# Conditions for Internal Group Direct Product

## Contents

## 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:

- $(1): \quad \forall h_1 \in H_1, h_2 \in H_2: h_1 \circ h_2 = h_2 \circ h_1$
- $(2): \quad G = H_1 \circ H_2$
- $(3): \quad H_1 \cap H_2 = \set e$

Condition $(1)$ can also be stated as:

- $(1): \quad$ Either $H_1$ or $H_2$ is normal in $G$

## Proof

### Necessary Condition

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

Then by definition the mapping:

- $C: H_1 \times H_2 \to G: \map C {h_1, h_2} = h_1 \circ h_2$

is a (group) isomorphism from the cartesian product $\struct {H_1, \circ {\restriction_{H_1} } } \times \struct {H_2, \circ {\restriction_{H_2} } }$ onto $\struct {G, \circ}$.

Let the symbol $\circ$ also be used for the operation induced on $H_1 \times H_2$ by $\circ {\restriction_{H_1} }$ and $\circ {\restriction_{H_2} }$.

$(1): \quad \forall h_1 \in H_1, h_2 \in H_2: h_1 \circ h_2 = h_2 \circ h_1$:

This follows directly from Internal Group Direct Product Commutativity.

From Factor of Group Inner Direct Product is Normal it follows that the other condition:

- $(1): \quad$ Either $H_1$ of $H_2$ is normal in $G$

is equivalent to this.

$\Box$

$(2): \quad G = H_1 \circ H_2$

This follows directly from Subgroup Product is Internal Group Direct Product iff Surjective.

$\Box$

$(3): \quad H_1 \cap H_2 = \set e$

Let $z \in H_1 \cap H_2$.

From Intersection of Subgroups is Subgroup, $z^{-1} \in H_1 \cap H_2$.

So $\tuple {z, z^{-1} } \in H_1 \times H_2$ and so:

- $\map C {z, z^{-1} } = z \circ z^{-1} = e = \map C {e, e}$

We have by definition that $C$ is a (group) isomorphism, therefore a bijection and so an injection.

So, as $C$ is injection, we have that:

- $\tuple {z, z^{-1} } = \tuple {e, e}$

and therefore $z = e$.

$\Box$

### Sufficient Condition

Suppose $H_1, H_2 \le G$ such that:

- $(1): \quad \forall h_1 \in H_1, h_2 \in H_2: h_1 \circ h_2 = h_2 \circ h_1$
- $(2): \quad G = H_1 \circ H_2$
- $(3): \quad H_1 \cap H_2 = \set e$

all apply.

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

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

Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in H_1 \times H_2$.

Then:

\(\displaystyle \map C {\tuple {x_1, x_2} \circ \tuple {y_1, y_2} }\) | \(=\) | \(\displaystyle \map C {\paren {x_1 \circ y_1}, \paren {x_2 \circ y_2} }\) | $\quad$ Definition of Operation Induced by Direct Product | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {x_1 \circ y_1} \circ \paren {x_2 \circ y_2}\) | $\quad$ Definition of $C$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {x_1 \circ \paren {y_1 \circ x_2} } \circ y_2\) | $\quad$ Associativity of $\circ$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {x_1 \circ \paren {x_2 \circ y_1} } \circ y_2\) | $\quad$ $(1)$: $x_2$ commutes with $y_2$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {x_1 \circ x_2} \circ \paren {y_1 \circ y_2}\) | $\quad$ Associativity of $\circ$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map C {x_1, x_2} \circ \map C {y_1, y_2}\) | $\quad$ Definition of $C$ | $\quad$ |

So $C$ is a (group) homomorphism.

It follows from $(2)$ that $C$ is a surjection and so, by definition, an epimorphism.

As $H_1$ and $H_2$ are subgroups of $G$, they are by definition groups.

Now let $h_1 \in H_1, h_2 \in H_2$ such that $h_1 \circ h_2 = e$.

That is, $h_2 = h_1^{-1}$.

By the Two-Step Subgroup Test it follows that $h_2 \in H_1$.

By a similar argument, $h_1 \in H_2$.

Thus by definition of set intersection, $h_1, h_2 \in H_1 \cap H_2$ and so $h_1 = e = h_2$.

By definition of $C$, that means:

- $\map C {h_1, h_2} = e \implies \tuple {h_1, h_2} = \tuple {e, e}$

That is:

- $\map \ker C = \set {\tuple {e, e} }$

From the Quotient Theorem for Group Epimorphisms it follows that $C$ is a monomorphism.

So $C$ is both an epimorphism and a monomorphism, and so by definition an isomorphism.

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

$\blacksquare$

## Also known as

Some authors give $H_1 \circ H_2$ as the **normal product of $H_1$ by $H_2$**.

Other sources use the term **semidirect product**.

## Used as Definition

Some authors use this set of conditions to *define* the internal group direct product, and from this definition deduce the isomorphism between $H_1 \times H_2$ and $G$.

## Also see

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 13$: Theorem $13.4$ - 1967: George McCarty:
*Topology: An Introduction with Application to Topological Groups*... (previous) ... (next): $\text{II}$: Problem $\text{DD}$