Order of Product of Abelian Group Elements Divides LCM of Orders of Elements
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Theorem
Let $G$ be an abelian group.
Let $a, b \in G$.
Then:
- $\order {a b} \divides \lcm \set {\order a, \order b}$
where:
- $\order a$ denotes the order of $a$
- $\divides$ denotes divisibility
- $\lcm$ denotes the lowest common multiple.
Proof
Let $\order a = m, \order b = n$.
Let $c = \lcm \set {m, n}$.
Then:
\(\ds c\) | \(=\) | \(\ds r m\) | for some $r \in \Z$ | |||||||||||
\(\ds \) | \(=\) | \(\ds s n\) | for some $s \in \Z$ |
So:
\(\ds \paren {a b}^c\) | \(=\) | \(\ds a^c b^c\) | Power of Product of Commuting Elements in Semigroup equals Product of Powers | |||||||||||
\(\ds \) | \(=\) | \(\ds a^{r m} b^{s n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a^m}^r \paren {b^n}^s\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^r e^s\) | Definition of Order of Group Element | |||||||||||
\(\ds \) | \(=\) | \(\ds e\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \order {a b}\) | \(\divides\) | \(\ds c\) | Element to Power of Multiple of Order is Identity |
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 41 \beta$