Group has Latin Square Property/Proof 4

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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:


\(\displaystyle b\) \(=\) \(\displaystyle a \circ g\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^{-1} \circ b\) \(=\) \(\displaystyle a^{-1} \circ \paren {a \circ g}\) 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.

$\blacksquare$


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