# Group has Latin Square Property/Proof 4

## 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:

 $\displaystyle b$ $=$ $\displaystyle a \circ g$ $\displaystyle \leadsto \ \$ $\displaystyle a^{-1} \circ b$ $=$ $\displaystyle a^{-1} \circ \paren {a \circ g}$ Group Axiom $\text G 3$: Existence of Inverse Element $\displaystyle$ $=$ $\displaystyle \paren {a^{-1} \circ a} \circ g$ Group Axiom $\text G 1$: Associativity $\displaystyle$ $=$ $\displaystyle e \circ g$ Group Axiom $\text G 3$: Existence of Inverse Element $\displaystyle$ $=$ $\displaystyle g$ Group Axiom $\text G 2$: Existence of Identity Element

Conversely:

 $\displaystyle g$ $=$ $\displaystyle a^{-1} \circ b$ $\displaystyle \leadsto \ \$ $\displaystyle a \circ g$ $=$ $\displaystyle a \circ \paren {a^{-1} \circ b}$ $\displaystyle$ $=$ $\displaystyle \paren {a \circ a^{-1} } \circ b$ Group Axiom $\text G 1$: Associativity $\displaystyle$ $=$ $\displaystyle e \circ b$ Group Axiom $\text G 3$: Existence of Inverse Element $\displaystyle$ $=$ $\displaystyle b$ Group Axiom $\text G 2$: Existence of Identity Element

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

$\blacksquare$