# 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:

 $\displaystyle a \circ g$ $=$ $\displaystyle b$ $\displaystyle \iff \ \$ $\displaystyle a^{-1} \circ \left({a \circ g}\right)$ $=$ $\displaystyle a^{-1} \circ b$ $\circ$ is a Cancellable Binary Operation $\displaystyle \iff \ \$ $\displaystyle \left({a^{-1} \circ a}\right) \circ g$ $=$ $\displaystyle a^{-1} \circ b$ Group axiom $G1$: Associativity $\displaystyle \iff \ \$ $\displaystyle e \circ g$ $=$ $\displaystyle a^{-1} \circ b$ Group axiom $G3$: property of Inverses $\displaystyle \iff \ \$ $\displaystyle g$ $=$ $\displaystyle a^{-1} \circ b$ Group axiom $G2$: property of Identity

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.

$\blacksquare$