Proving Alternating Groups in Sn: An Index 2 Subgroup

  • Thread starter Thread starter zcdfhn
  • Start date Start date
  • Tags Tags
    Groups
zcdfhn
Messages
23
Reaction score
0
Let An (the alternating group on n elements) consist of the set of all even permutations in Sn. Prove that An is indeed a subgroup of Sn and that it has index two in Sn and has order n!/2.

First of all, I need clarification on the definition of an alternating group. My book wasn't really good in explaining it.

My attempt at a solution is limited but I was thinking I can prove that An is a subgroup by showing that its elements are closed under composition and it is closed under inverses, but I need to better understand what an alternating group is before I start a proof. Also the order n!/2, I believe has to do with calculating permutations nPn-2 = n!/2! = n!/2.

Thanks in advance.
 
Physics news on Phys.org
A_n is simply the set of all even permutations. You know, i assume, what a permutation is right?

So, now you are asked to prove that actually the set of all even permutations, namely A_n, is a subgroup, when the opertaion is composition(multiplication). So all you need to show is that for any two elements of An, call them a,b, then ab is in A_n, that the inverse of every even permutaion is an even permutation, and you are done, siince there is a theorem that guarantees that any set that satisfies these conditions is a subgroup.
 
thank you so much, it makes sense now
 
Now the part of proving that the order of An is n!/2 requires more work.

First you probbably know that the order of Sn is n!. however we cannot rightaway conclude that An has n!/2 elements. We first need to show that there are as many even permutations as ther are odd permutations in Sn and then it is clear that the ord of An is n!/2.

Hint:

Let

S_n=(\alpha_1,\alpha_2,...,\alpha_k,\beta_1,\beta_2,...,\beta_r)

Be the set of all permutations where \alpha_i,\beta_j are even and odd permutations respectively.

Now, what you can do is show that \beta_1\alpha_i for i=1,..,k are all odd permutations...it requires a proof by contradiction somewhere along the lines.

Then again, you might want to show that \beta_1\beta_j for j=1,...,r are all even permutations.

What do these two things tell you?
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top