Group has Latin Square Property/Proof 2

Theorem

Let $\struct {G, \circ}$ be a group.

Then $G$ satisfies the Latin square property.

That is, for all $a, b \in G$, there exists a unique $g \in G$ such that $a \circ g = b$.

Similarly, there exists a unique $h \in G$ such that $h \circ a = b$.

Proof

We shall prove that this is true for the first equation:

\(\ds a \circ g\)

\(=\)

\(\ds b\)

\(\ds \leadstoandfrom \ \ \)

\(\ds a^{-1} \circ \paren {a \circ g}\)

\(=\)

\(\ds a^{-1} \circ b\)

$\circ$ is a Cancellable Binary Operation

\(\ds \leadstoandfrom \ \ \)

\(\ds \paren {a^{-1} \circ a} \circ g\)

\(=\)

\(\ds a^{-1} \circ b\)

Group Axiom $\text G 1$: Associativity

\(\ds \leadstoandfrom \ \ \)

\(\ds e \circ g\)

\(=\)

\(\ds a^{-1} \circ b\)

Group Axiom $\text G 3$: Existence of Inverse Element

\(\ds \leadstoandfrom \ \ \)

\(\ds g\)

\(=\)

\(\ds a^{-1} \circ b\)

Group Axiom $\text G 2$: Existence of Identity Element

Because the statements:

$a \circ g = b$

and

$g = a^{-1} \circ b$

are equivalent, we may conclude that $g$ is indeed the only solution of the equation.

The proof that the unique solution of $h$ is $b \circ a^{-1}$ in the second equation follows similarly.

$\blacksquare$

Sources

• 1974: Thomas W. Hungerford: Algebra ... (previous) ... (next): $\text{I}$: Groups: $\S 1$ Semigroups, Monoids and Groups