# 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 $\left({S, \circ}\right)$ be a group.

$\left({S, \circ}\right)$ 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 $\left({S, \circ}\right)$ be a semigroup on which the Latin square property holds.

Taking the group axioms in turn:

#### G0: Closure

$\left({S, \circ}\right)$ is semigroup.

Hence $\left({S, \circ}\right)$ is closed by definition.

$\Box$

#### G1: Associativity

$\left({S, \circ}\right)$ is semigroup.

Hence $\circ$ is associative by definition.

$\Box$

#### G2: Identity

Let $a \in S$.

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

 $(1):\quad$ $\displaystyle a \circ x$ $=$ $\displaystyle a$

Let $b \in S$.

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

 $(2):\quad$ $\displaystyle y \circ a$ $=$ $\displaystyle b$

Proving that $x$ is a right identity:

 $\displaystyle b \circ x$ $=$ $\displaystyle \left({y \circ a}\right) \circ x$ by $(2)$ $\displaystyle$ $=$ $\displaystyle y \circ \left({a \circ x}\right)$ Definition of Associativity $\displaystyle$ $=$ $\displaystyle y \circ a$ by $(1)$ $\displaystyle$ $=$ $\displaystyle b$ by $(2)$

Let $c \in S$.

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

 $(3):\quad$ $\displaystyle x \circ z$ $=$ $\displaystyle c$

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

 $\displaystyle x \circ c$ $=$ $\displaystyle x \circ \left({x \circ z}\right)$ by $(3)$ $\displaystyle$ $=$ $\displaystyle \left({x \circ x}\right) \circ z$ Definition of Associativity $\displaystyle$ $=$ $\displaystyle x \circ z$ $x$ is a right identity $\displaystyle$ $=$ $\displaystyle c$ by $(3)$

Relabel $x$ as $e$.

$\Box$

#### G3: Inverses

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:

 $(4):\quad$ $\displaystyle x \circ y$ $=$ $\displaystyle e$
 $(5):\quad$ $\displaystyle z \circ x$ $=$ $\displaystyle e$

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

 $\displaystyle y \circ x$ $=$ $\displaystyle \left({z \circ x}\right) \circ \left({y \circ x}\right)$ Definition of Identity $\displaystyle$ $=$ $\displaystyle z \circ \left({\left({x \circ y}\right) \circ x}\right)$ Definition of Associativity $\displaystyle$ $=$ $\displaystyle z \circ \left({e \circ x}\right)$ by $(4)$ $\displaystyle$ $=$ $\displaystyle z \circ x$ Definition of Identity $\displaystyle$ $=$ $\displaystyle e$ by $(5)$

Hence every element of $S$ has a inverse.

$\blacksquare$

## Proof 2

### Necessary Condition

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

$\left({S, \circ}\right)$ 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 $a \in G$.

By hypothesis:

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

and such an $e$ is unique.

Let $b \in G$.

By hypothesis:

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

and such an $x$ is unique.

Hence:

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

Also by hypothesis:

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

Thus $\left({S, \circ}\right)$ satisfies the right-hand Group Axioms

Thus $\left({S, \circ}\right)$ is a group.

$\blacksquare$