More Algebra

[headf1rst13]

I have been doing all right in my algebra course but now I have stumbled upon permutations. Sorry to bother you again but I have 2 more problems that I am stuck on. Thanks in advance.

1) Show that for any subgroup H of Sn either every element of H is an even permutation or else 1/2 the elements of H are even permuations.
Hints? Suggestions? Starting

Comments

[phlebas]

Some questions that might help:

1) What is the result of composing an even with an odd permutation? Odd and odd? Even and even?

2) What is required to prove a group is non-Abelian? Can an Abelian group contain a non-Abelian subgroup?

[headf1rst13]

1) Even and even is even. Odd and odd is even. Even and odd is odd. But I'm not seeing the connection...? Sorry.

2) I must show that the group is not cummutative. So I need two elements (or permuations here) such that ab is not ba. An Abelian group cannot have a non-Abelian subgroup since all the subgroups of Abelian groups are also Abelian... I think. Of course I could have totally made that up... I need to double check. But hypothetically, if that is true can I make the argument that Dn is a subgroup of Sn for all n >=3 and since Dn is non-Abelian then Sn is also non-Abelian. Nah. I don't like that.... Hmmm... still not sure where you are leading me. Sorry.

[phlebas]

1) If H contains only even permutations, then we're done.
We need to prove that if it contains any odd permutation then it contains the same number of odd as even.
Does that help now? You might need an observation about inverses.

If that still doesn't make sense we can try another approach to the proof: can you say anything about normal subgroups of H?

[headf1rst13]

Can I just say consider H a subgroup of Sn. Either H contains odd permuations or not. If it does not contain odd, then all are even and we are done. If it contains odd permutations then claim that the number of even = the number of odd. Then we need to construct a function f: Even --> Odd such that f is one to one and onto. Then this will show 1/2 the elements are even and half are odd. So f: Even --> Odd, f(T)=T(1 2). F is 1-1... T_1 (1 2) = T_2 (1 2). Then T_1 (1 2)(1 2) = T_2 (1 2)(1 2). Then T_1 = T_2. F is onto... T in Odd. Consider T(1 2) in even. f(T(1 2))=T(1 2)(1 2) = T.

I'm not sure if this is anywhere near complete...
Sorry. but thanks again for all of your help.

[rscholar]

Just find two elements a, b in S_3 s.t. a b ≠ b a.

Then you can treat them as permutations in S_n
where other (n - 3) positions remain unchanged.

[headf1rst13]

Yes I can find said a and b. But I don't understand the second part of your comment. I'm sorry.

[headf1rst13]

Or maybe my edit just now is something similar to what you were telling me...?