Group has Latin Square Property/Proof 1

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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

\(\displaystyle g\) \(=\) \(\displaystyle a^{-1} \circ b\)
\(\displaystyle \implies \ \ \) \(\displaystyle a \circ g\) \(=\) \(\displaystyle a \circ \left({a^{-1} \circ b}\right)\)
\(\displaystyle \implies \ \ \) \(\displaystyle a \circ g\) \(=\) \(\displaystyle \left({a \circ a^{-1} }\right) \circ b\) 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.

$\blacksquare$


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