Disjoint left translates of a function from a non-compact locally compact group to R with compact support. I'm having trouble trying to prove an unjustified (and probably obvious!) statement in an academic paper. $G$ is a (locally compact) topological group which is not compact. $f$ is a continuous function $G \rightarrow \mathbb{R} $ with compact support. I wish to prove that there exists $x_1, ..., x_m \in G$ such that $f, x_1 f, ..., x_m f$ have pairwise disjoint support. Here $xf(y) = f(xy)$ for each $x,y \in G$ My approach was to prove $m=2$, and use induction to get it for the general case. I've tried to get a contradiction for $m=2$, and various other things, but I am stuck. It is pretty easy to prove for $\mathbb{R}$, but I used the Heine–Borel theorem and boundedness to do so. Any pointers?

One needs to prove: if $K,L\subset G$ are compact subsets, there exist $x\in G$ such that $K\cap xL=\emptyset$. Consider the map $f:K\times L\to G$, $f(a,b)=ab^{-1}$. It can't be surjective ($K\times L$ is compact, hence the image of $f$ is compact). If $x\in G$ is not in the image of $f$ then $K\cap xL=\emptyset$.