Invertible Element of Associative Structure is Cancellable

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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:

\(\displaystyle x\) \(=\) \(\displaystyle e_S \circ x\) Behaviour of Identity
\(\displaystyle \) \(=\) \(\displaystyle \left({a^{-1} \circ a}\right) \circ x\) Behaviour of Inverse
\(\displaystyle \) \(=\) \(\displaystyle a^{-1} \circ \left({a \circ x}\right)\) Associativity of $\circ$
\(\displaystyle \) \(=\) \(\displaystyle a^{-1} \circ \left({a \circ y}\right)\) By Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \left({a^{-1} \circ a}\right) \circ y\) [Definition:Associative Operation
\(\displaystyle \) \(=\) \(\displaystyle e_S \circ y\) Behaviour of Inverse
\(\displaystyle \) \(=\) \(\displaystyle y\) Behaviour of Identity


A similar argument shows that $x \circ a = y \circ a \implies x = y$.

$\blacksquare$


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