ord((a,b)) in group theory

Question

$G,H$ are groups and $a \in G$ and $b\in H.$ $\operatorname{ord}(a)=n$ and $\operatorname{ord}(b)= m.$ I need to find the order of $(a,b) \in G\times H.$ I know it supposed to be something with lcm but doesn't have a clue how to get to there.

Answer 1

We have that $(a,b)^k=(a^k,b^k)$. We want to show that the minimum positive integer $k$ such that $(a^k,b^k)=(e_G,e_H)$ is $k=\text{lcm}(n,m)$.

Fact $1$: $a^k=e_G$ iff $k\equiv 0 \pmod n$ and, similarly, $b^k=e_H$ iff $k\equiv 0 \pmod m$.

For fact $1$, we write $k=qn+r$ with $q\in \mathbb{Z}$ and $0\leqslant r<n$. Then $a^k=(a^n)^q a^r=a^r$, which equals $e_G$ iff $r=0$.

Fact $2$: The smallest positive integer $k$ which satisfies $k\equiv 0 \pmod n$ and $k\equiv 0\pmod m$ is $\text{lcm}(n,m)$.