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.
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 4.6$. Elementary theorems on groups: Theorem $\text{(ii)}$
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms