Invertible Elements of Semigroup Also Cancellable

Theorem

Let $\left({S, \circ}\right)$ be an 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:

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

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

$\blacksquare$

Sources

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