Direct product of groups

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.

If a group has carrier $A$ and a disjoint set $B$ is in bijection with $A$, identify $A\cup B$ with $A\times C_2$, sending the original carrier to $A\times \{0\}$. Transporting the direct-product multiplication gives a larger group operation that agrees exactly with the old one on $A$. A maximal partial group on a fixed carrier universe therefore has complement of cardinality strictly below its own, using cardinal comparability.

This abelian group has eight elements $c^r{a}^s$, with $r$ taken modulo four and $s$ modulo two. Its group operation adds these exponent pairs separately. The elements of order two are $c^2,a,c^{2}a$, and the other four nonidentity elements $c,c^3,ca,c^{3}a$ have order four. It is a direct product of groups rather than a cyclic group, since no element has order eight.

With the generators $c,a$ of the direct product of cyclic groups of orders four and two, the subgroups are the trivial group, the whole group, the three order-two groups $⟨c^2⟩$, $⟨a⟩$, $⟨c^{2}a⟩$, and the three order-four groups $⟨c⟩$, $⟨ca⟩$, $⟨c^2,a⟩$. These are exhaustive by Lagrange's theorem: a proper nontrivial subgroup has order two or four; an order-four subgroup either contains an element of order four and is one of the two cyclic examples, or is precisely the four-element set killed by squaring. The latter is a Klein four-group.

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

If the order of a group element is $m$ for $g$ and $n$ for $h$, with both finite, thenexactly when both $m$ and $n$ divide $k$. The smallest positive such $k$ is their least common multiple, givingThis concerns the orders of the selected elements, which need not equal the orders of their entire groups. If either component has infinite order of a group element, no positive power can return the pair to the identity element, so the pair also has infinite order.

For example, in $C_4\times C_6$, the additive element $(2,3)$ has component orders $2,2$, hence pair order $2$. The element $(1,1)$ instead has order $\mathrm{lcm}(4,6)=12$. The product has $24$ elements, so even this latter element does not generate it. This distinguishes group order from order of a group element while exposing the simultaneous-return interpretation.

The restricted direct sum of a family of groups $G_i$ is the subgroup of their direct product of groups consisting of tuples with only finitely many nonidentity coordinates. Multiplication is coordinatewise. If every factor is a finite group, each subgroup that is a finitely generated group lies in a finite subproduct, so the restricted direct sum is a locally finite group. This gives concrete infinite torsion groups without a uniform bound on element orders.

An internal direct product finds independent coordinates already present inside a group. Let $H,K$ be normal subgroups of $G$ withThen the multiplication mapis a group isomorphism. Thus every element of $G$ has a unique factorization into one element of each subgroup. Conversely, the embedded factors $G\times \{e_H\}$ and $\{e_G\}\times H$ inside an external direct product of groups satisfy these conditions.

For a concrete example, take the additive cyclic group $C_6=\mathbb{Z}/6\mathbb{Z}$ and its subgroups $H=\{0,3\}$ and $K=\{0,2,4\}$. They are normal subgroups because $C_6$ is an abelian group. Their intersection is $\{0\}$, and the sums $h+k$ give every residue class exactly once. For instance, $1=3+4\mathrm{mod}6$. The table makes the unique decomposition visible.

The proof explains each hypothesis. Normality puts the group commutator $[h,k]=h^{-1}{k}^{-1}hk$ in both $H$ and $K$, so it is the identity element. Therefore the factors commute, and $\Phi$ preserves the group operation. The condition $HK=G$ makes it surjective. If $hk=e$, then $h=k^{-1}$ belongs to the intersection, forcing both to be the identity element; thus its group kernel is trivial and it is injective.

Normality cannot simply be omitted. In the symmetric group $S_3$, take $H=⟨(123)⟩$ and $K=⟨(12)⟩$. Their intersection is trivial and $HK=S_3$, but $K$ is not a normal subgroup. With permutations composed right to left, $(123)(12)=(13)$ while $(12)(123)=(23)$, so the factors do not commute. The setwise decomposition is unique, but its multiplication map is not a group homomorphism from $H\times K$. An independent-coordinate interpretation requires the group operations to agree, not merely a bijection of the underlying sets.