# 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.

$\blacksquare$