# Conditions under which Commutative Semigroup is Group

## 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:

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

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