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 $h \circ a = b$.
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
\(\ds g\) | \(=\) | \(\ds a^{-1} \circ b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a \circ g\) | \(=\) | \(\ds a \circ \paren {a^{-1} \circ b}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a \circ g\) | \(=\) | \(\ds \paren {a \circ a^{-1} } \circ b\) | 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.
$\blacksquare$