Cyclic subgroups, flaw in proof Let $G$ be a group, denote for $g \in G:$ $\langle g \rangle=\{g^0,g^1,\ldots \}$ 
I should show that for $x,y \in G$ with $xy=yx$ if $ord(\langle x \rangle)=l<\infty$ and $ord(\langle y \rangle)=m<\infty$ then $ord(\langle xy \rangle)<\infty$
To show it I assumed $ord(\langle xy \rangle)=\infty$, then especially
$$((xy)^l)^m=(x^l)^m\cdot (y^m)^l=e^me^l=e\neq e$$
But that is a contradiction. However I did not use that $xy=yx$ so the proof should be incorrect, but where is my mistake and how to show it correctly?

You actually did use that property when you said 
$$((xy)^l)^m=(x^l)^m⋅(y^m)^l$$
since, for example
$$(xy)^3=xyxyxy$$
which is not necessarily the same as $xxxyyy$ if $x$ and $y$ do not commute.
