Equality of left and right inverses in monoid
This article gives a statement (possibly with proof) of how, if a left-based construction and a right-based construction both exist, they must be equal.
View other such statements
Contents
Statement
Verbal statement
Suppose is the associative binary operation of a monoid, and 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 is a monoid with binary operation and neutral element . If an element has a left inverse (i.e., )and a right inverse (i.e., ), then .
Related facts
Similar facts
Statement | Elaboration | Relation with existing statement |
---|---|---|
Flexible and cancellative and existence of neutral element implies equality of left and right inverses | Flexible means that , cancellative means cancellation property holds on both left and right | We sacrifice associativity for flexibility, but need cancellation to make up for it. |
Left inverse property implies two-sided inverses exist | In a loop, if a left inverse exists and satisfies the left inverse property, then it must also be the unique right inverse (though it need not satisfy the right inverse property) | The left inverse property allows us to use associativity as required in the proof. Ssince we are dealing with a loop, the existence of a right inverse is automatic. |
Corollaries
Some easy corollaries:
- 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).
More indirect corollaries:
Proof
Proof idea
The idea is to pit the left inverse of an element against its right inverse. Starting with an element , whose left inverse is and whose right inverse is , we need to form an expression that pits against , and can be simplified both to and to .
The only relation known between and is their relation with : is the neutral element and is the neutral element. To use both these facts, we construct the expression . The two ways of parenthesizing this expression allow us to simplify the expression in different ways.
The key idea here is that since and are related through , we need to put 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 with associative binary operation and neutral element . An element of with left inverse and right inverse .
To prove:
Proof: We consider two ways of associating the expression .
by associativity. The left side simplifies to while the right side simplifies to . Hence, .