Factors of Internal Direct Product of Subsemigroups are Normal Subgroups

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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:

$G = H \odot K$

$\Box$


Suppose $x \in H \cap K$.

We have:

\(\ds x = x \odot e\) \(:\) \(\ds x \in H, e \in K\)
\(\ds x = e \odot x\) \(:\) \(\ds e \in H, x \in K\)


Thus we see that:

\(\ds \map \phi {x, e}\) \(=\) \(\ds \map \phi {e, x}\)
\(\ds \leadsto \ \ \) \(\ds \tuple {x, e}\) \(=\) \(\ds \tuple {e, x}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds e\)


Thus:

$H \cap K = \set e$

$\Box$


In the following, the term unique representation of $x \in G$ will be used specifically to mean:

the a priori 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:

\(\ds h\) \(\in\) \(\ds H\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds h \odot e\) \(=\) \(\ds h\) is the unique representation of $h \in G$, as $e \in K$
\(\ds e\) \(=\) \(\ds \paren {h \odot e} \odot \paren {x \odot y}\) Definition of Inverse Element
\(\ds \) \(=\) \(\ds \paren {h \odot x} \odot y\) Definition of Identity Element
\(\ds \leadsto \ \ \) \(\ds \paren {h \odot x}\) \(=\) \(\ds e\) a priori: $e \odot e$ is the unique representation of $e$
\(\ds \land \ \ \) \(\ds y\) \(=\) \(\ds e\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds h^{-1}\)
\(\ds \) \(\in\) \(\ds H\)

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:

\(\ds k\) \(\in\) \(\ds K\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds e \odot k\) \(=\) \(\ds k\) is the unique representation of $k \in G$, as $e \in K$
\(\ds e\) \(=\) \(\ds \paren {x \odot y} \odot \paren {e \odot k}\) Definition of Inverse Element
\(\ds \) \(=\) \(\ds x \odot \paren {y \odot k}\) Definition of Identity Element
\(\ds \leadsto \ \ \) \(\ds \paren {y \odot k}\) \(=\) \(\ds e\) a priori: $e \odot e$ is the unique representation of $e$
\(\ds \land \ \ \) \(\ds x\) \(=\) \(\ds e\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds k^{-1}\)
\(\ds \) \(\in\) \(\ds K\)

Thus we see that:

$k \in K \implies k^{-1} \in K$


Now suppose $a, b \in H$.

We have:

\(\ds a \odot b\) \(=\) \(\ds \paren {a \odot e} \odot \paren {b \odot e}\) where $a \odot e$ and $b \odot e$ are the unique representations of $a$ and $b$
\(\ds \) \(=\) \(\ds \paren {a \odot \paren {e \odot b} } \odot e\) Group Axiom $\text G 1$: Associativity
\(\ds \) \(=\) \(\ds \paren {a \odot b} \odot e\) Group Axiom $\text G 2$: Existence of Identity Element

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.

$\Box$


Now suppose $c, d \in K$.

We have:

\(\ds c \odot d\) \(=\) \(\ds \paren {e \odot c} \odot \paren {e \odot d}\) where $e \odot c$ and $e \odot d$ are the unique representations of $c$ and $d$
\(\ds \) \(=\) \(\ds e \odot \paren {\paren {c \odot e} \odot d}\) Group Axiom $\text G 1$: Associativity
\(\ds \) \(=\) \(\ds e \odot \paren {c \odot d}\) Group Axiom $\text G 2$: Existence of Identity Element

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.

$\Box$


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.

$\blacksquare$


Sources

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