$G$ is a group with order $mn$ where $m$ and $n$ are coprime. Then for any $g$ there are $x$ and $y$ such that $xy=g=yx$ and $x^m=1=y^n$.

Question

Let $g$ be an element of a group $G$, and suppose that $|G|=mn$ where $m$ and $n$ are coprime. Prove that there are unique elements $x$ and $y$ of $G$ such that $xy=g=yx$ and $x^m=1=y^n$.

Answer 1

Suppose such $x,y $ exist. Then $g=xy = (xy)^1$.

$m, n $ are coprime so according to Bezout's identity there are integers $a,b $ such that $am+bn=1$.

Replacing the $1$ in the previous equality, $g=(xy)^{am+bn} $.

Given that $x,y $ commute and $x^m=y^n=1$, we get $g=x^{bn}y^{am} $.

We get the two equations $g=xy $ and $g=x^{bn}y^{am} $.

So $y = x^{-1}g $ and substituting $y $ in the second equality, $g = x^{bn}(x^{-1}g)^{am} = x^{bn}g^{am} $, so $x^{bn} = g^{1-am}$.

Using $am+bn=1$ twice, we get $x=g^{bn}$. This gives $y=g^{am} $.

You can verify that $x=g^{bn}, y=g^{am} $ verify your hypothesis, hence the existence.

Now, for uniqueness, I'll let you prove that $a $ is unique mod $n $ and $b $ is unique mod $m $ and why this is enough to assert uniqueness.
Hint: for the first part, try solving $tm+un =1$ for integers $t, u $, knowing that $am+bn=1$.