Question on Symmetric group

Discussion in 'Undergraduate Math' started by Jones Wu, Feb 21, 2009.

  1. Jones Wu

    Jones Wu Guest

    Let H be a subgroup of S_n, the symmetric group. How to prove that H = A_n, the alternating group, given that the order of H is n!/2?

    I thought about the set H intersect A_n. Since A_n is a normal subgroup of G, H intersect A_n is a normal subgroup of G too. For n>=5, A_n is simple. Since H intersect A_n cannot be trivial(if so, then any element in H is odd except 1, which is not possible), H has to be A_n. But I am really having trouble handling the case where n=4.

    Is this the correct approach to this question?
    Thanks a lot for any help!
     
    Jones Wu, Feb 21, 2009
    #1
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  2. Jones Wu

    Jim Heckman Guest

    This question is essentially identical to that in a current thread
    Can you show that A_4 has no subgroup of index 2?
    It's *a* correct approach, yes.
     
    Jim Heckman, Feb 22, 2009
    #2
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  3. Jones Wu

    Jones Wu Guest

    Jones Wu, Feb 22, 2009
    #3
  4. Jones Wu

    Jim Heckman Guest

    Jim Heckman, Feb 23, 2009
    #4
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