# Group has Latin Square Property

## 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 $h \circ a = b$.

This result can also be written in additive notation as follows:

Let $\struct {G, +}$ 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 $a + g = b$.

Similarly, there exists a unique $h \in G$ such that $h + a = b$.

### Corollary

The Cayley table for any finite group is a Latin square.

## Proof 1

 $\ds g$ $=$ $\ds a^{-1} \circ b$ $\ds \leadsto \ \$ $\ds a \circ g$ $=$ $\ds a \circ \paren {a^{-1} \circ b}$ $\ds \leadsto \ \$ $\ds a \circ g$ $=$ $\ds \paren {a \circ a^{-1} } \circ b$ Group Axiom $\text G 1$: Associativity $\ds \leadsto \ \$ $\ds a \circ g$ $=$ $\ds e \circ b$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds \leadsto \ \$ $\ds a \circ g$ $=$ $\ds b$ 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:

 $\ds g$ $=$ $\ds e \circ g$ Group Axiom $\text G 2$: Existence of Identity Element $\ds$ $=$ $\ds \paren {a^{-1} \circ a} \circ g$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds$ $=$ $\ds a^{-1} \circ \paren {a \circ g}$ Group Axiom $\text G 1$: Associativity $\ds$ $=$ $\ds a^{-1} \circ b$ Substitution for $a \circ g$ $\ds$ $=$ $\ds a^{-1} \circ \paren {a \circ g'}$ Substitution for $a \circ g'$ $\ds$ $=$ $\ds \paren {a^{-1} \circ a} \circ g'$ Group Axiom $\text G 1$: Associativity $\ds$ $=$ $\ds e \circ g'$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds$ $=$ $\ds g'$ 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.

$\blacksquare$

## Proof 2

We shall prove that this is true for the first equation:

 $\ds a \circ g$ $=$ $\ds b$ $\ds \leadstoandfrom \ \$ $\ds a^{-1} \circ \paren {a \circ g}$ $=$ $\ds a^{-1} \circ b$ $\circ$ is a Cancellable Binary Operation $\ds \leadstoandfrom \ \$ $\ds \paren {a^{-1} \circ a} \circ g$ $=$ $\ds a^{-1} \circ b$ Group Axiom $\text G 1$: Associativity $\ds \leadstoandfrom \ \$ $\ds e \circ g$ $=$ $\ds a^{-1} \circ b$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds \leadstoandfrom \ \$ $\ds g$ $=$ $\ds a^{-1} \circ b$ 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.

$\blacksquare$

## Proof 3

Suppose that $\exists x, y \in G: a \circ x = b = a \circ y$.

 $\ds a \circ x$ $=$ $\ds a \circ y$ $\ds \leadsto \ \$ $\ds a^{-1} \circ \paren {a \circ x}$ $=$ $\ds a^{-1} \circ \paren {a \circ y}$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds \leadsto \ \$ $\ds \paren {a^{-1} \circ a} \circ x$ $=$ $\ds \paren {a^{-1} \circ a} \circ y$ Group Axiom $\text G 1$: Associativity $\ds \leadsto \ \$ $\ds e \circ x$ $=$ $\ds e \circ y$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds \leadsto \ \$ $\ds x$ $=$ $\ds y$ Group Axiom $\text G 2$: Existence of Identity Element

So such an element, if it exists, is unique.

Now it is demonstrated that $g = a^{-1} b$ satisfies the requirement for $a \circ g = b$

Since $a \in G$, it follows by group axiom $G3$: existence of inverses that $a^{-1} \in G$.

 $\ds a$ $\in$ $\ds G$ $\ds \leadsto \ \$ $\ds a^{-1}$ $\in$ $\ds G$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds \leadsto \ \$ $\ds a^{-1} \circ b$ $\in$ $\ds G$ Group Axiom $\text G 0$: Closure

Then:

 $\ds a \circ g$ $=$ $\ds a \circ \paren {a^{-1} \circ b}$ $\ds$ $=$ $\ds \paren {a \circ a^{-1} } \circ b$ Group Axiom $\text G 1$: Associativity $\ds$ $=$ $\ds e \circ b$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds$ $=$ $\ds b$ Group Axiom $\text G 2$: Existence of Identity Element

Thus, such a $g$ exists.

The properties of $h$ are proved similarly.

$\blacksquare$

## Proof 4

We shall prove that this is true for the first equation:

 $\ds b$ $=$ $\ds a \circ g$ $\ds \leadsto \ \$ $\ds a^{-1} \circ b$ $=$ $\ds a^{-1} \circ \paren {a \circ g}$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds$ $=$ $\ds \paren {a^{-1} \circ a} \circ g$ Group Axiom $\text G 1$: Associativity $\ds$ $=$ $\ds e \circ g$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds$ $=$ $\ds g$ Group Axiom $\text G 2$: Existence of Identity Element

Conversely:

 $\ds g$ $=$ $\ds a^{-1} \circ b$ $\ds \leadsto \ \$ $\ds a \circ g$ $=$ $\ds a \circ \paren {a^{-1} \circ b}$ $\ds$ $=$ $\ds \paren {a \circ a^{-1} } \circ b$ Group Axiom $\text G 1$: Associativity $\ds$ $=$ $\ds e \circ b$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds$ $=$ $\ds b$ Group Axiom $\text G 2$: Existence of Identity Element

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

$\blacksquare$