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


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

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 $a \in G$ be the identity element of $G$.

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

We have that:
 * $\forall g \in G: g \odot a = g = a \odot g$

This also applies to $a$:
 * $a \odot a = a$

But we have:
 * $\forall z \in G: a = z \odot z^{-1}$

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

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

where $a \in G \cap H$.

We have that:
 * $\forall g \in G: g \odot a = g$

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

Then we have by definition of right operation:
 * $\forall h \in H: a \odot h = h$

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

We have by definition of the identity element of $G$ that:
 * $\forall g \in G: a \odot g = g$

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:
 * $\forall s \in S: a \odot s = s$

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

It remains to be shown that:
 * $\forall x \in S: \exists y \in S: y \odot x = a$

Now by, for all $g \in G$:
 * $\exists g^{-1} \in G: g^{-1} \odot g = a$

and so for elements $g$ of $G$, $g^{-1}$ fulfils the condition of $y$.