# 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:

\(\displaystyle a \circ g\) | \(=\) | \(\displaystyle b\) | |||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle a^{-1} \circ \paren {a \circ g}\) | \(=\) | \(\displaystyle a^{-1} \circ b\) | $\circ$ is a Cancellable Binary Operation | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \paren {a^{-1} \circ a} \circ g\) | \(=\) | \(\displaystyle a^{-1} \circ b\) | Group Axiom $\text G 1$: Associativity | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle e \circ g\) | \(=\) | \(\displaystyle a^{-1} \circ b\) | Group Axiom $\text G 3$: Existence of Inverse Element | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle g\) | \(=\) | \(\displaystyle a^{-1} \circ b\) | Group Axiom $\text G 2$: Existence of Identity Element |

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