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A direct product combines two groups into one while keeping their operations in separate coordinates. Its underlying set is the Cartesian product $G\times H$, and its group operation isThe identity element is $(e_G,e_H)$, and the inverse element of $(g,h)$ is $(g^{-1},h^{-1})$. Associativity follows coordinate by coordinate. For finite groups, the multiplication principle gives $|G\times H|=|G||H|$.

For example, write the cyclic group $C_m$ as the additive group of residue classes modulo $m$. In $C_2\times C_3$,since each coordinate reduces with its own integer modulus. A pair can track two cyclic counters that advance independently. Updating the first coordinate does not alter the second, and vice versa.

This independence has an algebraic form. The embedded factor elements always commute:The coordinate maps $(g,h)\mapsto g$ and $(g,h)\mapsto h$ are group homomorphisms, so one can analyze either component separately. This does not make the full product an abelian group when one of its factors is nonabelian: two arbitrary pairs need not commute.

The construction is useful in both directions. It builds a larger group from familiar pieces, while the internal direct product theorem recognizes when an existing group can be separated into such pieces. A cyclic group can sometimes be split this way; whether two finite cyclic factors combine into one cyclic group is decided by the cyclicity of a product of two finite cyclic groups. Definitions and the basic recognition theorems are developed in Thomas Judson, Abstract Algebra: Theory and Applications, §9.2, Direct Products.

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.