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Proof on Normal Subgroups and Cosets in Group Theory

  1. Feb 23, 2010 #1
    This is a proof I am struggling on ...

    Let H be a subgroup of the permutation of n and let A equal the intersection of H and the alternating group of permutation n. Prove that if A is not equal to H, than A is a normal subgroup of H having index two in H.

    My professor gave me the hint to begin by letting g be in H but not in A and then showing that gA and A are two cosets of A in H.
     
  2. jcsd
  3. Feb 23, 2010 #2

    Dick

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    The alternating group consists of all even permutations, right? If A is not equal to H, then H must contain an odd permutation, g, right? What you want to show is that half the elements of H are even permutations and half are odd. Hint: gH=H. That makes A a subgroup of H of index 2.
     
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