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:
\(\ds a \circ g\) | \(=\) | \(\ds b\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds a^{-1} \circ \paren {a \circ g}\) | \(=\) | \(\ds a^{-1} \circ b\) | $\circ$ is a Cancellable Binary Operation | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \paren {a^{-1} \circ a} \circ g\) | \(=\) | \(\ds a^{-1} \circ b\) | Group Axiom $\text G 1$: Associativity | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds e \circ g\) | \(=\) | \(\ds a^{-1} \circ b\) | Group Axiom $\text G 3$: Existence of Inverse Element | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds g\) | \(=\) | \(\ds a^{-1} \circ b\) | 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.
$\blacksquare$
Sources
- 1974: Thomas W. Hungerford: Algebra ... (previous) ... (next): $\text{I}$: Groups: $\S 1$ Semigroups, Monoids and Groups