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.
$\blacksquare$