# Tour:Equivalence of definitions of group action

**This article adapts material from the main article:** equivalence of definitions of group action

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In an earlier page of this tour, you saw one definition of group action. There are actually two definitions of group action: the one you saw and another one in terms of homomorphism to a symmetric group. This page gives both definitions and proves their equivalence.WHAT YOU NEED TO DO: Understand both definitions, and the proof of their equivalence.

## Contents

## The definitions we have to prove as equivalent

### Definition as a map from a product

A group action of a group on a set is a map satisfying:

Here denotes the identity element.

Further, the group action is termed faithful or effective if for every non-identity element , there exists such that .

### Definition as a homomorphism to a symmetric group

A group action of a group on a set is a homomorphism of groups (where denotes the symmetric group on ).

Further, the action is termed faithful or effective if is injective.

## Proof of equivalence

### The first definition implies the second

Suppose we are given a map satisfying the conditions:

and:

We now construct the map . For every , is defined as the map:

Clearly, is a function from to , and is the identity map (by the second assumption on . The fact that is in for every holds because:

Hence is a left inverse for , and a similar argument shows that it is a right inverse. Thus, is an invertible map from to , hence an element of .

Finally the fact that is a homomorphism follows from the first condition:

Further, suppose is faithful. Consider any two distinct elements of . Consider . Since are distinct, is a non-identity element, so by faithfulness, there exists such that . Using the definition of group action, we obtain that while . In particular, and are permutations that differ on , hence are not equal. Thus, distinct elements of map to distinct permutations, and is thus injective.

### The second definition implies the first

Given a homomorphism , the map is given by:

Let us check that satisfies both the specified conditions:

The latter is because a homomorphism of groups takes the identity element to the identity element.

Further, if is injective, then for any non-identity element , , hence is not the identity permutation. Thus, there exists such that does not fix . Thus yields , as desired.

This page is part of the Groupprops guided tour for beginners (Jump to beginning of tour)PREVIOUS: Understanding the definition of a homomorphism|UP: Introduction five (beginners)|NEXT: Cayley's theorem

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