Group has Latin Square Property/Proof 3

From ProofWiki
Jump to navigation Jump to 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

Suppose that $\exists x, y \in G: a \circ x = b = a \circ y$.

\(\ds a \circ x\) \(=\) \(\ds a \circ y\)
\(\ds \leadsto \ \ \) \(\ds a^{-1} \circ \paren {a \circ x}\) \(=\) \(\ds a^{-1} \circ \paren {a \circ y}\) Group Axiom $\text G 3$: Existence of Inverse Element
\(\ds \leadsto \ \ \) \(\ds \paren {a^{-1} \circ a} \circ x\) \(=\) \(\ds \paren {a^{-1} \circ a} \circ y\) Group Axiom $\text G 1$: Associativity
\(\ds \leadsto \ \ \) \(\ds e \circ x\) \(=\) \(\ds e \circ y\) Group Axiom $\text G 3$: Existence of Inverse Element
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds y\) Group Axiom $\text G 2$: Existence of Identity Element

So such an element, if it exists, is unique.


Now it is demonstrated that $g = a^{-1} b$ satisfies the requirement for $a \circ g = b$

Since $a \in G$, it follows by group axiom $G3$: existence of inverses that $a^{-1} \in G$.

\(\ds a\) \(\in\) \(\ds G\)
\(\ds \leadsto \ \ \) \(\ds a^{-1}\) \(\in\) \(\ds G\) Group Axiom $\text G 3$: Existence of Inverse Element
\(\ds \leadsto \ \ \) \(\ds a^{-1} \circ b\) \(\in\) \(\ds G\) Group Axiom $\text G 0$: Closure


Then:

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

Thus, such a $g$ exists.


The properties of $h$ are proved similarly.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Group_has_Latin_Square_Property/Proof_3&oldid=528040"