# Tour:Equality of left and right inverses

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General instructions for the tour | Pedagogical notes for the tour | Pedagogical notes for this part
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, $b = c$.