On the existence of an element of a group whose order is the LCM of orders of given two elements which commute

Question

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.

Answer 1

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 $G$.

Answer 2

can you prove that if $x$ has order $m$ ,$y$ has order $n$ and $xy=yx$ then order of $xy$ is $lcm(m,n)$ if $gcd (m,n) =1$.