Internal Direct Product Theorem

Theorem

Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $H_1, H_2 \le G$.

Then $G$ is the internal group direct product of $H_1$ and $H_2$ iff:

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

$(2): \quad H_1 \cap H_2 = \left\{{e}\right\}$

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

General Result

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

Let $\left \langle {H_k} \right \rangle_{1 \le k \le n}$ be a sequence of subgroups of $G$.

Then $G$ is the internal group direct product of $\left \langle {H_k} \right \rangle_{1 \le k \le n}$ iff:

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

$(2): \quad \left \langle {H_k} \right \rangle_{1 \le k \le n}$ is a sequence of independent subgroups

$(3): \quad$ For each $k \in \left[{1 \,.\,.\, n}\right]$, $H_k \triangleleft G$.

Proof

Sufficient Condition

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

$(1): \quad$ From Internal Group Direct Product Surjective, $G = H_1 \circ H_2$.

$(2): \quad$ From Internal Group Direct Product Injective, $H_1$ and $H_2$ are independent subgroups of $G$.

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

$\Box$

Necessary Condition

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

$\forall \left({h_1, h_2}\right) \in H_1 \times H_2: C \left({\left({h_1, h_2}\right)}\right) = h_1 \circ h_2$

Suppose the three conditions hold.

$(1): \quad$ From Internal Group Direct Product Surjective, $C$ is surjective.

$(2): \quad$ From Internal Group Direct Product Injective, $C$ is injective.

$(3): \quad$ From Internal Group Direct Product of Normal Subgroups, $C$ is a group homomorphism.

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 $\left \langle {H_k} \right \rangle_{1 \le k \le n}$.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 13$: Theorem $13.5$