= Cyclicity of a product of two finite cyclic groups
{title2=$C_m\times C_n\text{ cyclic}\ \Longleftrightarrow\ \gcd(m,n)=1$}
For positive <integers> $m,n$, two finite <cyclic groups> combine into one <cyclic group> precisely when their <group orders> are <coprime>:
$$
C_m\times C_n\cong C_{mn}\quad\Longleftrightarrow\quad\gcd(m,n)=1.
$$
The reason is that both cyclic counters must return to their <identity elements> simultaneously. If $a,b$ are <generators of a group> for the two factors, the <order of a direct-product element> gives $\operatorname{ord}(a,b)=\operatorname{lcm}(m,n)$. The pair generates all $mn$ elements exactly when this <least common multiple> is $mn$.
For $C_2\times C_3$, repeated addition of $(1,1)$ visits
$$
(0,0),(1,1),(0,2),(1,0),(0,1),(1,2),(0,0).
$$
All six pairs appear before the return, so $(1,1)$ is a generator. In contrast, in $C_2\times C_2$ the same step gives only $(0,0),(1,1),(0,0)$. Every nonidentity element has <order of a group element> two, so none can generate all four pairs. This is the <Klein four-group>; there is no <group isomorphism> to $C_4$ even though both have four elements.
\Image[/direct-product-cyclic-orbits.png]
{title=Adding the same pair generates all of C2 times C3 but only a two-element subgroup of C2 times C2}
For a brief proof of the converse, each component <order of a group element> divides the corresponding factor <group order>. Every pair's order therefore divides $\operatorname{lcm}(m,n)$. If $\gcd(m,n)>1$, this is smaller than $mn$, so no pair can generate the whole product. The <Chinese remainder theorem> gives the explicit isomorphism in the <coprime> case: $t\bmod mn\mapsto(t\bmod m,t\bmod n)$.
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