Tour:Equivalence of definitions of group action

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

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

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.

The definitions we have to prove as equivalent

Definition as a map from a product

A group action of a group $G$ on a set $S$ is a map $\alpha :G\times S\to S$ satisfying:

1. $\alpha (gh,s)=\alpha (g,\alpha (h,s))$
2. $\alpha (e,s)=s$

Here $e$ denotes the identity element.

Further, the group action is termed faithful or effective if for every non-identity element $g\in G$, there exists $s\in S$ such that $\alpha (g,s)\neq s$.

Definition as a homomorphism to a symmetric group

A group action of a group $G$ on a set $S$ is a homomorphism of groups $\rho :G\to \operatorname {Sym} (S)$ (where $\operatorname {Sym} (S)$ denotes the symmetric group on $S$).

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

Proof of equivalence

The first definition implies the second

Suppose we are given a map $\alpha :G\times S\to S$ satisfying the conditions:

$\alpha (gh,s)=\alpha (g,\alpha (h,s))$

and:

$\alpha (e,s)=s$

We now construct the map $\rho$. For every $g\in G$, $\rho (g)$ is defined as the map:

$s\mapsto \alpha (g,s)$

Clearly, $\rho (g)$ is a function from $S$ to $S$, and $\rho (e)$ is the identity map (by the second assumption on $\alpha$. The fact that $\rho (g)$ is in $Sym(S)$ for every $g$ holds because:

$\alpha (e,s)=\alpha (g^{-1}g,s)=\alpha (g^{-1}\alpha (g,s))=\rho (g^{-1})(\rho (g)s)$

Hence $\rho (g^{-1})$ is a left inverse for $\rho (g)$, and a similar argument shows that it is a right inverse. Thus, $\rho (g)$ is an invertible map from $S$ to $S$, hence an element of $Sym(S)$.

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

$\rho (gh)(s)=\alpha (gh,s)=\alpha (g,\alpha (h,s))=\alpha (g,\rho (h)s)=(\rho (g)\rho (h))(s)$

Further, suppose $\alpha$ is faithful. Consider any two distinct elements $g,h$ of $G$. Consider $k=hg^{-1}$. Since $g,h$ are distinct, $k$ is a non-identity element, so by faithfulness, there exists $s\in S$ such that $\alpha (k,s)=t\neq s$. Using the definition of group action, we obtain that $\alpha (h,\alpha (g^{-1},s))=t$ while $\alpha (g,\alpha (g^{-1},s))=s$. In particular, $\rho (h)$ and $\rho (g)$ are permutations that differ on $\alpha (g^{-1},s)$, hence are not equal. Thus, distinct elements of $G$ map to distinct permutations, and $\rho$ is thus injective.

The second definition implies the first

Given a homomorphism $\rho :G\to \operatorname {Sym} (S)$, the map $\alpha$ is given by:

$\alpha (g,s)=\rho (g)s$

Let us check that $\alpha$ satisfies both the specified conditions:

1. $\alpha (gh,s)=\rho (gh)s=(\rho (g)\rho (h))(s)=\rho (g)(\rho (h)s)=\rho (g)\alpha (h,s)=\alpha (g,\alpha (h,s))$
2. $\alpha (e,s)=\rho (e)s=s$

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

Further, if $\rho$ is injective, then for any non-identity element $g\in G$, $\rho (g)\neq \rho (e)$, hence is not the identity permutation. Thus, there exists $s\in S$ such that $\rho (g)$ does not fix $s$. Thus yields $\alpha (g,s)\neq s$, 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