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

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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.

$\blacksquare$


Also see


Sources