Finding permutations of a stabilizer subgroup of An

[goalieplayer]

Alright, I understand what a stabilizer is in a group, and I know how to find the permutations of An for any small integer n, but for a stabilizer, since it just maps every element to 1, would all permutations just be (1 2) (1 3) ... (1 n) for An?

 

[AcidRainLiTE]

I think you may be confused about what the stabilizer is.

Suppose you have

1. a group G
2. a set S (this set is not necessarily a group, it is just a bunch of elements)

There need be no connection between the elements of S and the elements of G. That is, S is just a collection of elements that can be entirely distinct from the elements of G.

An action of G on S is a map from G \times S \rightarrow S. (We also place two requirements on the behavior of this map, but just ignore this for the moment). In other words, "G acts on S" means that given any g \in G and any x \in S we can "apply" g to x and get a new element, y, of S. Symbolically, gx = y (though this looks like we are operating g and x using the group operation of G, this is not what we are doing. x is not even in G; it is in S).

The group of S_n of permutations provides a very natural example of all this. Take for instance S₃. Here G = S₃ and S = {1,2,3}. Given a particular permutation \sigma \in S_3, we can talk about what the permutation does to any element of S. Take \sigma = 'the permutation that transposes 1 and 2'. Then \sigma2 = 1. So \sigma "acts" on the element 2 and gives the element 1.

Are there any permutations in S_3 which act on 2 and just give 2? Yes, there are two of them:

1. \sigma_1 = 'the permutation that transposes 1 and 3'
2. \sigma_2= 'the permuation that leaves all the elements fixed' (identity)

The stablilzer of 2 is the set of both these permutations: Stab(2) = \{\sigma_1, \sigma_2\}. In general, the stablilizer of an element x \in S is:
Stab(x) = \{g \in G | gx = x\}.

...since it just maps every element to 1.

Here is where I think you are somewhat confused. The stablilzer does not map every element to 1. The stablilzer of x \in S consists of all the elements of G that send x to x. The stablilzer does not send anything to 1 because there isn't really a 1 in S. 1 is in the group G.