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 $\struct {K, \odot_K}$:


 * $\struct {G, \odot} = \struct {H, \odot_H} \times \struct {K, \odot_K}$

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 $\struct {G, \odot}$.

Then by definition 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 cartesian 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 $\odot {\restriction_K}$.

As $\phi$ is an isomorphism, it is a bijection.

As $\phi$ is a bijection, it is surjective, so:


 * $G \subseteq H \times K$

From Set of Finite Subsets under Induced Operation is Closed:
 * $H \times K \subseteq G$

Thus:
 * $G = H \times K$

As $\phi$ is a bijection, it is injective.

Let $\map \phi {h_1, k_1} = \map \phi {h_2, k_2}$.

Then by definition of injective:
 * $\tuple {h_1, k_1} = \tuple {h_2, k_2}$

and thus:
 * $h_1 = h_2, k_1 = k_2$

From the definition of $\phi$, this means:
 * $h_1 \odot k_1 = h_2 \odot k_2$

Thus, each element of $G$ that can be expressed as a product of the form $h \odot k$ can be thus expressed uniquely.

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

and so by definition of subset product:


 * $G = H \odot K$

As $\phi$ is a bijection, there exists an inverse mapping $\phi^{-1}$ such that:
 * $\forall \tuple {h, k} \in H \times K: \map {\phi^{-1} } {\map \phi {h, k} } = \tuple {h, k}$

Then we have, for all $h_1, h_2 \in H$ and $k_1, k_2 \in K$:

Suppose $x \in H \cap K$.

We have:

Thus we see that:

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

Thus as $e \in H$ and $e \in K$,