Semigroup is Group Iff Latin Square Property Holds

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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.

$\blacksquare$