# Group has Latin Square Property/Proof 4

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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:

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

\(\displaystyle \leadsto \ \ \) | \(\displaystyle a^{-1} \circ b\) | \(=\) | \(\displaystyle a^{-1} \circ \paren {a \circ g}\) | Group Axiom $G \, 3$: Inverses | |||||||||

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

\(\displaystyle \) | \(=\) | \(\displaystyle e \circ g\) | Group Axiom $G \, 3$: Inverses | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle g\) | Group Axiom $G \, 2$: Identity |

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 $G1$: Associativity | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e \circ b\) | Group Axiom $G3$: Property of Inverse | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle b\) | Group Axiom $G2$: Property of Identity |

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

$\blacksquare$

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 4.6$. Elementary theorems on groups: Theorem $\text{(ii)}$ - 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms