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 $$\mathrm{lcm} \left\{{m, n}\right\}$$.

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

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

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

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

Thus:

$$ $$ $$ $$

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

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 $$m \backslash k$$.

Similarly it follows that $$n \backslash k$$.

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 $$\left({g, h}\right)^l = \left({e_G, e_H}\right)$$ and the result follows by Order of an Element.