Condition for Semigroup to be Internal Direct Product of Subgroup and Subsemigroup with Right Operation

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Theorem

Let $\struct {S, \odot}$ be a semigroup.

Then:

$\struct {S, \odot}$ is the internal direct product of a subgroup $\struct {G, \odot_G}$ and subsemigroup $\struct {H, \odot_H}$ such that $\odot_H$ is the right operation

if and only if:

for all $x \in S$ there exists a left identity $a$ such that $x \odot a = x$, and element $y$ such that $y \odot x = a$.


Proof

Sufficient Condition

Let $\struct {S, \odot}$ be the internal direct product of a subgroup $\struct {G, \odot_G}$ and subsemigroup $\struct {H, \odot_H}$ such that $\odot_H$ is the right operation.

Hence, by definition, the mapping $\phi: G \times H \to S$ defined as:

$\forall g \in H, h \in H: \map \phi {g, h} = g \odot h$

is an isomorphism from the (external) direct product $\struct {G, \odot_G} \times \struct {H, \odot_H}$ onto $\struct {S, \odot}$.

Thus a fortiori $\phi$ is a bijection.


From Condition for Mapping between Structure and Cartesian Product of Substructures to be Bijection:

for all $s \in S$: there exists a unique $\tuple {x, y} \in A \times B$ such that $x \odot y = s$.



So, let $e \in G$ be the identity element of $G$.

As $e \in G \implies e \in S$, the above condition therefore applies to $e$.

We have that:

$\forall g \in G: g \odot e = g = e \odot g$

This also applies to $e$:

$e \odot e = e$

But we have:

$\forall z \in G: e = z \odot z^{-1}$

where $z^{-1} \in G$.

So, as $e$ can be represented uniquely in the form $e = x \odot y$, it follows that that unique representation is:

$e = e \odot e$

where $e \in G \cap H$.

We have that:

$\forall g \in G: g \odot e = g$

is the unique representation of $g$ such that $\tuple {g, e} \in G \odot H$.

Then we have by definition of right operation:

$\forall h \in H: e \odot h = h$

which is the unique representation of $h$ such that $\tuple {e, h} \in G \odot H$.

We have by definition of the identity element of $G$ that:

$\forall g \in G: e \odot g = g$


Now let $z \in G \odot H$ such that $z \notin G$ and $z \notin H$.

Then we have that:

$z = g \odot h$

where $g \in G$ and $h \in H$ uniquely.

Thus:

\(\ds e \odot z\) \(=\) \(\ds e \odot \paren {g \odot h}\) Definition of $z$
\(\ds \) \(=\) \(\ds \paren {e \odot g} \odot h\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds g \odot h\) as $e$ is the identity element of $G$
\(\ds \) \(=\) \(\ds z\) Definition of $z$

So we have shown that:

$\forall s \in S: e \odot s = s$

Hence, by definition, $e$ is a left identity in $S$.

Hence there exists at least one left identity in $S$.

$\Box$


Let $x \in S$ be arbitrary.

It is to be shown that:

there exists a left identity $a$ in $S$ such that $x \odot a = x$
$\exists y \in S: y \odot x = a$


By Group Axiom $\text G 2$: Existence of Identity Element, for all $g \in G$:

$\exists e \in G: g \odot e = g$

By Group Axiom $\text G 3$: Existence of Inverse Element, for all $g \in G$:

$\exists g^{-1} \in G: g^{-1} \odot g = e$

and so for elements $g$ of $G$:

$e$ fulfils the condition of $a$
$g^{-1}$ fulfils the condition of $y$.


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Necessary Condition

Let $\struct {S, \odot}$ be a semigroup such that for all $x \in S$:

there exists a left identity $a$ such that $x \odot a = x$
there exists an element $y$ such that $y \odot x = a$.

Then $\struct {S, \odot}$ is not empty, by definition.

Let $H \subseteq S$ be the set of all left identities of $S$.

Let $e \in H$ be arbitrary.

Let $G = S \odot \set e$ be the subset product of $S$ with $\set e$.


This theorem requires a proof.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.


Sources

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