Assume $a \in G$, where $G$ is a group and element $a$'s order is $mn$ with $(m,n) = 1$, the problem is to prove there exist elements $b$ and $c$ in the group where $a = bc = cb$ , $b$'s order is $m$ and $c$'s order is $n$ and $b,c$ are unique.
I have proven that the existence of b and c using $ms+nt = 1$, but still don't know how to prove the uniqueness.
If $a=bc$ and $b$ and $c$ commute, then$$a^m=b^mc^m=c^m$$and$$c=c^{ms+nt}=(c^m)^s=a^{ms}.$$By the same argument, $b=a^{nt}$.
Now, suppose that $m',n'\in\mathbb Z$ are such that $m's+n't$ is also $1$. Then $m'=m+kmn$ and $n'=n-kmn$ for some $k\in\mathbb Z$. Therefore, $a^{ms}=a^{m's}$ and $a^{nt}=a^{n't}$. So, $b$ and $c$ are unique.