If $G$ is a group, $x,y, \in G$ and $xy =yx$ then prove that $|xy| = \operatorname{lcm}\{|x|,|y|\}$ given that $|x|$ and $|y|$ are coprime

Question

Well I know that since $|x|,|y|$ are both positive integers then we have that $$|x||y| = \gcd(|x|,|y|)\operatorname{lcm}(|x|,|y|)$$

And since we are given that $|x|,|y|$ are coprime then we have $gcd(|x|,|y|)=1$ and so $$|x||y| = \operatorname{lcm}(|x|,|y|)$$

Now all I need to show is that $|x||y| = |xy|$.

Now let $|x| =n,|y| = m$ and also let $|xy|=k$ and consider $(xy)^{mn}$.

$(xy)^{mn} = (x)^{mn}(y)^{mn} = (x^n)^m(y^m)^n = e^me^n = ee= e$. However, since we have $|xy|=k$ then we know that $$k\leq mn.$$ Now I need to show that $mn \leq k$ to conclude that $k=mn$ and so $|x||y| = |xy|$. But I am stuck in here. I know for sure that I need to use the fact that $xy=yx$. But I dunno how

Answer 1

If $k$ is a common multiple of $m$ and $n$, then $k=ma=nb$ for some $a,b\in\Bbb N^+$. So $$(xy)^k=x^ky^k=(x^n)^b(y^m)^a=e$$.

Therefore, $(xy)^k=e$ for every common multiple of $m$ and $n$.

Now suppose that $(xy)^k=e$, then $x^k=y^{-k}=y^{m-k}$. $|x^k|=n/d$ where $d=(n,k)$ and $|y^{m-k}|=n/d$ as well. But $|y^{m-k}|=m/d'$ with $d'=(m-k,m)$.

So, $$\frac nd= \frac md'\implies nd'=md.$$ Since $(m,n)=1$, $m|d'$ and $n|d$. So $n=d$ and $m=d'$, and so $x^k=e=y^k$.