Showing that a topological group is compact under certain conditions

Question

Let $G$ be a topological group and suppose there exists a compact subset $K$ of $G$ such that $xK ∩ K \neq ∅$ for every $x ∈ G$. Show that $G$ is compact.

I'm new to this subject and I'm really stuck on this question. My thinking so far is this:

As $K$ is a compact subset of $G$ it must contain $1_G$, the identity element(?)

$\Rightarrow \bigcup _{x \in G} xK = G$

So what I need to show is that if $K$ compact and $xK ∩ K \neq ∅ \Rightarrow xK$ is compact.

$xK ∩ K \neq ∅$ for every $x ∈ G \Rightarrow 0 \in K$

Is the above correct and am I at all heading in the right direction? I also have from Tychonoff’s theorem that if $A,B$ compact then $A\times B$ is compact which implies that $AB$ is compact too from definition of a topological group.

Answer 1

Let $x\in G$, since $xK\cap K$ is not empty, there exists $k_1,k_2\in K$ such that $xk_1=k_2$, this implies that $x=k_2k_1^{-1}$. We deduce that $G\subset K.K^{-1}\subset G$. Where $K^{-1}=\{k^{-1},k\in K\}$. Since the inverse map $i:G\rightarrow G$ defined by $i(x)=x^{-1}$ is continuous, we deduce that $K^{-1}$ is compact since the image of a compact set by a continuous map is compact. Consider the map $f:K\times K^{-1}\rightarrow G$ defined by $f(x,y)=xy$. It is the restriction of the product of $G$ so it is continuous, we deduce that $f(K\times K^{-1})=G$ is compact since $K\times K^{-1}$ is compact (Tychonoff) and the image of a compact set by a continuous map is compact.