I'm trying to prove this special case of a well-known result. I'm actually very happy because the entire proof is by myself. Could you please confirm if it is correct?
Let $G$ be an abelian group with elements $x, y$ of orders $m$ and $n$ respectively. Moreover, $\gcd(m,n)=1$. Then the order of $xy$ is $mn$.
My attempt:
WLOG, we assume $m<n$. Clearly, $(xy)^{mn}=x^{mn} y^{mn} = (x^m)^n (y^n)^m = 1$. Assume the contrary that $o(xy)=t$ for some $1<t <mn$. By Lagrange's theorem, $t | mn$. Because $\gcd(m,n)=1$, either $t | m$ or $t | n$.
- If $tq = m$ for some $1<q<m$
Then $(xy)^t = 1=x^m = x^{tq}$. Then $y^t = x^{t(q-1)}$ and thus $o(\langle y^t \rangle) = o(\langle x^{t(q-1)}\rangle)$. Because $\gcd(t,n) = 1$, $o(\langle y^t \rangle) = n$. It follows that $n = o(\langle x^{t(q-1)}\rangle) \le o(\langle x\rangle) = m$, which is a contradiction.
- If $tq = n$ for some $1<q<n$
Then $(xy)^t = 1=y^n = y^{tq}$. Then $x^t = y^{t(q-1)}$ and thus $o(\langle x^t \rangle) = o(\langle y^{t(q-1)}\rangle)$. Because $\gcd(t,m) = 1$, $o(\langle x^t \rangle) = m$. It follows that $m = o(\langle y^{t(q-1)}\rangle)$. On the other hand, $\langle y^{t(q-1)}\rangle$ is a subgroup of $\langle y \rangle$. By Lagrange's theorem, $m | n$, which is a contradiction.
Update: I've just found a simple fix
Let $t = o(xy)$. We have $(xy)^t = 1$ and thus $x^t = y^{-t}$ Then $k := o(\langle x^t\rangle) = o(\langle y^{-t} \rangle)$. This means $k| m$ and $k | n$. It follows that $k = 1$. Hence $\langle x^t\rangle = \langle y^{-t}\rangle = \{1\}$. Then $x^t = y^{-t}=1$ and thus $m|t$ and $n|t$. Because $\gcd(m,n)=1$, $t$ is a multiple of $mn$. Moreover, $(xy)^{mn}=1$. As such, $t=mn$.
