# Semigroup is Group Iff Latin Square Property Holds

## Contents

## 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.

$\blacksquare$