Conditions for Internal Group Direct Product

Theorem
Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $H, K \le G$.

Let the mapping $\phi: H \times K \to G$ defined as:
 * $\forall h \in H, k \in K: \map \phi {h, k} = h \circ k$

be a group isomorphism from the cartesian product $\struct {H, \circ {\restriction_H} } \times \struct {K, \circ {\restriction_K} }$ onto $\struct {G, \circ}$.

Then $G$ is the internal group direct product of $H$ and $K$ :


 * $(1): \quad \forall h \in H, k \in K: h \circ k = k \circ h$
 * $(2): \quad G = H \circ K$
 * $(3): \quad H \cap K = \set e$

Necessary Condition
Let the mapping $\phi: H \times K \to G$ defined as:
 * $\forall h \in H, k \in K: \map \phi {h, k} = h \circ k$

be a group isomorphism from the cartesian product $\struct {H, \circ {\restriction_H} } \times \struct {K, \circ {\restriction_K} }$ onto $\struct {G, \circ}$.

Let the symbol $\circ$ also be used for the operation induced on $H \times K$ by $\circ {\restriction_H}$ and $\circ {\restriction_K}$.

$(1): \quad \forall h \in H, k \in K: h \circ k = k \circ h$:

This follows directly from Internal Group Direct Product Commutativity.

$(2): \quad G = H \circ K$

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

$(3): \quad H \cap K = \set e$

Let $z \in H \cap K$.

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

So $\tuple {z, z^{-1} } \in H \times K$ and so:
 * $\map \phi {z, z^{-1} } = z \circ z^{-1} = e = \map \phi {e, e}$

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

So, as $\phi$ is injection, we have that:
 * $\tuple {z, z^{-1} } = \tuple {e, e}$

and therefore $z = e$.

Sufficient Condition
Suppose $H, K \le G$ such that:
 * $(1): \quad \forall h \in H, k \in K: h \circ k = k \circ h$
 * $(2): \quad G = H \circ K$
 * $(3): \quad H \cap K = \set e$

all apply.

Let $\phi: H \times K \to G$ be the mapping defined as:
 * $\forall \tuple {h, k} \in H \times K: \map \phi {h, k} = h \circ k$

Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in H \times K$.

Then:

So $\phi$ is a (group) homomorphism.

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

As $H$ and $K$ are subgroups of $G$, they are by definition groups.

Now let $h \in H, k \in K$ such that $h \circ k = e$.

That is, $k = h^{-1}$.

By the Two-Step Subgroup Test it follows that $k \in H$.

By a similar argument, $h \in K$.

Thus by definition of set intersection, $h, k \in H \cap K$ and so $h = e = k$.

By definition of $C$, that means:
 * $\map \phi {h, k} = e \implies \tuple {h, k} = \tuple {e, e}$

That is:
 * $\map \ker \phi = \set {\tuple {e, e} }$

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

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

Thus, by definition, $G$ is the internal group direct product of $H$ and $K$.