Internal direct product theorem

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.