The product of two elements with finite order which commute in a group has finite order which divides the product of the orders of the elements.

If $G$ is finite and $a b = b a$ for fixed $a, b \in G$ , then $| a b |$ divides $(| a | | b |)$ .

Proof

Let $| a | = n$ , $| b | = n$ .
Since $a$ commutes with $b$ , $(a b)^{m n} = (a^{n})^{m} (b^{m})^{n} = e^{m} e^{n} = e$
Therefore, $| a b |$ divides $(| a | | b |)$ .

QED