Suppose $G$ is an abelian group. Define $ord(a)=m$ and $ord(b)=n$ where $a,b \in G$. Show that $ord(ab) | \dfrac{mn}{\gcd(m,n)}$ and $\dfrac{mn}{\gcd(m,n)^2}|ord(ab)$.
Since $(ab)^{\dfrac{mn}{\gcd(m,n)}}=((a^m)^n(b^n)^m)^{\dfrac{1}{\gcd(m,n)}}=1$, we have $ord(ab) | \dfrac{mn}{\gcd(m,n)}$
For the second part, let $ord(ab)=\dfrac{mn}{\gcd(m,n)^2}q+r$, where $0 \leq r \leq \dfrac{mn}{\gcd(m,n)^2}-1$.
Notice that $1=(ab)^{ord(ab)}=(ab)^{\dfrac{mn}{\gcd(m,n)^2}q+r}=(ab)^r$
But $r < \dfrac{mn}{\gcd(m,n)^2}$, Hence, $r=0$ and proven.
Does this proof work?
