# 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.
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## 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 $b = c$.

## 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 $x * (y * x) = (x * y) * x$, 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 $a$, 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, $b = c$.