Tour:Invertible implies cancellative in monoid

This article adapts material from the main article: invertible implies cancellative in monoid

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WHAT YOU NEED TO DO:

• Understand the statement below; try proving it yourself
• Understand the proof, and the crucial way in which it relies on associativity

PONDER:

• What happens if we remove the assumption of associativity? Can you cook up a magma (set with binary operation) having a neutral element, where an element has a left inverse but is not cancellative?

Statement

In a monoid (a set with an associative binary operation possessing a multiplicative identity element) the following are true:

• Any left invertible element (element having a left inverse) is left cancellative.
• Any right invertible element (element having a right inverse) is right cancellative.
• Any invertible element is cancellative.

Proof

We'll give here the proof for left invertible and left cancellative. An analogous proof works for right invertible and right cancellative.

Given: A monoid $M$ with binary operation $*$, and identity element (also called neutral element) $e$. $a \in M$ has a left inverse $b$ (i.e. an element $b * a = e$)

To prove: $a$ is left-cancellative: whenever $c,d \in M$ are such that $a * c = a * d$, then $c = d$.

Proof: We start with:

$a * c = a * d$

Left-multiply both sides by $b$:

$b * (a * c) = b * (a * d)$

Use associativity:

$(b * a) * c = (b * a) * d$

We now use that $b * a = e$ is the identity element, to conclude that $c = d$.

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