Equivalence of definitions of group action

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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.