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Group Theory -

  1. Feb 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove: If H is a subgroup with finite index in G
    Then there is a normal subgroup K of G such that K is a subgroup of H and K has index less than n! in G.

    2. Relevant equations
    Note: |G:H| represents the index of H in G

    |G:H| is the number of left cosets of H in G, ie # of elements in {gH: g in G}

    3. The attempt at a solution

    I haven't had much progress in this proof at all.
    The only thing that I can think of using is that |G : K| = |G : H||H : K| for a subgroup K of H

    But i don't know what to try.
  2. jcsd
  3. Feb 25, 2012 #2
    Have you tried going from what you have and then using the first iso theorem?
  4. Feb 26, 2012 #3


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    suppose that [G:H] = n.

    consider the mapping φg: G/H→G/H given by xH→(gx)H for any given g in G.

    prove this mapping is a bijection on G/H, for every g in G (hint: it has an inverse, what is it?).

    conclude g→φg is a homomorphism of G into Sym(G/H).

    what is |Sym(G/H)| (hint: G/H is a set with n elements)?

    what can you say about the kernel of the homomorphism g→φg?


    note that the theorem, strictly speaking, isn't true. for example, let G = S5, and let H = S4 = {elements of S5 that fix 5}. then [G:H] = 5, but the ONLY normal subgroups of S5 are A5 and {e}, and of these two, only {e} is a subgroup of S4, and {e} has index 5!, in other words, the inequality [G:K] < n! isn't true, but the inequality [G:K] ≤ n! is.
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