# Group has Latin Square Property/Proof 2

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## 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.

$\blacksquare$

## Sources

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