Prove that there exist integers $k, l \in \mathbb{N}$ such that $\mathrm{ord}(x^ky^l) = \mathrm{lcm}(\mathrm{ord}(x), \mathrm{ord}(y))$ Let $x, y \in G$, where $G$ is a group, $\operatorname{ord}(x), \operatorname{ord}(y) \in \mathbb{N} < \infty$, and $xy=yx$ (seems like it's very important).
I want to prove that there exist integers $k, l \in \mathbb{N}$ such that $\operatorname{ord}(x^ky^l) = \operatorname{lcm}(\operatorname{ord}(x), \operatorname{ord}(y))$.
I know these two facts:

*

*If $\operatorname{gcd}(\operatorname{ord}(x), \operatorname{ord}(y)) = 1 \Rightarrow \operatorname{ord}(xy) = \operatorname{ord}(x)\operatorname{ord}(y)$.

*For $g \in G$ and $n \in \mathbb{N}$, $\operatorname{ord}(g^n) = \frac{\operatorname{ord}(g)}{\operatorname{gcd}(\operatorname{ord}(g), n)}$.

Looks like these two facts can help me, but I have no idea how.
And why is $xy=yx$ so important? I tried to consider infinite non-abelian groups (For example $2\times2$ matrices), but I couldn't find any counterexamples. Maybe you can help me.
 A: Let: $ord(x)=m$, $ord(y)=n$, $d=gcd(m,n)$, then:
$$m=\overline{m}d, n=\overline{n}d,$$ with $gcd(\overline{m},\overline{n})=1$.
The idea is to find a partition $d=d_1d_2$, such that $gcd(d_1\overline{m},d_2\overline{n})=1$.
From fact $2)$, there exist $\overline{x} \in G$, with $ord(\overline{x})=d_1\overline{m}$, and $\overline{y} \in G$, with $ord(\overline{y})=d_2\overline{n}$. Due to fact $1)$, it gives us: $$ord(\overline{x}\overline{y})=ord(\overline{x})ord(\overline{y})=d\overline{m}\overline{n}=lcm(m,n).$$
In fact, $\overline{x}=x^{d/d_1}$ and $\overline{y}=y^{d/d_2}$, so we can take $k:=d/d_1$ and $l:=d/d_2$, and we are done.
Now, returning to the multiplicative partition of $d$:
Take $d=p_1^{\alpha_1}...p_h^{\alpha_h}$ its decomposition into primes. For each $p_i^{\alpha_i}$, $1\leq i \leq h$, we have three cases:
$i)$ $p_i\mid \overline{m}$, then $d_1$ will contain $p_i^{\alpha_i}$, whereas $p_i\nmid \overline{n}$;
$ii)$ $p_i\mid \overline{n}$, then $d_2$ will contain $p_i^{\alpha_i}$, whereas $p_i\nmid \overline{m}$;
$iii)$ $p_i\nmid \overline{m}$ and $p_i\nmid \overline{n}$, then either $d_1$ or $d_2$ will contain $p_i^{\alpha_i}$.
