Order of Group Element in Group Direct Product

Theorem
Let $G$ and $H$ be finite groups.

Let $g \in G: \left|{g}\right| = m, h \in H: \left|{h}\right| = n$.

Then $\left|{\left({g, h}\right)}\right|$ in $G \times H$ is $\operatorname{lcm} \left\{{m, n}\right\}$.

Proof
Let $G$ and $H$ be a groups whose identities are $e_G$ and $e_H$.

Let $l = \operatorname{lcm} \left\{{m, n}\right\}$.

From the definition of lowest common multiple:
 * $\exists x, y \in Z: l = m x = n y$

From the definition of order of an element:
 * $g^m = e_G, h^n = e_H$

Thus:

Now suppose $\exists k \in \Z: \left({g, h}\right)^k = \left({e_G, e_H}\right)$.

It follows that:
 * $\left({g, h}\right)^k = \left({g^k, h^k}\right) = \left({e_G, e_H}\right)$

Thus:
 * $g^k = e_G, h^k = e_H$

It follows from $g^k = e_G$ and $\left|{g}\right| = m$, by Integer Absolute Value Greater than Divisors, that:
 * $m \mathrel \backslash k$.

Similarly it follows that:
 * $n \mathrel \backslash k$.

So $k$ is a positive common multiple of both $m$ and $n$.

Since $l$ is the least common multiple:
 * $l \le k$

Therefore $l$ is the smallest such that $\left({g, h}\right)^l = \left({e_G, e_H}\right)$.

The result follows by definition of the order of an element.