Conditions for Internal Group Direct Product

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:


 * $(1): \quad \forall h_1 \in H_1, h_2 \in H_2: h_1 \circ h_2 = h_2 \circ h_1$
 * $(2): \quad G = H_1 \circ H_2$
 * $(3): \quad H_1 \cap H_2 = \left\{{e}\right\}$

Condition $(1)$ can also be stated as:
 * $(1): \quad$ Either $H_1$ of $H_2$ is normal in $G$

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

Then by definition the mapping:
 * $C: H_1 \times H_2 \to G: C \left({\left({h_1, h_2}\right)}\right) = h_1 \circ h_2$

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

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.

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

This follows directly from Internal Group Direct Product Surjective.

$(3): \quad H_1 \cap H_2 = \left\{{e}\right\}$

Let $z \in H_1 \cap H_2$.

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

So $\left({z, z^{-1}}\right) \in H_1 \times H_2$ and so:
 * $C \left({\left({z, z^{-1}}\right)}\right) = z \circ z^{-1} = e = C \left({\left({e, e}\right)}\right)$

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:
 * $\left({z, z^{-1}}\right) = \left({e, e}\right)$

and therefore $z = e$.

Sufficient Condition
Suppose $H_1, H_2 \le G$ such that:
 * $(1): \quad \forall h_1 \in H_1, h_2 \in H_2: h_1 \circ h_2 = h_2 \circ h_1$
 * $(2): \quad G = H_1 \circ H_2$
 * $(3): \quad H_1 \cap H_2 = \left\{{e}\right\}$

all apply.

Let $C: H_1 \times H_2 \to G$ be the mapping defined as:
 * $\forall \left({h_1, h_2}\right) \in H_1 \times H_2: C \left({\left({h_1, h_2}\right)}\right) = h_1 \circ h_2$

Let $\left({x_1, x_2}\right), \left({y_1, y_2}\right) \in H_1 \times H_2$.

Then:

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:
 * $C \left({h_1, h_2}\right) = e \implies \left({h_1, h_2}\right) = \left({e, e}\right)$

That is:
 * $\ker \left({C}\right) = \left\{{\left({e, e}\right)}\right\}$

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 $H_2$

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

 * Factor of Group Inner Direct Product is Normal