# 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 $x \circ a = y \circ a \implies x = y$.

$\blacksquare$