Subsemigroup of a topological group I'm working in a problem of topological groups a few days and I'm not successful. Here is the problem:

Let $G$ be a topological group such that $G = SK$ for subsets $S$ and $K$ of $G$ with compact $K$. Show that if $S$ is closed and $SS \subset S$, then $S$ is a subgroup of $G$.

Honestly, I didn't have any good idea. I cannot see how I should use the compactness of $K$ and how connect the hypothesis. My approaches were in the following way: take $s \in S$ and suppose that $s^{-1} \not\in S$, then $s^{-1} \in S^c$ with $S^c$ open, but $S^c$ is not necessary a subgroup so, I don't know if it is useful to find a smart cover of $K$. Anyway.. I appreciate any hints.
 A: Let's consider a particular case: $S$ closed subset, $S\cdot S \subset S$, of $G$ compact topological group ( so $K=G$).
It is enough to show that for every $g \in G$ there exists a subsequence $g^{n_k} \to e$.
I will prove this only in the case $G$ compact connected Lie group. Then every $g\in G$ is contained in a maximal torus. Now we can reduce to the case $G = (\mathbb{R}/\mathbb{Z})^d$. So let $\alpha = (\alpha_1, \ldots, \alpha_d) \in (\mathbb{R}/\mathbb{Z})^d$. Now all will follow from the Dirichlet's principle: consider several multiples $k \alpha$, $1 \le k \le N$. Now divide $[0,1]^d$ into small boxes. If $N$ is large enough two distinct $k \alpha$, $l \alpha$ will be in the same box, and this means $(l-k) \alpha \simeq 0$.
We could probably extend this to $G$ compact Lie group ( so finitely many connected components).
Now, we know that every compact topological group is a projective limit of compact Lie groups. Can this be applied to get proof in general?
Maybe we could try to prove the statement for $G$ a countable product of finite groups ( so a totally disconnected group).
$\bf{Added:}$ The argument with boxes seems to be working for any compact group $G$. First, a simple lemma: given $U$ a neighborhood of $e$ in $G$, there exists $V$ a neighborhood of $e$ such that $g V g^{-1} \subset U$ for all $g \in G$.
Now, take any $U$, and consider $V$ such as above. Now take $W$ another neighborhood of $e$ with $W \cdot W^{-1} \subset V$. Now finitely many left translates of $W$ cover $G$.  Therefore, there exist powers $g^m$, $g^n$ in the same translate $a \cdot W$. We get
$$g^{n-m} =  a w_2\cdot (a w_1)^{-1} = a w_2 w_1^{-1} a^{-1} \in a V a^{-1} \subset U$$
$\bf{Added:}$  @kabenyuk: pointed out the paper of Numakura that appears very interesting. It seems that with the argument above one can show the following: if $S$ is a compact semigroup inside a topological group $G$, then $S$ is a subgroup.
