Semigroup is Group Iff Latin Square Property Holds/Proof 2

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 $a \in G$.

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.

$\blacksquare$

Sources

• 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms: Theorem $1$