Prove that the closure $\overline{H}$ of a subgroup $H \subseteq G$ of a topological group, is a subgroup.

Question

I'm tackling the problems in `Introduction to Topological Manifolds' by Lee, and this is the first time I've been tackling with topological groups. This is problem 13-9:

"Let $G$ be a topological group and let $H \subseteq G$ be a subgroup, then show that it's closure $\overline{H}$ is also a subgroup."

Intuitively, I can sort of visualize this for $\mathbb{R}$ but it's hard to imagine (at least for me) this being true in general for groups since there's so many different kinds of groups. Because of this, I just don't trust my proof. My attempt at a proof goes like this:

"By Proposition 2.30, and since $+ : G^2 \to G$ is a continuous map by definition of a topological group, it follows that for any $A \subseteq G^2$ that $+ \overline{(A)} \subseteq \overline{+(H)}$.

Consider the restriction of the group operation, $+|_{\overline{H}} : \overline{H}^2 \to G$. Then, $$x+y \in +(\overline{H^2}) \implies x+y \in \overline{+(H^2)}.$$ But since $+(H^2) = H$ by the closure of the subgroup operation, then it follows that $\overline{+(H^2)} = \overline{H}$ and hence our restricted group operation only targets $\overline{H}$, granting the property of closure and proving it's a subgroup."

My notation might be a little strange, but because topological groups require you to operate on the images and domains when dealing with continuity, I don't really have a better notation than to just treat $+$ as a function and write $+(A)$, so apologies if that makes this hard to read.

Answer 1

I felt like something was missing, and as Alex Ortiz said, it was the lack of inverses. I'll post my own answer, but leave the question open just to verify my proof is correct.

Part two of the proof is as follows:

Let $i|_{\overline{H}}$ be the restriction of the inverse map to $\overline{H}$. Then since $G$ is a topological group, the inverse mapping is continuous and hence $i({\overline{H}} ) \subseteq \overline{i({H})}$ which is equal to $\overline{H}$ since the inverse mapping is closed under $H$ as a subgroup. Therefore, our restriction of the inverse map only targets the closure of $H$, granting closure of inverse elements.

Answer 2

Don't forget that to justify $\overline H$ is a subgroup, you also need to address taking inverses of elements in $\overline H$.

Here is perhaps a more natural way to write a proof using sequences. Of course, not every topological group is first countable (and hence a proof using sequences is not always available), but you can update the following argument to work with nets instead of sequences if you are really curious. The essence of the argument is the same if you replace sequences by nets.

If $x,y\in \overline H$, then let $x_n\to x$ and $y_n\to y$ with $x_n,y_n\in H$ for every $n$. The group operation $\cdot$ is continuous, so $x_n\cdot y_n\to x\cdot y$. As $x_n\cdot y_n\in H$ for every $n$ ($H$ is a subgroup), this shows $x\cdot y\in \overline H$.

A similar argument works with inverses.