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For positive integers $m,n$, two finite cyclic groups combine into one cyclic group precisely when their group orders are coprime: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 $\mathrm{ord}(a,b)=\mathrm{lcm}(m,n)$. The pair generates all $mn$ elements exactly when this least common multiple is $mn$.

For $C_2\times C_3$, repeated addition of $(1,1)$ visitsAll 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.

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 $\mathrm{lcm}(m,n)$. 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: $t\mathrm{mod}mn\mapsto (t\mathrm{mod}m,t\mathrm{mod}n)$.

The order of a group element $x$ is the least positive integer $r$ with $x^r=e$, or infinity if none exists. In a finite group, some two powers among $e,x,\dots ,x^{|G|}$ coincide; cancellation then gives a positive power equal to $e$, so $r$ exists. The generated cyclic group $H=⟨x⟩$ consists of the distinct elements $e,x,\dots ,x^{r-1}$ and has size $r$.

We now prove the needed divisibility directly by cosets. Multiplication by $g$ bijects $H$ with the left coset $gH$, so every left coset has $r$ elements. If $gH$ and $hH$ intersect, write $ga=hb$ with $a,b\in H$. Then $h^{-1}g=b{a}^{-1}\in H$, which implies $gH=hH$. Thus distinct left cosets are disjoint, and every element of $G$ belongs to its own left coset. The finite set $G$ is partitioned into, say, $k$ such cosets. Hence $|G|=kr$ andThis supplies the counting argument underlying Lagrange's theorem, rather than appealing to it or to an unproved orbit formula.

A divisor of $|G|$ need not occur as an element order. Use $G=C_2\times C_2\times C_2$, of order eight, and $d=4<8$. Every element has square equal to the identity, so there is no element of order four. Thus the proposed converse is false, even for an abelian group.

For the direct product of groups $C_m\times C_n$, let $a,b$ generate the factors. The least $k>0$ for which $(a,b{)}^k=(e,e)$ is the least common multiple of $m,n$. If $m,n$ are coprime, this is $mn$, the size of the whole product, so $(a,b)$ generates every element. Conversely, for any pair $(u,v)$ its order is $\mathrm{lcm}(\mathrm{ord}u,\mathrm{ord}v)$ and divides $\mathrm{lcm}(m,n)$. If $\gcd (m,n)>1$, that bound is $mn/\gcd (m,n)<mn$, so no element can generate the product. ThereforeThis is the cyclicity of a product of two finite cyclic groups; the general pair-order criterion is the order of a direct-product element.