# Equivalence of definitions of group action

This article gives a proof/explanation of the equivalence of multiple definitions for the term group action
View a complete list of pages giving proofs of equivalence of definitions

## 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 $\alpha(g,s) \ne s$.

### 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 $\alpha(k,s) = t \ne s$. 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 $g \in G$, $\rho(g) \ne \rho(e)$, hence is not the identity permutation. Thus, there exists $s \in S$ such that $\rho(g)$ does not fix $s$. Thus yields $\alpha(g,s) \ne s$, as desired.