Invertible Element of Associative Structure is Cancellable
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Theorem
Let $\struct {S, \circ}$ be an algebraic structure where $\circ$ is associative.
Let $\struct {S, \circ}$ have an identity element $e_S$.
An element of $\struct {S, \circ}$ which is invertible is also cancellable.
Corollary
Let $\struct {S, \circ}$ be a monoid whose identity element is $e_S$.
An element of $\struct {S, \circ}$ which is invertible is also cancellable.
Proof
Let $a \in S$ be invertible.
Suppose $a \circ x = a \circ y$.
Then:
\(\ds x\) | \(=\) | \(\ds e_S \circ x\) | Definition of Identity Element | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a^{-1} \circ a} \circ x\) | Definition of Inverse Element | |||||||||||
\(\ds \) | \(=\) | \(\ds a^{-1} \circ \paren {a \circ x}\) | Associativity of $\circ$ | |||||||||||
\(\ds \) | \(=\) | \(\ds a^{-1} \circ \paren {a \circ y}\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a^{-1} \circ a} \circ y\) | Associativity of $\circ$ | |||||||||||
\(\ds \) | \(=\) | \(\ds e_S \circ y\) | Definition of Inverse Element | |||||||||||
\(\ds \) | \(=\) | \(\ds y\) | Definition of Identity Element |
A similar argument shows that $x \circ a = y \circ a \implies x = y$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Theorem $7.1$