Factors of Internal Direct Product of Subsemigroups are Normal Subgroups

Theorem
Let $\struct {G, \odot}$ be a group.

Let $\struct {H, \odot_H}$ and $\struct {K, \odot_K}$ be subsemigroups of $\struct {G, \odot}$.

Let $\struct {G, \odot}$ be the internal direct product of $\struct {H, \odot_H}$ and $\struct {K, \odot_K}$.

Then $\struct {H, \odot_H}$ and $\struct {K, \odot_K}$ are normal subgroups of $\struct {G, \odot}$.

Proof
Let $\struct {G, \odot}$ be the internal direct product of $\struct {H, \odot_H}$ and $\struct {K, \odot_K}$.

Let $e$ denote the identity element of $\struct {G, \odot}$.

By definition of internal direct product, the mapping $\phi: H \times K \to G$ defined as:
 * $\forall h \in H, k \in K: \map \phi {h, k} = h \odot k$

is an isomorphism from the (external) direct product $\struct {H, \odot {\restriction_H} } \times \struct {K, \odot {\restriction_K} }$ onto $\struct {G, \odot}$.

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

From Condition for Mapping between Structure and Cartesian Product of Substructures to be Bijection:
 * every element of $G$ can be expressed uniquely as a product of the form $h \odot k$ such that $\tuple{h, k} \in H \times K$.

Hence:
 * $G = \set {h \odot k: h \in H, k \in K}$

and so by definition of subset product:


 * $G = H \odot K$

Suppose $x \in H \cap K$.

We have:

Thus we see that:

Thus:
 * $H \cap K = \set e$

In the following, the term unique representation of $x \in G$ will be used specifically to mean:
 * the unique representation $x = a \odot b$ where $a \in H$ and $b \in K$.

First we note that:
 * $e \odot e = e$

is the unique representation of $e \in G$.

Let $h \in H$ be arbitrary.

Now consider $h^{-1} \in G$.

Let its unique representation be $x \odot y$.

Then we have:

Thus we see that:
 * $h \in H \implies h^{-1} \in H$

Similarly, let $k \in K$ be arbitrary.

Now consider $k^{-1} \in G$.

Let its unique representation be $x \odot y$.

Then we have:

Thus we see that:
 * $k \in K \implies k^{-1} \in K$

Now suppose $a, b \in H$.

We have:

Hence $\paren {a \odot b} \odot e$ is the unique representation of $a \odot b \in G$.

It follows that $a \odot b \in H$.

Hence:
 * $a, b \in H \implies a \odot b \in H$

and $H$ is a subgroup of $G$ by the Two-Step Subgroup Test.

Now suppose $c, d \in K$.

We have:

Hence $e \odot \paren {c \odot d}$ is the unique representation of $c \odot d \in G$.

It follows that $c \odot d \in K$.

Hence:
 * $c, d \in K \implies c \odot d \in K$

and $K$ is a subgroup of $G$ by the Two-Step Subgroup Test.

Consolidating our gains, we have that:


 * $(1): \quad \struct {H, \odot {\restriction_H} }$ and $\struct {K, \odot {\restriction_K} }$ are subgroups of the group $\struct {G, \odot}$


 * $(2): \quad$ The mapping $\phi: H \times K \to G$ defined as:
 * $\forall h \in H, k \in K: \map \phi {h, k} = h \odot k$
 * is an isomorphism from the (external) direct product $\struct {H, \odot {\restriction_H} } \times \struct {K, \odot {\restriction_K} }$ onto $\struct {G, \odot}$.

It follows from the Internal Direct Product Theorem that:


 * $(1): \quad \struct {H, \circ {\restriction_H} }$ and $\struct {K, \circ {\restriction_K} }$ are both normal subgroups of $\struct {G, \circ}$


 * $(2): \quad G$ is the subset product of $H$ and $K$, that is: $G = H \circ K$


 * $(3): \quad$ $H \cap K = \set e$ where $e$ is the identity element of $G$.

and the result follows.