Let $S$ be a compact subset of metrizable topological group $G$, such that $xy \in S$ if $x$ and $y$ are in $S$. Show that for each $x \in S$, $xS=S$.
I have the following suggestions. Let $y$ be a cluster point of the sequence $x,x^2,x^3, \cdots$ in $S$ show that $yS=\bigcap_{n=1}^{\infty} x^nS$; deduce that $yxS=yS$.
I tried to prove that $yS=\bigcap_{n=1}^{\infty} x^nS$, let $a \in yS$ then $a=ys$, where $s \in S$ and since $y$ is cluster point $x,x^2,x^3, \cdots$, for all $\varepsilon>0$, $ \ \exists b \in S$ such that $0<d(y,b)<\varepsilon$ and $b=x^k$ for some $k \in \mathbb{N}$, but I don't know to prove that $a \in x^nS$ for all $n \in \mathbb{N}$ using the previous part.
For any fixed $n$, multiplying by $x^{-n}$ is a homeomorphism, so $x^{-n}y$ will be a cluster point of $x^{-n+1},x^{-n+2}\dots,$, thus also of its subsequence $x,x^2,x^3,\dots$.
Since $S$ is compact and all $x^k\in S$ ($k>0$), we have $x^{-n}y\in S$.
Then $a=ys$ implies $x^{-n}a=(x^{-n}y)s\in S$, so $a\in x^n S$.
Conversely, if for all $n$ we have $x^{-n}a\in S$ then, as the mapping $g\mapsto g^{-1}a$ is also a homeomorphism, $y^{-1}a$ is a cluster point of $x^{-n}a$, so that $y^{-1}a\in S$, or put another way, $a\in yS$.