Cancellation Property of Product Inverse Operator

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:

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

Proof

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

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

\(=\)

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

Definition of Product Inverse Operation

\(\ds \)

\(=\)

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

Definition of Product Inverse Operation

\(\ds \)

\(=\)

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

Inverse of Group Product

\(\ds \)

\(=\)

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

Group Axiom $\text G 1$: Associativity

\(\ds \)

\(=\)

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

Group Axiom $\text G 3$: Existence of Inverse Element

\(\ds \)

\(=\)

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

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

\(\ds \)

\(=\)

\(\ds x \oplus y\)

Definition of Product Inverse Operation

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.7 \ \text {(a)}: 4^\circ$