# Open Profinite Subgroup $\Rightarrow$ Profinite Group

I'm currently struggling with profinite groups. I'm not able to solve many practice exercises. I was wondering if someone could give me some hints on the following: if $G$ is a compact group and $H$ is an open profinite subgroup show $G$ is profinite.

You have to show that $G$ is Hausdorff, compact and totally disconnected
Hausdorff, let $x,y\in G$, $L_x$ the map defined by $L_x(u)=xu$. $L_x(H)$ is an open neighborhood of $x$. If $y\in L_x(H)$, $y=xh,h\in H$ since $H$ is Hausdorff, there exists neighborhoods $U$ of $1$ and $V$ of $h$ such that $U\cap V$ is empty. $U$ and $V$ are also open neighborhood of $1$ and $h$ in $G$ since $U$ is open. This implies that $L_x(U)$ is an open neighborhood of $x$ and $L_x(V)$ is an open neighborhood of $y$ such that $L_x(U)\cap L_x(V)$ is empty. If $y\neq L_x(H)$, since $H$ is profinite, it is compact, so $L_x(H)$ is compact as the image of a compact set by a continuous map. This implies that $W$ the complementary subset of $L_x(H)$ is open. We have $y\in W$ and $W\cap L_x(H)$ is open.
Totally disconnected. Let $x\in G$ and $C$ the connected component of $x$ in $U$. $C'=L_{x^{-1}}(C)\cap H$ is open and closed since $H$ and $C$ are open and closed, it implies that $L_{x^{-1}}(C)\cap H$ is the connected component of $1$ in $H$ and in $G$, so $C'=\{1\}$ since $H$ is totally disconnect. This implies that $C=\{x\}$.