Internal Direct Product Theorem

Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $H_1, H_2 \le G$.

$(1): \quad G = H_1 \circ H_2$

$(2): \quad H_1 \cap H_2 = \set e$

$(3): \quad H_1, H_2 \lhd G$

where $H_1 \lhd G$ denotes that $H_1$ is a normal subgroup of $G$.

Let $G$ be a group whose identity is $e$.

Let $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of subgroups of $G$.

Then $G$ is the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ if and only if:

$(1): \quad G = H_1 H_2 \cdots H_n$

$(2): \quad \sequence {H_k}_{1 \mathop \le k \mathop \le n}$ is a sequence of independent subgroups

$(3): \quad \forall k \in \set {1, 2, \ldots, n}: H_k \lhd G$

where $H_k \lhd G$ denotes that $H_k$ is a normal subgroup of $G$.

Let $G$ be the internal group direct product of $H_1$ and $H_2$.

$(3): \quad$ From Internal Group Direct Product Isomorphism, $H_1 \lhd G$ and $H_2 \lhd G$.

$\Box$

Let $C: H_1 \times H_2 \to G$ be the mapping defined as:

$\forall \tuple {h_1, h_2} \in H_1 \times H_2: \map C {h_1, h_2} = h_1 \circ h_2$

Suppose the three conditions hold.

Putting these together, we see that $C$ is a bijective homomorphism, and therefore an isomorphism.

So by definition, $G$ is the internal group direct product of $H_1$ and $H_2$.

$\blacksquare$

A specific instance of the general result, with $n = 2$.

$\blacksquare$