Product Inverse Operation is Self-Inverse
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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 $\oplus$ is self inverse in the sense that:
- $\forall x \in G: x \oplus x = e$
Proof
\(\ds \forall x \in G: \, \) | \(\ds x \oplus x\) | \(=\) | \(\ds x \circ x^{-1}\) | Definition of Product Inverse Operation | ||||||||||
\(\ds \) | \(=\) | \(\ds e\) | Group Axiom $\text G 3$: Existence of Inverse Element |
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.7 \ \text {(a)}: 1^\circ$