Group Product Inverse Operation with Identity

Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $\oplus: G \times G \to G$ be the product inverse of $\circ$ on $G$.

Then:

$\forall x, y \in G: e \oplus \paren {x \oplus y} = y \oplus x$

Proof

\(\ds \forall x, y \in G: \, \)

\(\ds e \oplus \paren {x \oplus y}\)

\(=\)

\(\ds e \oplus \paren {x \circ y^{-1} }\)

Definition of Product Inverse Operation

\(\ds \)

\(=\)

\(\ds e \circ \paren {x \circ y^{-1} }^{-1}\)

Definition of Product Inverse Operation

\(\ds \)

\(=\)

\(\ds e \circ \paren {y \circ x^{-1} }\)

Inverse of Group Product

\(\ds \)

\(=\)

\(\ds e \circ \paren {y \oplus x}\)

Definition of Product Inverse Operation

\(\ds \)

\(=\)

\(\ds y \oplus x\)

Group Axiom $\text G 2$: Existence of Identity Element

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.7 \ \text {(a)}: 3^\circ$