Invertible Element of Associative Structure is Cancellable
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Contents
Theorem
Let $\left({S, \circ}\right)$ be an algebraic structure where $\circ$ is associative.
Let $\left({S, \circ}\right)$ have an identity element $e_S$.
An element of $\left({S, \circ}\right)$ which is invertible is also cancellable.
Corollary
Let $\left({S, \circ}\right)$ be a monoid whose identity is $e_S$.
An element of $\left({S, \circ}\right)$ which is invertible is also cancellable.
Proof
Let $a \in S$ be invertible.
Suppose $a \circ x = a \circ y$.
Then:
| \(\displaystyle x\) | \(=\) | \(\displaystyle e_S \circ x\) | $\quad$ Behaviour of Identity | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \left({a^{-1} \circ a}\right) \circ x\) | $\quad$ Behaviour of Inverse | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a^{-1} \circ \left({a \circ x}\right)\) | $\quad$ Associativity of $\circ$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a^{-1} \circ \left({a \circ y}\right)\) | $\quad$ By Hypothesis | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \left({a^{-1} \circ a}\right) \circ y\) | $\quad$ [Definition:Associative Operation | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle e_S \circ y\) | $\quad$ Behaviour of Inverse | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle y\) | $\quad$ Behaviour of Identity | $\quad$ |
A similar argument shows that $x \circ a = y \circ a \implies x = y$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): $\S 7$: Theorem $7.1$
