# Order of Product of Commuting Group Elements of Coprime Order is Product of Orders

## Theorem

Let $G$ be a group.

Let $g_1, g_2 \in G$ be commuting elements such that:

 $\displaystyle \order {g_1}$ $=$ $\displaystyle n_1$ $\displaystyle \order {g_1}$ $=$ $\displaystyle n_2$

where $\order {g_1}$ denotes the order of $g_1$ in $G$.

Let $n_1$ and $n_2$ be coprime.

Then:

$\order {g_1 g_2} = n_1 n_2$

## Proof

Let $g_1 g_2 = g_2 g_1$.

We have:

$\paren {g_1 g_2}^{n_1 n_2} = e$

Thus:

$\order {g_1 g_2} \le n_1 n_2$

Suppose $\order {g_1 g_2}^r = e$.

Then:

 $\displaystyle {g_1}^r$ $=$ $\displaystyle {g_2}^{-r}$ $\displaystyle$ $\in$ $\displaystyle \gen {g_1} \cap \gen {g_2}$ $\displaystyle \leadsto \ \$ $\displaystyle {g_1}^r = {g_2}^r$ $=$ $\displaystyle e$

Thus $r$ is divisible by both $m$ and $n$.

The result follows.

$\blacksquare$