Unique Composition of Group Element whose Order is Product of Coprime Integers

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $g \in G$ be an element of $g$.

Let:
 * $\left\lvert{g}\right\rvert = m n$

where:
 * $\left\lvert{g}\right\rvert$ denotes the order of $g$ in $G$
 * $m$ and $n$ are coprime integers.

Then $g$ can be expressed uniquely as the product of two commutative elements $a$ and $b$ of order $m$ and $n$ respectively.

Proof
Let $g_1 = g^n$ and $g_2 = g^m$.

By Powers of Group Element Commute, $g_1$ and $g_2$ commute.

We have:
 * ${g_1}^m = g^{n m} = e$
 * ${g_2}^n = g^{m n} = e$

It follows that:
 * the order of $g_1$ is $m$
 * the order of $g_2$ is $n$

otherwise if either $g_1$ or $g_2$ were of a smaller order then $g$ would also be of a smaller order.

By Bézout's Lemma:
 * $\exists u, v \in \Z: u n + v m = 1$

as $m \perp n$.

Thus:
 * $g = g^{u n + v m} = \left({g^n}\right)^u \left({g^m}\right)^v = {g_1}^u {g_2}^v$

Also by Bézout's Lemma:
 * $u \perp m$

and:
 * $v \perp n$

Thus by Order of Group Element equals Order of Coprime Power:
 * $\left\lvert{ {g_1}^u}\right\rvert = m$

and:
 * $\left\lvert{ {g_2}^v}\right\rvert = n$

We have that $g_1$ and $g_2$ commute.

So by Commutativity of Powers in Group, ${g_1}^u$ and ${g_2}^v$ also commute.

Putting $a = g_1^u$ and $b = g_2^v$, we have $a$ and $b$ which satisfy the required conditions.

It remains to prove uniqueness.

Suppose that:
 * $(1): \quad g = r_1 r_2 = s_1 s_2$

where:
 * $r_1$ and $r_2$ commute
 * $s_1$ and $s_2$ commute
 * $\left\lvert{r_1}\right\rvert = \left\lvert{s_1}\right\rvert = m$
 * $\left\lvert{r_2}\right\rvert = \left\lvert{s_2}\right\rvert = n$

Raising $(1)$ to the $n u$th power:
 * $g^{n u} = {r_1}^{n u} {r_2}^{n u} = {s_1}^{n u} {s_2}^{n u}$

and so:

It follows directly from $(1)$ that $r_2 = s_2$.

Hence the result.