Group Identity is Right Identity for Product Inverse Operation

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 $7.7 \ \text {(a)}: 2^\circ$