\
&\iff \varphi_k = \mathrm{id}_H\
&\iff k \in \ker \varphi\
&\iff (h,k) \in H 	imes \ker(\varphi),
\end{align*}
($H 	imes \ker(\varphi)$ is here the {\bf set} $\{(h,k) \mid h \in H,\  k\in \ker(\varphi)\}$)

so that 
$$C_{G}(	ilde{H}) = H 	imes \ker(\varphi),$$
where the law on the subgroup $H 	imes \ker(\varphi)$ is the restriction of the law of $G$, so that
$$C_{G}(	ilde{H}) = H times_{\varphi'} \ker(\varphi),$$
where $\varphi' = \varphi \mid _{\ker(\varphi)}$ is the restriction of $\varphi$ to $\ker(\varphi)$, so $\varphi'$ is the trivial homomorphism, and the law on $H 	imes \ker(\varphi)$ is the direct product.
$$C_{G}(	ilde{H}) = H times_{\varphi'} \ker(\varphi) \simeq 	ilde{H} 	imes \ker(\varphi).$$

Hence
$$C_G(	ilde{H})/	ilde{H} \simeq \ker(\varphi).$$
(See the Appendix of Problem 7 for an application.)

Problem 5.5.1 (If $G = H \rtimes_\varphi K$, then $C_K(H) = \ker(\varphi)$)

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Problem 5.5.1 (If $G = H \rtimes_\varphi K$, then $C_K(H) = \ker(\varphi)$)

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