Semigroup is Group Iff Latin Square Property Holds/Proof 1

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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

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 $(2)$
\(\ds \) \(=\) \(\ds y \circ \paren {a \circ x}\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds y \circ a\) 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 $(3)$
\(\ds \) \(=\) \(\ds \paren {x \circ x} \circ z\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds x \circ z\) $x$ 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 $x$:

\(\ds y \circ x\) \(=\) \(\ds e \circ \paren {y \circ x}\) Definition of Identity Element
\(\ds \) \(=\) \(\ds \paren {z \circ x} \circ \paren {y \circ x}\) by $(5)$
\(\ds \) \(=\) \(\ds z \circ \paren {\paren {x \circ y} \circ x}\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds z \circ \paren {e \circ x}\) by $(4)$
\(\ds \) \(=\) \(\ds z \circ x\) Definition of Identity Element
\(\ds \) \(=\) \(\ds e\) by $(5)$

Hence every element of $S$ has a inverse.

$\blacksquare$


Sources

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