# Permutation groups

1. Jan 14, 2008

### jacobrhcp

Hehe, I'm working through the complete groups books right now, so don't think I ask you all my homework questions... I'm doing alot myself too =).

1. The problem statement, all variables and given/known data

1) If H is a subgroup of $$S_n$$, and is not contained in $$A_n$$, show that precisely half of the elements in H are even permutations
2) show that for n>3, all elements of $$S_n$$, can be written as a product of two permutations, each of which has order 2.

3. The attempt at a solution

1) If there is an element x that's not in $$A_n$$, then x=yz, where either y or z is odd.
call this element x'. x'=y'z', and either y' or z' is odd again.
If you continue this process, you eventually have to get back to x, because $$S_n$$ is finite. If you can show this is when you repeated the process n/2 times, you're done... because the other half are the even ones you encountered when you wrote it as a product every time.

But this is a step I'm having problems with. And I realize this argument has loads of holes in it.

2) I have only a vague clue on this one. I thought I might write the element as a product of disjoint cyclic permutations. Now cyclic permutations are of order 2.

because this is grammatically close to what I want I thought it might help =P, but I really don't know how get from here to the point where you have two elements of order 2.

Last edited: Jan 14, 2008
2. Jan 14, 2008

### NateTG

If $x$ is an odd element of $H$, then you can show that each coset of $<x>$ has the same number of even and odd elements, but you can skip the middleman.

Let's say:
$$H=\{h_1,h_2...h_k\}$$
and $x \in H$ is some odd permutation what can you say about
$$\{xh_1,xh_2...xh_k\}$$
relative to $H$?

For part 2:
Cyclic permutations aren't of order 2 - but they are all even permutations...

3. Jan 17, 2008

### jacobrhcp

1) oh my. I've thought about it for quite some time before I realized. that set is again H, ofcourse. Because odd * odd=odd and odd * even=even, by symmetry it can only be that half the elements are odd. Thanks!

2) cyclic permutations are indeed not of order 2, but are they even? I'm not sure I get this. If you know this yourselves, can you help me a little bit further?

4. Jan 18, 2008

### NateTG

Sorry, I made a mistake -- cyclic permutations can be odd. Cycles of odd length are even.

The order of element of a group $<x>$ is the number of elements in the group that is generated by $x$. (This group ends up looking like $x^0,x^1...x^n$ if it is finite and $...x^{-1},x^{0},x^1...$ if it is infinite.) For elements of finite order, it is also the smallest exponent so that $x^n$ is the identity.

An even permutation is a permutation that, can be written as the composition of an even number of transpositions. (http://en.wikipedia.org/wiki/Even_permutation).

You might consider the following as a warm-up for part 2:
Show that any cycle of length n can be decomposed into n-1 transpositions.
Show that any permutation can be written as the product of two permutations that are either the identity or have order 2.