# 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

 $\displaystyle g$ $=$ $\displaystyle a^{-1} \circ b$ $\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.

$\blacksquare$

## Proof 2

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.

$\blacksquare$

## Proof 3

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

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

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

Then:

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

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

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 $\text G 1$: Associativity $\displaystyle$ $=$ $\displaystyle e \circ b$ Group Axiom $\text G 3$: Existence of Inverse Element $\displaystyle$ $=$ $\displaystyle b$ 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.

$\blacksquare$