# Difference between revisions of "Tour:Inverse map is involutive"

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## Latest revision as of 17:41, 29 August 2008

**This article adapts material from the main article:** inverse map is involutive

This page is part of the Groupprops guided tour for beginners (Jump to beginning of tour)PREVIOUS: Associative binary operation|UP: Introduction two (beginners)|NEXT: Order of a groupExpected time for this page: 5 minutes

General instructions for the tour | Pedagogical notes for the tour | Pedagogical notes for this part

WHAT YOU NEED TO DO:

- Understand the statements below
- Try proving them, using what you understand about associativity and inverses
- Check the proof given in the text
PONDER: What weakening of associativity would be sufficient for the reversal law to continue to hold? (See where precisely associativity is used in the proof)

## Statement

The inverse map in a group, i.e. the map sending any element of the group, to its inverse element, is an *involutive map*, in the sense that it has the following two properties:

- It satisfies the reversal law:

- Applying it twice sends an element to itself:

It fact, both these are true in the greater generality of a monoid, under the condition that all the s have two-sided inverse (note: we still need a monoid to guarantee that two-sided inverses, when they exist, are unique).

## Proof

### Proof of reversal law

In order to show that the element is a two-sided inverse of , it suffices to show that their product both ways is the identity element. Consider first the product:

Due to associativity, we can drop the parentheses and we get:

Now, consider the middle product . This is the identity element, and since the identity element has no effect on the remaining product, it can be removed, giving the product:

We now repeat the argument with the middle product and *cancel* them. Proceeding this way, we are able to cancel all terms and eventually get the identity element.

A similar argument follows for the product the other way around:

Thus, the elements are two-sided inverses of each other.

**Note**: In fact, it suffices to check only one of the two inverse conditions, i.e., check only that the first product is the identity element. This is because, in a group, every element has a two-sided inverse. Further, equality of left and right inverses in monoid forces any *one-sided* (left *or* right) inverse to be equal to the two-sided inverse.

### Proof for applying it twice

This is direct from the definition. let . Then, by the inherent symmetry in the definition of inverse element, we see that .

More explicitly, if , that means that . But this is precisely the condition for stating that .

This page is part of the Groupprops guided tour for beginners. Make notes of any doubts, confusions or comments you have about this page before proceeding.PREVIOUS: Associative binary operation|UP: Introduction two (beginners)|NEXT: Order of a group