Conditions for Internal Group Direct Product

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H$ and $K$ be subgroups of $G$.
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 $K$ :
:$(1): \quad \forall h \in H, k \in K: h \circ k = k \circ h$
:$(2): \quad G = H \circ K$
:$(3): \quad H \cap K = \set e$
== Proof ==
=== 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 Codomain of Internal Direct Isomorphism is Subset Product of Factors.
;$(3): \quad H \cap K = \set e$
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 $K$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.4$
* : Chapter $\text{II}$: Groups: Problem $\text{DD}$
Category:Internal Group Direct Products