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Homework Help: Intermediate subgroups between symmetric groups

  1. Jun 28, 2011 #1
    1. The problem statement, all variables and given/known data

    For n>1, show that the subgroup H of [itex] S_n [/itex] (the symmetric group on n-letters) consisting of permutations that fix 1 is isomorphic to [itex] S_{n-1} [/itex]. Prove that there are no proper subgroups of [itex] S_n [/itex] that properly contain H.

    3. The attempt at a solution

    The first part is fairly straightforward. Using cycle notation, we ignore the trivial cycle on 1, then just reduce all other components by 1. It can be shown to be bijective hence giving the desired isomorphism.

    Now for the second part, I figure this simply amounts to showing that there is no group N such that [itex] S_{n-1} \subsetneq N \subsetneq S_n [/itex]. My first thought is to attack the problem via contradiction. If such a subgroup existed, it would be necessary that [itex] (n-1)! \big| |N| \big| n! [/itex]. Unfortunately, I cannot find a reason this can't be true, except when n is prime.
  2. jcsd
  3. Jun 28, 2011 #2
    Hi Kreizhn! :smile:

    Did you see the orbit-stabilizer theorem? Your group N acts on {1,...,n}, so by orbit-stabilizer, we have that


    for all x. Now take x=1.
  4. Jun 29, 2011 #3
    It seems like the stabilizer of identity should be trivial, and the orbit of identity is the whole group. So then [itex] |N| = |S_n| [/itex] ? And this would be a contradiction since we assumed N was proper.
  5. Jun 29, 2011 #4
    No, I didn't mean 1 to be the identity. I meant 1 to be an element of the set {1,...,n}.

    We know that N acts on {1,...,n}. So we can calculate G(1), G(2),...,G(n). And we can also calculate the stabilizers of these elements.
    So, what is the orbit and what is the stabilizer of 1? (or 2,...,or n)
  6. Jun 29, 2011 #5
    Oh yes, silly me. It's a group action on the set {1,..., n} so it wouldn't make sense for x to be identity.

    Okay, now H would be a subset of the stabilizer of 1, since elements of H fix 1. Though since H is properly contained in N, the orbit is not just {1}.

    If the orbit were "full" (I guess some say trivial), so that G(1) = {1,..., n} then we would get
    [tex] |N| = |G(x)||G_x| \geq |H| n = (n-1)! n = n! [/tex]
    which implies [itex] N = S_n. [/itex]. We assumed that N was a proper subset so the orbit cannot be trivial, and in particular [itex] \exists m \in \{ 1,\ldots, n \} [/itex] such that [itex] \forall \sigma \in N, \{ 1 \} \sigma \neq \{ m \} [/itex] (where I'm using right-actions).

    Is this the right path? Or is there a simple "order argument" that is obvious that I'm just missing?
  7. Jun 29, 2011 #6
    This is indeed the right path. So you must prove that if [itex]S_{n-1}\subset N[/itex], then the orbit is everything. That is

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