TITLE: On the existentence of an element of a group whose order is the LCM of orders of given two elements which are commutaive
QUESTION [6 upvotes]: I came up with the following proposition.
Proposition
Let $G$ be a group.
Let $x, y$ be elements of finite order in $G$ such that $xy = yx$.
Let $n$ be the order of $x$.
Let $m$ be the order of $y$.
Let $l =$ lcm$(n, m)$.
Then there exists an element of order $l$ in $G$.
Outline of my proof
There exist a divisor $a$ of $n$ and a divisor $b$ of $m$ such that gcd$(a, b) = 1$ and $l = ab$.
There exist $x, y \in G$ whose orders are $a, b$ respectively.
Then $z = xy$ is of order $l$.
My question
How do you prove the proposition?
I would like to know other proofs based on different ideas from mine.
I welcome you to provide as many different proofs as possible.

REPLY [6 votes]: Suppose $m$ and $n$ are relatively prime. Let $\circ(xy) = s$. Since $(xy)^l =e$, we have $s \mid l$. Now $(xy)^s = e$, implies $x^sy^s =e$ and therefore $x^s=y^{-s}$. Now $x^{sn}=e$ and $y^{sm}=e$ which yield $m \mid ns$ and $n\mid sm$. Since $gcd(m,n)=1$, we have $m\mid s$ and $n\mid s$ and thus $mn=lcm(m,n)=s$.
Now we assume that $m$ and $n$ are not relatively prime. We write $l=p^{r1}_{1}⋯p^{rs}_{s}$, $p_i$ are primes and $ri \geq 1$. Using part 1, it is enough to show that for each $i$, there is an element of order $p^{ri}_{i}$.  Note that $p^{ri}_{i}$ divides either $m$ or $n$. Thus $y^{m/p^{ri}_{i}}$ or $x^{n/p^{ri}_{i}}$ (whichever one divides evenly) has order $p^{ri}_{i}$. Therefore we have shown that an element of order $l$, the least common multiple of $m$ and $n$, is in $G$.