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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.

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 $\mathrm{lcm}(2,3)=6$, so it generates all six pairs. An explicit group isomorphism isIt 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, soFor positive integers $n,m$, the maximum element order in $C_n\times C_m$ is $\mathrm{lcm}(n,m)=nm/\gcd (n,m)$, 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 givesFor coprime factors the map $[k{]}_{nm}\mapsto ([k{]}_n,[k{]}_m)$ gives the isomorphism, the group form of the Chinese remainder theorem.

The group is the direct product of groups ${\mathbb{Z}}_p\times {\mathbb{Z}}_q$. Repeatedly adding $(1,1)$ returns to $(0,0)$ for the first time when both $p$ and $q$ divide the step number. Its order of a group element is therefore $\mathrm{lcm}(p,q)$. If $p,q$ are coprime, this is $pq$, so it generates every element:Explicitly, $[t{]}_{pq}\mapsto ([t{]}_p,[t{]}_q)$ is the Chinese remainder theorem group isomorphism.

If $d=\gcd (p,q)>1$, every pair satisfiesThus every element order divides $pq/d<pq$, and no element can generate the whole group. HenceThis is the cyclicity of a product of two finite cyclic groups.

In additive notation every element $(a,b)\in {\mathbb{Z}}_2\times {\mathbb{Z}}_4$ obeys $4(a,b)=(0,0)$. Its order of a group element therefore divides four, whereas $1\in {\mathbb{Z}}_8$ has order eight. A group isomorphism preserves element orders, soFor the second product, define the group homomorphismIt is well-defined because multiples of six are divisible by both two and three. Its kernel contains only zero: simultaneous divisibility by two and three implies divisibility by six. Thus $\Psi$ is injective, and equal finite group orders make it surjective. Explicitly the six images areThereforeFor general positive integers $n,m$, the order of a direct-product element is the least common multiple of its component orders. Hence every element order divides $\mathrm{lcm}(n,m)=nm/\gcd (n,m)$, while $(1,1)$ attains this order. It generates the whole product precisely when $n,m$ are coprime:This is the cyclicity of a product of two finite cyclic groups, with the explicit isomorphism furnished by the Chinese remainder theorem.

Put $N=HK=⟨r⟩$, the normal subgroup of order $35$. By Cauchy's theorem for finite groups, $G$ contains an element $t$ of order two. Such an element cannot belong to the odd-order subgroup $N$. Thus $N\cap ⟨t⟩=\{1\}$ and $G=N⟨t⟩$, giving the semidirect productEvery automorphism of a finite cyclic group sends $r$ to $r^a$ for a unit $a$ modulo $35$. The relation $t^2=1$ forces the conjugation action of $t$ to square to the identity:The Chinese remainder theorem identifies a congruence class modulo $35$ with its pair of classes modulo $5$ and $7$. In each of these prime fields, the factorization $(a-1)(a+1)=0$ forces $a=1$ or $a=-1$. The choices are independent, giving exactly four homomorphisms from a finite cyclic group $C_2\to \mathrm{Aut}(C_{35})$. Explicitly,The factors $C_5$ and $C_7$ of $N$ are preserved by every automorphism, since each is its unique subgroup of that order. The four conjugation actions therefore give:

Here $D_n$ has group order $n$, as in the paper. The center of a group distinguishes these four groups of order seventy: its orders are $70,1,5,7$, respectively, so the four groups are pairwise nonisomorphic. Each construction is a semidirect product with $70$ elements, so every listed possibility occurs.