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

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## Contents

## 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

- 1978: John S. Rose:
*A Course on Group Theory*... (previous) ... (next): $0$: Some Conventions and some Basic Facts: Exercise $6$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $17$