TITLE: closed subgroup on a topological group
QUESTION [10 upvotes]: Today  a student ask me the following question regarding topological groups in the tutorial centre. 

Let $H$ be a subgroup of a topological group $G$. Suppose that $U$ is an open containing the identity such that $\overline{U} \cap H$ is closed. Prove that $H$ is closed.

I am a bit rusty on point set topology so I couldn't figure out a proper solution. I tried a straightforward approach without success:

Let $x \in \overline{H}$, we want to show that $x \in H$. Since $e \in U$ then $x \in xU$, and we can take an element $y \in xU \cap H$, therefore $y = xa = h$ for some $a \in U$, $h\in H$, so $x = ha^{-1}$

Trying to prove that $a^{-1} \in H$, I want to use somehow that $\overline{U} \cap H$ is closed; but I'm stuck at this point. I wish I could dedicate more time to solve the problem and review the basic properties of topological groups, but unfortunely I am overhelmed with my own stuff.

REPLY [3 votes]: First, a bit of point set topology: $\bar U \cap H$ closed is equivalent to $U \cap H$ is closed in $U$. 
Second, by homogeneity, for every $h\in H$, there exists $U_h$ neighborhood of $h$ so that $U_h \cap H$ is closed in $U_h$ ( take $U_h = h\cdot U$)
The above means that $H$ is locally closed in $G$. This is equivalent to : $H$ is the intersection between a closed set and an open set, or, what we are really after: $H$ is open in its closure $\bar H$. 
Now, $\bar H$ is a topological group ( with the induced topology. Let's now recall: an open subgroup of a topological group is also closed. Indeed, any coset will be open, so any union of cosets. Therefore $H$ is also closed in $\bar H$, and so equal to $\bar H$, closed.
Take home message: a locally closed subgroup is closed.