# 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.

$\blacksquare$