I am asked to prove that if we have a topological group $G$ then if $H$ is open subgroup of $G$ we have that $H$ contains identity component of $G$. I dont see how those two relates. I know that identity component of $H$ is containted in identity component of $G$ but I dont know how to complete the proof, if its possible to do it this way anyway.
But this makes me think, what about the case when $G$ is connected topological group? Isnt in that case $G=G_0$, and doesnt this imply that $G$ doesnt have any open subgroup?
Since $H$ is open, for each $g\in G$, $gH$ is also an open set. Now, suppose that $g\notin H$. Then $gH\cap H=\emptyset$. So, $G\setminus H=\bigcup_{g\in G\setminus H}gH$, which is an union of open sets. Therefore, $G\setminus H$ is open, which means that $H$ is closed. So, since $e_G\in H$ and since the identity component $G_0$ of $G$ is the largest connected subset of $G$ to which $e_G$ belongs, $G_0\subset H$.
And it follows from this that if $G$ is connected, then its only open subgroup is $G$ itself.