# Internal Direct Product Theorem/Proof 1

Jump to navigation
Jump to search

## Theorem

Let $\struct {G, \circ}$ 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 $H_2$ if and only if:

- $(1): \quad G = H_1 \circ H_2$
- $(2): \quad H_1 \cap H_2 = \set e$
- $(3): \quad H_1, H_2 \lhd G$

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

## Proof

### Sufficient Condition

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

- $(1): \quad$ From Subgroup Product is Internal Group Direct Product iff Surjective, $G = H_1 \circ H_2$.
- $(2): \quad$ From Internal Group Direct Product is Injective, $H_1$ and $H_2$ are independent subgroups of $G$.
- $(3): \quad$ From Internal Group Direct Product Isomorphism, $H_1 \lhd G$ and $H_2 \lhd G$.

$\Box$

### Necessary Condition

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.

- $(1): \quad$ From Subgroup Product is Internal Group Direct Product iff Surjective, $C$ is surjective.
- $(2): \quad$ From Internal Group Direct Product is 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 $H_1$ and $H_2$.

$\blacksquare$

## Sources

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