Conditions under which Commutative Semigroup is Group/Warning

Warning concerning Conditions under which Commutative Semigroup is Group

Consider an algebraic structure $\struct {S, \circ}$ fulfilling the following conditions:

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

Let $\struct {S, \circ}$ have the following properties:

$(1)$	$:$			$\ds \forall x \in S: \exists y \in S:$	$\ds y \circ x = x $
$(2)$	$:$			$\ds \forall x, y \in S:$	$\ds y \circ x = x \implies \exists z \in S: z \circ x = y $

While it is the case that such an algebraic structure $\struct {S, \circ}$ is a group, if $\struct {S, \circ}$ is a semigroup which is not commutative, this does not necessarily follow.

Let $S$ be a set with more than $1$ element.

Let $\struct {S, \circ}$ be a semigroup such that $\circ$ is the left operation.

From Right Operation is Anticommutative we have that $\circ$ is specifically not commutative.

By definition of the left operation:

$\forall x \in S: x \circ x = x$

Thus $x$ fulfils the role of $y$ in condition $(1)$:

$\forall x \in S: \exists y \in S: y \circ x = x$

and so $(1)$ is satisfied.

By definition of the left operation:

$\forall x, y \in S: y \circ x = y$

That is:

$\forall x, y \in S: \exists z \in S: z \circ x = y$

where here $z = y$.

Hence $x$ fulfils the role of $y$ and $y$ fulfils the role of $z$ in:

$\forall x, y \in S: y \circ x = x \implies \exists z \in S: z \circ x = y$

and so $(2)$ is satisfied.

But $\struct {S, \circ}$ is not a group.

$\blacksquare$

1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.16 \ \text {(e)}$

\((1)\)	$:$			\(\ds \forall x \in S: \exists y \in S:\)	\(\ds y \circ x = x \)
\((2)\)	$:$			\(\ds \forall x, y \in S:\)	\(\ds y \circ x = x \implies \exists z \in S: z \circ x = y \)