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
 A: 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 $\mathbb{Z}_m$
A: You have that the order of an element $\left(g,h\right)$ in a product $G\times H$ is the minimum common multiple of the two orders. Note that the product of two cyclic groups is cyclic if and only if their orders are coprime. Then, under this assumption, your statement follows, since the element $\left(1,1\right)$ has order the product of the two cardinalities, which is the cardinality of the product.
A: Your generalization is true, but it's not harder to prove than the fact that $(1,1)$ generates $\mathbb{Z}_m\times\mathbb{Z}_n$ (assuming the product is cyclic). Indeed, if $g$ is a generator of the finite cyclic group $G$, there is a unique isomorphism $\varphi_g\colon\mathbb{Z}_m\to G$ such that $\varphi_g(1)=g$ (where $m=|G|$).
The existence is easy to prove: there is a unique homomorphism $\psi_g\colon\mathbb{Z}\to G$ such that $\psi_g(1)=g$, which is surjective because $g$ is a generator. Therefore $\psi_g$ induces an isomorphism $\varphi_g\colon\mathbb{Z}/\ker\psi_g\to G$ and, by counting orders, $\ker\psi_g=m\mathbb{Z}$. Uniqueness is obvious, because homomorphisms that coincide on a set of generators of the domain coincide everywhere.
So, if you have $G=\langle g\rangle$ and $H=\langle h\rangle$ cyclic, the map
$$
(\varphi_g,\varphi_h)\colon\mathbb{Z}_m\times\mathbb{Z}_n\to G\times H,
\quad
(x,y)\mapsto(\varphi_g(x),\varphi_h(y))
$$
is an isomorphism. Therefore $G\times H$ is cyclic if and only if $\mathbb{Z}_m\times\mathbb{Z}_n$ is cyclic and, in this case, a generator is mapped to a generator.
Finally, if $m$ and $n$ are coprime, the order of $(1,1)$ in $\mathbb{Z}_m\times\mathbb{Z}_n$ is $mn$: the order of $(x,y)$ is, in general, the lowest common multiple of the orders of $x$ and $y$.
