# Product of Orders of Abelian Group Elements Divides LCM of Order of Product

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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:

\(\displaystyle c\) | \(=\) | \(\displaystyle r m\) | for some $r \in \Z$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle s n\) | for some $s \in \Z$ |

So:

\(\displaystyle \paren {a b}^c\) | \(=\) | \(\displaystyle a^c b^c\) | Power of Product of Commuting Elements in Semigroup equals Product of Powers | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a^{r m} b^{s n}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a^m}^r \paren {b^n}^s\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e^r e^s\) | Definition of Order of Group Element | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \order {a b}\) | \(\divides\) | \(\displaystyle c\) | 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: $\S 41 \beta$