Could someone please help me understand this proof?
If $(g,h)\in G\times H$ with $|g|=r$ and $|h|=s$, then $|(g,h)|=\operatorname{lcm}(r,s)$.
Let $m = lcm(r, s)$ and $n = |(g, h)|$. Then $(g, h)^{m} = (g^{m}, h^{m}) = (e_{G}, e_{H})$ since $m$ is the order of both $g$ and $h$.
I don't understand this part. How is $m$ the order of both $g$ and $h$? Is this a fundamental theorem?
Then $n \leq m$.
I also don't understand this part. If $m$ is the order of both $g$ and $h$, why is is greater than $n$?
Also, $(g, h)^{n} = (g^{n}, h^{n}) = (e_{G}, e_{H})$ since $n$ is the order of $(g, h)$. Then $g^{n} = e_{G}$ and $h^{n} = e_{H}$. Then $|g| = r$ divides $n$ and $|h| = s$ divides $n$.
I don't understand this division part, but I am sure I would understand if I understood the above confusions.
Then $n$ is a common multiple of $r$ and $s$. Since $m$ is the least common multiple of $r$ and $s$, $m \leq n$. Then $\operatorname{lcm}(r, s) = m = n = |(g, h)|$.