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

== 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
:
: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 $\phi$ 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:
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 $S$.
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 , for all $g \in G$:
:$\exists e \in G: g \odot e = g$
By , 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 $y$.
=== 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 $\set e$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.16$
Category:Semigroups
Category:Internal Direct Products
Category:Right Operation