 Solution

The group is the direct product of groups ${\mathbb{Z}}_p\times {\mathbb{Z}}_q$. Repeatedly adding $(1,1)$ returns to $(0,0)$ for the first time when both $p$ and $q$ divide the step number. Its order of a group element is therefore $\mathrm{lcm}(p,q)$. If $p,q$ are coprime, this is $pq$, so it generates every element:Explicitly, $[t{]}_{pq}\mapsto ([t{]}_p,[t{]}_q)$ is the Chinese remainder theorem group isomorphism.

If $d=\gcd (p,q)>1$, every pair satisfiesThus every element order divides $pq/d<pq$, and no element can generate the whole group. HenceThis is the cyclicity of a product of two finite cyclic groups.