# Semigroup is Group Iff Latin Square Property Holds

## Theorem

Let $\left({S, \circ}\right)$ be a semigroup.

Then $\left({S, \circ}\right)$ is a group if and only if for all $a, b \in S$ the Latin square property holds in $S$:

$a \circ x = b$
$y \circ a = b$

for $x$ and $y$ each unique in $S$.

## Proof 1

### Necessary Condition

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

$\struct {S, \circ}$ is a semigroup by the definition of a group.

By Group has Latin Square Property, the Latin square property holds in $S$.

$\Box$

### Sufficient Condition

Let $\struct {S, \circ}$ be a semigroup on which the Latin square property holds.

Taking the group axioms in turn:

#### Group Axiom $\text G 0$: Closure

$\struct {S, \circ}$ is semigroup.

Hence $\struct {S, \circ}$ is closed by definition.

$\Box$

#### Group Axiom $\text G 1$: Associativity

$\struct {S, \circ}$ is semigroup.

Hence $\circ$ is associative by definition.

$\Box$

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

Let $a \in S$.

Then there is an $x \in S$ such that:

 $\text {(1)}: \quad$ $\ds a \circ x$ $=$ $\ds a$

Let $b \in S$.

Then there is a $y \in S$ such that:

 $\text {(2)}: \quad$ $\ds y \circ a$ $=$ $\ds b$

Proving that $x$ is a right identity:

 $\ds b \circ x$ $=$ $\ds \paren {y \circ a} \circ x$ by $(2)$ $\ds$ $=$ $\ds y \circ \paren {a \circ x}$ Definition of Associative Operation $\ds$ $=$ $\ds y \circ a$ by $(1)$ $\ds$ $=$ $\ds b$ by $(2)$

Let $c \in S$.

Then there is a $z \in S$ such that:

 $\text {(3)}: \quad$ $\ds x \circ z$ $=$ $\ds c$

Proving that $x$ is a left identity and thus an identity:

 $\ds x \circ c$ $=$ $\ds x \circ \paren {x \circ z}$ by $(3)$ $\ds$ $=$ $\ds \paren {x \circ x} \circ z$ Definition of Associative Operation $\ds$ $=$ $\ds x \circ z$ $x$ is a right identity $\ds$ $=$ $\ds c$ by $(3)$

Relabel $x$ as $e$.

$\Box$

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

Let $x \in S$.

Then $x$ has a right and left inverse by the property as there exists a $y$ and $z$ in $S$ such that:

 $\text {(4)}: \quad$ $\ds x \circ y$ $=$ $\ds e$
 $\text {(5)}: \quad$ $\ds z \circ x$ $=$ $\ds e$

We shall prove that $y$ is also the left inverse of $x$:

 $\ds y \circ x$ $=$ $\ds \paren {z \circ x} \circ \paren {y \circ x}$ Definition of Identity Element $\ds$ $=$ $\ds z \circ \paren {\paren {x \circ y} \circ x}$ Definition of Associative Operation $\ds$ $=$ $\ds z \circ \paren {e \circ x}$ by $(4)$ $\ds$ $=$ $\ds z \circ x$ Definition of Identity Element $\ds$ $=$ $\ds e$ by $(5)$

Hence every element of $S$ has a inverse.

$\blacksquare$

## Proof 2

### Necessary Condition

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

$\struct {S, \circ}$ is a semigroup by the definition of a group.

By Group has Latin Square Property, the Latin square property holds in $S$.

$\Box$

### Sufficient Condition

Let $a \in G$.

We have by hypothesis:

$\exists e \in S: a \circ e = a$

and such an $e$ is unique.

Let $b \in G$.

Also by hypothesis:

$\exists x \in G: x \circ a = b$

and such an $x$ is unique.

Hence:

 $\ds b$ $=$ $\ds x \circ a$ $\ds$ $=$ $\ds x \circ a \circ e$ $\ds$ $=$ $\ds b \circ e$

Also by hypothesis:

$\exists a' \in G: a' \circ a = e$

Thus $\struct {S, \circ}$ satisfies the right-hand Group Axioms

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

$\blacksquare$