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