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Permutation groups

  1. Jan 14, 2008 #1
    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 [tex]S_n[/tex], and is not contained in [tex]A_n[/tex], show that precisely half of the elements in H are even permutations
    2) show that for n>3, all elements of [tex]S_n[/tex], 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 [tex]A_n[/tex], 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 [tex]S_n[/tex] 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. jcsd
  3. Jan 14, 2008 #2


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    Cleaning up your argument:
    If [itex]x[/itex] is an odd element of [itex]H[/itex], then you can show that each coset of [itex]<x>[/itex] has the same number of even and odd elements, but you can skip the middleman.

    Let's say:
    and [itex]x \in H[/itex] is some odd permutation what can you say about
    relative to [itex]H[/itex]?

    For part 2:
    Cyclic permutations aren't of order 2 - but they are all even permutations...
  4. Jan 17, 2008 #3
    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?
  5. Jan 18, 2008 #4


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

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