= Order of a direct-product element
{title2=$\operatorname{ord}(g,h)=\operatorname{lcm}(\operatorname{ord}g,\operatorname{ord}h)$}
If the <order of a group element> is $m$ for $g$ and $n$ for $h$, with both finite, then
$$
(g,h)^k=(g^k,h^k)=(e_G,e_H)
$$
exactly when both $m$ and $n$ divide $k$. The smallest positive such $k$ is their <least common multiple>, giving
$$
\operatorname{ord}(g,h)=\operatorname{lcm}(m,n).
$$
This concerns the orders of the selected elements, which need not equal the orders of their entire <groups>. If either component has infinite <order of a group element>, no positive power can return the pair to the <identity element>, so the pair also has infinite order.
For example, in $C_4\times C_6$, the additive element $(2,3)$ has component orders $2,2$, hence pair order $2$. The element $(1,1)$ instead has order $\operatorname{lcm}(4,6)=12$. The product has $24$ elements, so even this latter element does not generate it. This distinguishes <group order> from <order of a group element> while exposing the simultaneous-return interpretation.
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