= 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 $\operatorname{lcm}(2,3)=6$, so it generates all six pairs. An explicit <group isomorphism> is
$$
\boxed{C_6\longrightarrow C_2\times C_3,\qquad[k]_6\longmapsto([k]_2,[k]_3).}
$$
It 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, so
$$
\boxed{C_2\times C_4\not\cong C_8.}
$$
For positive integers $n,m$, the maximum element order in $C_n\times C_m$ is $\operatorname{lcm}(n,m)=nm/\gcd(n,m)$, 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> gives
$$
\boxed{C_n\times C_m\cong C_{nm}\quad\Longleftrightarrow\quad\gcd(n,m)=1.}
$$
For coprime factors the map $[k]_{nm}\mapsto([k]_n,[k]_m)$ gives the isomorphism, the group form of the <Chinese remainder theorem>.
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