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?
 A: You used the fact that $xy=yx$ when you split up exponents over $xy$ - in general, for any $d>1$,
$$(xy)^d=\underbrace{(xy)(xy)\cdots(xy)}_{d\text{ times}}\neq \left(\underbrace{ x\cdot x\cdots x}_{d\text{ times}}\right)\left(\underbrace{ y\cdot y\cdots y}_{d\text{ times}}\right)=x^dy^d$$
unless you can turn the successive $yx$'s on the left into $xy$'s: 
$$\underbrace{(xy)(xy)\cdots (xy)}_{d\text{ times}}=x\underbrace{(yx)(yx)\cdots(yx)}_{(d-1)\text{ times}}y=x\underbrace{(xy)(xy)\cdots(xy)}_{(d-1)\text{ times}}y=x^2\underbrace{(yx)(yx)\cdots(yx)}_{(d-2)\text{ times}}y^2$$
$$(\text{ repeat })\cdots = x^dy^d$$

Also, note that you did not need to use a proof by contradiction - you have directly proven that 
$$(xy)^{lm}=x^{lm}y^{lm}=(x^l)^m(y^m)^l=e^me^l=ee=e$$
so there was no need to assume $\text{ord}(xy)=\infty$ and then show it was false. It is not wrong, but it is much cleaner to say  

Here is a proof of claim $P$.  

than to say 

Suppose claim $P$ were false. But here is a proof that it is true; contradiction. Thus, our assumption that claim $P$ is false must have been false, i.e. the claim $P$ is true.

A: 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.
