# Group has Latin Square Property/Proof 1

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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

\(\displaystyle g\) | \(=\) | \(\displaystyle a^{-1} \circ b\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle a \circ g\) | \(=\) | \(\displaystyle a \circ \paren {a^{-1} \circ b}\) | ||||||||||

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

\(\displaystyle \leadsto \ \ \) | \(\displaystyle a \circ g\) | \(=\) | \(\displaystyle e \circ b\) | Group Axiom $\text G 3$: Existence of Inverse Element | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle a \circ g\) | \(=\) | \(\displaystyle b\) | Group Axiom $\text G 2$: Existence of Identity Element |

Thus, such a $g$ exists.

Suppose $g, g' \in G$ where $a \circ g = b = a \circ g'$.

Then:

\(\displaystyle g\) | \(=\) | \(\displaystyle e \circ g\) | Group Axiom $\text G 2$: Existence of Identity Element | ||||||||||

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

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

\(\displaystyle \) | \(=\) | \(\displaystyle a^{-1} \circ b\) | Substitution for $a \circ g$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a^{-1} \circ \paren {a \circ g'}\) | Substitution for $a \circ g'$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a^{-1} \circ a} \circ g'\) | 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 |

Thus uniqueness holds.

To prove the second part of the theorem, let $h = b \circ a^{-1}$.

The remainder of the proof follows a similar procedure to the above.

$\blacksquare$

## Sources

- 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 1.4$: Theorem $1$