Group has Latin Square Property/Proof 3

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

1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory: $(1.9)$
1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $3$: Elementary consequences of the definitions: Proposition $3.3$