Order and Least Common Multiple Abelian Question

Question

\item Let $G$ be an abelian group and let $x, y\in G$ be elements so that $o(x)=m$ and $o(y)=n$. Show that $o(xy)=\frac{mn}{(m,n)}$. (Note that this is the least common multiple of $m$ and $n$) Is this true if $G$ is non-abelian? Give an example.

My Solution

Let $r$ be the least common multiple of $m,n$ then $r = zm =yn$ for some integers $y,z$ so then we can write $(ab)^{r} = a^r b^r = a^{(m)z}b^{(n)y}= e^ze^y = e$ Since $(ab)^r =e$ then the order of $ab$ must divide $r$

My question lies in where this fails if $G$ is non-abelian. I know it fails, but what is a good example for this?

Answer 1

take the non-abelian group on two generators, $x$ and $y$ with $x^2 = y^2 = e$. in this case $xy$ generates an infinite cyclic subgroup

Answer 2

The first statement is wrong even in abelian groups. Take $o(x)=2$. We have $o(xx)=1 \neq 2= \frac{2 \cdot 2}{(2,2)}$.

It becomes true, when you require $(m,n)=1$.