Conditions under which Commutative Semigroup is Group/Lemma 3

Lemma for Conditions under which Commutative Semigroup is Group

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 $

Then:

If $y \circ x = x$ and $z \circ w = w$, then $y = z$.

$\ds \paren {y \circ x} \circ w$

$=$

$\ds x \circ w$

by hypothesis: $y \circ x = x$

$\ds \leadsto \ \ $

$\ds y \circ \paren {x \circ w}$

$=$

$\ds x \circ w$

$\ds \leadsto \ \ $

$\ds y \circ \paren {w \circ x}$

$=$

$\ds w \circ x$

as $\struct {S, \circ}$ is a commutative semigroup: $x \circ w = w \circ x$

Then:

$\ds \paren {z \circ w} \circ x$

$=$

$\ds w \circ x$

by hypothesis: $z \circ w = w$

$\ds \leadsto \ \ $

$\ds z \circ \paren {w \circ x}$

$=$

$\ds w \circ x$

Thus:

$y \circ \paren {w \circ x} = w \circ x = z \circ \paren {w \circ x}$

and $y = z$ follows from Lemma 1.

$\blacksquare$

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

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