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:


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

$\blacksquare$


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