If $\mathbb Z_m\times\mathbb Z_n$ is cyclic, then it's generated by $(\mathrm{gen}(F),\mathrm{gen}(G))$ My question is quite simple, I'm trying to formalize if $\mathbb Z_m\times\mathbb Z_n$ is a cyclic group, then it's generated by $(1,1)$. Is that true, the generalization? If $F\oplus G$ is the direct sum of the groups $F$ e $G$, which is cyclic, then $F\oplus G$ is generated by $(\mathrm{gen}(F),\mathrm{gen}(G))$? Thanks in advance

It is a theorem that $\mathbb{Z}_m\times\mathbb{Z}_n$ is cyclic iff $\gcd(m,n)=1$. Given that $\gcd(m,n)=1$, it's very straight forward to compute that the order of $(1,1)=mn$, since $mn/\gcd(m,n)=\operatorname{lcm}(m,n)$.

Your generalization is false, but can be fixed. Let $\langle g\rangle=G$ and let $\langle h\rangle=H$. Then $\gcd(o(g),o(h))=1$ iff $\langle(g,h)\rangle=G\times H$, and the proof is not harder than my comments on $\mathbb{Z}_m$