Group has Latin Square Property/Proof 1

From ProofWiki
Jump to: navigation, search

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

\(\displaystyle g\) \(=\) \(\displaystyle a^{-1} \circ b\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle a \circ g\) \(=\) \(\displaystyle a \circ \left({a^{-1} \circ b}\right)\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle a \circ g\) \(=\) \(\displaystyle \left({a \circ a^{-1} }\right) \circ b\) $\quad$ Group axioms: $G1$: associativity $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle a \circ g\) \(=\) \(\displaystyle e \circ b\) $\quad$ Group axioms: $G3$: existence of inverse element $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle a \circ g\) \(=\) \(\displaystyle b\) $\quad$ Group axioms: $G2$: existence of identity element $\quad$

Thus, such a $g$ exists.


Suppose $g, g' \in G$ where $a \circ g = b = a \circ g'$.

Then:

\(\displaystyle g\) \(=\) \(\displaystyle e \circ g\) $\quad$ Group axioms: $G2$: existence of identity element $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({a^{-1} \circ a}\right) \circ g\) $\quad$ Group axioms: $G3$: existence of inverse element $\quad$
\(\displaystyle \) \(=\) \(\displaystyle a^{-1} \circ \left({a \circ g}\right)\) $\quad$ Group axioms: $G1$: associativity $\quad$
\(\displaystyle \) \(=\) \(\displaystyle a^{-1} \circ b\) $\quad$ Substitution for $a \circ g$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle a^{-1} \circ \left({a \circ g'}\right)\) $\quad$ Substitution for $a \circ g'$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({a^{-1} \circ a}\right) \circ g'\) $\quad$ Group axioms: $G1$: associativity $\quad$
\(\displaystyle \) \(=\) \(\displaystyle e \circ g'\) $\quad$ Group axioms: $G3$: existence of inverse element $\quad$
\(\displaystyle \) \(=\) \(\displaystyle g'\) $\quad$ Group axioms: $G2$: existence of identity element $\quad$

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.

$\blacksquare$


Sources