# Finding permutations of a stabilizer subgroup of An

1. Oct 15, 2012

### 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?

2. Oct 25, 2012

### 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 Sn 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:
$$Stab(x) = \{g \in G | gx = x\}.$$

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.

Last edited: Oct 25, 2012