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Tour:Invertible implies cancellative in monoid

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

This page is part of the Groupprops guided tour for beginners (Jump to beginning of tour)
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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:

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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PREVIOUS: Equivalence of definitions of group| UP: Introduction two (beginners)| NEXT: Equivalence of definitions of subgroup