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:

\(\ds \order {g_1}\)

\(=\)

\(\ds n_1\)

\(\ds \order {g_1}\)

\(=\)

\(\ds n_2\)

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:

\(\ds {g_1}^r\)

\(=\)

\(\ds {g_2}^{-r}\)

\(\ds \)

\(\in\)

\(\ds \gen {g_1} \cap \gen {g_2}\)

\(\ds \leadsto \ \ \)

\(\ds {g_1}^r = {g_2}^r\)

\(=\)

\(\ds e\)

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

The result follows.

$\blacksquare$

Also see

• Unique Composition of Group Element whose Order is Product of Coprime Integers, a converse of this

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 $17$