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:


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

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.

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

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 So $h_2 \in H_1$.

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 $H_2$