 Solution

Write the cyclic factors additively. The order of a group element $(r,s)$ in a direct product of groups is the least common multiple of its component orders, since a power is the identity exactly when it is the identity in both coordinates.

In $C_2\times C_3$, $(1,1)$ has order $\mathrm{lcm}(2,3)=6$, so it generates all six pairs. An explicit group isomorphism isIt is well defined and multiplicative in additive notation; its kernel is zero because an integer divisible by both two and three is divisible by six. Equal finite orders then give bijectivity.

In $C_2\times C_4$, every element has order dividing four. A generator of $C_8$ has order eight, and isomorphisms preserve element orders, soFor positive integers $n,m$, the maximum element order in $C_n\times C_m$ is $\mathrm{lcm}(n,m)=nm/\gcd (n,m)$, attained by a pair of generators. The product is cyclic of order $nm$ exactly when that number is $nm$. Thus cyclicity of a product of two finite cyclic groups givesFor coprime factors the map $[k{]}_{nm}\mapsto ([k{]}_n,[k{]}_m)$ gives the isomorphism, the group form of the Chinese remainder theorem.