Group has Latin Square Property/Proof 1

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

\(\ds g\)

\(=\)

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

\(\ds \leadsto \ \ \)

\(\ds a \circ g\)

\(=\)

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

\(\ds \leadsto \ \ \)

\(\ds a \circ g\)

\(=\)

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

Group Axiom $\text G 1$: Associativity

\(\ds \leadsto \ \ \)

\(\ds a \circ g\)

\(=\)

\(\ds e \circ b\)

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

\(\ds \leadsto \ \ \)

\(\ds a \circ g\)

\(=\)

\(\ds b\)

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

Thus, such a $g$ exists.

Suppose $g, g' \in G$ where $a \circ g = b = a \circ g'$.

Then:

\(\ds g\)

\(=\)

\(\ds e \circ g\)

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

\(\ds \)

\(=\)

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

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

\(\ds \)

\(=\)

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

Group Axiom $\text G 1$: Associativity

\(\ds \)

\(=\)

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

Substitution for $a \circ g$

\(\ds \)

\(=\)

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

Substitution for $a \circ g'$

\(\ds \)

\(=\)

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

Group Axiom $\text G 1$: Associativity

\(\ds \)

\(=\)

\(\ds e \circ g'\)

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

\(\ds \)

\(=\)

\(\ds g'\)

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

Thus uniqueness holds.

To prove the second part of the theorem, let $h = b \circ a^{-1}$.

The remainder of the proof follows a similar procedure to the above.

$\blacksquare$

Sources

• 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.4$: Theorem $1$