Group has Latin Square Property/Proof 2

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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

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.

$\blacksquare$


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