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:
\(\ds \order {g_1}\) | \(=\) | \(\ds n_1\) | ||||||||||||
\(\ds \order {g_1}\) | \(=\) | \(\ds n_2\) |
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
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 $17$