Group Identity is Right Identity for Product Inverse Operation
Jump to navigation
Jump to search
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 $e$ is also the right identity of $\struct {G, \oplus}$, in the sense that:
- $\forall x \in G: x \oplus e = x$
Proof
\(\ds \forall x \in G: \, \) | \(\ds x \oplus e\) | \(=\) | \(\ds x \circ e^{-1}\) | Definition of Product Inverse Operation | ||||||||||
\(\ds \) | \(=\) | \(\ds x \circ e\) | Inverse of Identity Element is Itself | |||||||||||
\(\ds \) | \(=\) | \(\ds x\) | 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 $7.7 \ \text {(a)}: 2^\circ$