Conditions under which Commutative Semigroup is Group

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Theorem

Suppose the following:

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 $\struct {S, \circ}$ is a group.


Proof

Some lemmata:

Lemma 1

If $y \circ x = x = y' \circ x$, then $y = y'$.

$\Box$


Lemma 2

If $y \circ x = x$, then $y \circ y = y$.

$\Box$


Lemma 3

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

$\Box$


First we note that if $S = \set e$ is a singleton, $\struct {S, e}$ is the trivial group.

So in the following, it is assumed that there are at least two elements of $S$.


Taking the group axioms in turn:


Group Axiom $\text G 0$: Closure and Group Axiom $\text G 1$: Associativity

We have that $\struct {S, \circ}$ is a semigroup.

Hence the semigroup axioms are fulfilled:

Semigroup Axiom $\text S 0$: Closure
Semigroup Axiom $\text S 1$: Associativity

Hence Group Axiom $\text G 0$: Closure and Group Axiom $\text G 1$: Associativity are fulfilled by $\struct {S, \circ}$.

$\Box$


Group Axiom $\text G 2$: Existence of Identity Element


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By $(1)$:

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

Hence $S$ is not empty, by definition.

Let $x \in S$.

Let $e \in S$ be the element of $S$ asserted by $(1)$ such that $e \circ x = x$.

From Lemma $1$, this is the only such element of $S$ with this property.

Let $z \in S$ such that $z \ne x$ be arbitrary.

By $(1)$ there exists $e' \in S$ such that $e' \circ z = z$.

By Lemma $3$ we have that $e' = e$.

As $z$ is arbitrary, it follows that:

$\exists e \in S: \forall z \in S: e \circ z = z$

As $\struct {S, \circ}$ is a commutative semigroup, it follows also that:

$\forall z \in S: z \circ e = z$

Thus $e$ has been shown to be an identity of $\struct {S, \circ}$.

From Identity is Unique, $e$ is the only such element of $S$ which is an identity.

Hence $e$ is the identity of $\struct {S, \circ}$.

$\Box$


Group Axiom $\text G 3$: Existence of Inverse Element

We have that $e$ is the identity element of $\struct {S, \circ}$:

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

It follows directly from $(2)$ that:

$\forall x \in S: \exists z \in S: z \circ x = e$

As $\struct {S, \circ}$ is a commutative semigroup, it follows also that:

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

Thus every element $x$ of $\struct {S, \circ}$ has an inverse $z$.

$\Box$


All the group axioms are thus seen to be fulfilled, and so $\struct {S, \circ}$ is a group.

$\blacksquare$


Warning

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.


Sources

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