Group has Latin Square Property

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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 $h \circ a = b$.


Additive Notation

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

\(\displaystyle g\) \(=\) \(\displaystyle a^{-1} \circ b\)
\(\displaystyle \implies \ \ \) \(\displaystyle a \circ g\) \(=\) \(\displaystyle a \circ \left({a^{-1} \circ b}\right)\)
\(\displaystyle \implies \ \ \) \(\displaystyle a \circ g\) \(=\) \(\displaystyle \left({a \circ a^{-1} }\right) \circ b\) Group axioms: $G1$: associativity
\(\displaystyle \implies \ \ \) \(\displaystyle a \circ g\) \(=\) \(\displaystyle e \circ b\) Group axioms: $G3$: existence of inverse element
\(\displaystyle \implies \ \ \) \(\displaystyle a \circ g\) \(=\) \(\displaystyle b\) Group axioms: $G2$: 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 axioms: $G2$: existence of identity element
\(\displaystyle \) \(=\) \(\displaystyle \left({a^{-1} \circ a}\right) \circ g\) Group axioms: $G3$: existence of inverse element
\(\displaystyle \) \(=\) \(\displaystyle a^{-1} \circ \left({a \circ g}\right)\) Group axioms: $G1$: associativity
\(\displaystyle \) \(=\) \(\displaystyle a^{-1} \circ b\) Substitution for $a \circ g$
\(\displaystyle \) \(=\) \(\displaystyle a^{-1} \circ \left({a \circ g'}\right)\) Substitution for $a \circ g'$
\(\displaystyle \) \(=\) \(\displaystyle \left({a^{-1} \circ a}\right) \circ g'\) Group axioms: $G1$: associativity
\(\displaystyle \) \(=\) \(\displaystyle e \circ g'\) Group axioms: $G3$: existence of inverse element
\(\displaystyle \) \(=\) \(\displaystyle g'\) Group axioms: $G2$: 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:


\(\displaystyle a \circ g\) \(=\) \(\displaystyle b\)
\(\displaystyle \iff \ \ \) \(\displaystyle a^{-1} \circ \left({a \circ g}\right)\) \(=\) \(\displaystyle a^{-1} \circ b\) $\circ$ is a Cancellable Binary Operation
\(\displaystyle \iff \ \ \) \(\displaystyle \left({a^{-1} \circ a}\right) \circ g\) \(=\) \(\displaystyle a^{-1} \circ b\) Group axiom $G1$: Associativity
\(\displaystyle \iff \ \ \) \(\displaystyle e \circ g\) \(=\) \(\displaystyle a^{-1} \circ b\) Group axiom $G3$: property of Inverses
\(\displaystyle \iff \ \ \) \(\displaystyle g\) \(=\) \(\displaystyle a^{-1} \circ b\) Group axiom $G2$: property of Identity


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 $G \, 3$: Inverses
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {a^{-1} \circ a} \circ x\) \(=\) \(\displaystyle \paren {a^{-1} \circ a} \circ y\) Group Axiom $G \, 1$: Associativity
\(\displaystyle \leadsto \ \ \) \(\displaystyle e \circ x\) \(=\) \(\displaystyle e \circ y\) Group Axiom $G \, 3$: Inverses
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle y\) Group Axiom $G \, 2$: Identity

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 $G \, 3$: Inverses
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^{-1} \circ b\) \(\in\) \(\displaystyle G\) Group Axiom $G \, 0$: 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 $G \, 1$: Associativity
\(\displaystyle \) \(=\) \(\displaystyle e \circ b\) Group Axiom $G \, 3$: Inverses
\(\displaystyle \) \(=\) \(\displaystyle b\) Group Axiom $G \, 2$: Identity

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 $G \, 3$: 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.

$\blacksquare$