Tour:Equality of left and right inverses

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This article adapts material from the main article: equality of left and right inverses 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:
  • Read, and understand, the statement below, and try proving it.
  • Read the proof and make sure you understand it, as well as the significance of associativity.
    PONDER (WILL BE EXPLORED LATER): What happens when we remove associativity? Can you cook up binary operations where left and right inverses exist but are no longer equal?

Statement

Verbal statement

Suppose * is the associative binary operation of a monoid, and e is its neutral element (or identity element). If an element has both a left and a right inverse with respect to *, then the left and right inverse are equal.

Statement with symbols

Suppose S is a monoid with binary operation * and neutral element e. If an element a \in S has a left inverse b (i.e., b * a = e)and a right inverse c (i.e., a * c = e), then b = c.


  • Two-sided inverse is unique if it exists in monoid
  • In a monoid, if an element has a left inverse, it can have at most one right inverse; moreover, if the right inverse exists, it must be equal to the left inverse, and is thus a two-sided inverse.
  • In a monoid, if an element has a right inverse, it can have at most one left inverse; moreover, if the left inverse exists, it must be equal to the right inverse, and is thus a two-sided inverse.
  • In a monoid, if an element has two distinct left inverses, it cannot have a right inverse, and hence cannot have a two-sided inverse.
  • In a monoid, if an element has two distinct right inverses, it cannot have a left inverse, and hence cannot have a two-sided inverse.
  • In a group, every element has a unique left inverse (same as its two-sided inverse) and a unique right inverse (same as its two-sided inverse).


Proof

Proof idea

The idea is to pit the left inverse of an element against its right inverse. Starting with an element a, whose left inverse is b and whose right inverse is c, we need to form an expression that pits b against c, and can be simplified both to b and to c.

The only relation known between b and c is their relation with a: b * a is the neutral element and a * c is the neutral element. To use both these facts, we construct the expression b * a * c. The two ways of parenthesizing this expression allow us to simplify the expression in different ways.

The key idea here is that since b and c are related through a, we need to put a in between them in the expression. Then, we need associativity to interpret the expression in different ways and simplify to obtain the result.

Formal proof

Given: A monoid S with associative binary operation * and neutral element e. An element a of S with left inverse b and right inverse c.

To prove: b = c

Proof: We consider two ways of associating the expression b * a * c.

(b * a) * c = b * (a * c) by associativity. The left side simplifies to e * c  = c while the right side simplifies to b * e = b. Hence, b = c.


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