= Solution
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 $e,x,\ldots,x^{|G|}$ coincide; cancellation then gives a positive power equal to $e$, so $r$ exists. The generated <cyclic group> $H=\langle x\rangle$ consists of the distinct elements $e,x,\ldots,x^{r-1}$ 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 $h^{-1}g=ba^{-1}\in H$, 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 $|G|=kr$ and
$$
\boxed{\operatorname{ord}(x)\mid |G|.}
$$
This 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 \b[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 $\operatorname{lcm}(\operatorname{ord}u,\operatorname{ord}v)$ and divides $\operatorname{lcm}(m,n)$. If $\gcd(m,n)>1$, that bound is $mn/\gcd(m,n)<mn$, so no element can generate the product. Therefore
$$
\boxed{C_m\times C_n\text{ is cyclic exactly when }\gcd(m,n)=1.}
$$
This is the <cyclicity of a product of two finite cyclic groups>; the general pair-order criterion is the <order of a direct-product element>.
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