Equivalence of definitions of group action

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.

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.