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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?

MotionMath
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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.

José Carlos Santos
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  • Hmmm, I proved that every open subgroup is also closed, which is proved here too. But last part I dont understant. First why does whole $H$ has to be in $G_0$? We dont know that $H$ is coonected? And if that is, how does that shows $G_0\subset H$? – MotionMath Jul 24 '21 at 17:10
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    Since $H$ is open and closed, $H\cap G_0$ s a subset of $G_0$ which is open and closed in $G_0$. But $G_0$ is connected. Therefore, $H\cap G_0=\emptyset$ or $H\cap G_0=G_0$. But $e_G\in H\cap G_0$. Therefore, $H\cap G_0=G_0$, and this is equivalent to asserting that $G_0\subset H$. – José Carlos Santos Jul 24 '21 at 17:13