Invertible Element of Associative Structure is Cancellable

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

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): $\S 7$: Theorem $7.1$