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Composition Series and simple groups

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data
    For each m >= 2, find a group with a composition series of length 1 with a subgroup of length m.



    2. Relevant equations
    Simple groups iff length 1.

    If G is abelian of order p1^k1....pr^kr, then length G = k1 + ... + kr

    If G has a composition series and K is normal subgroup of G, then Length G = Length K + Length G/K



    3. The attempt at a solution

    I don't really see how this is even possible considering the fact that simple groups have length 1. So this means that G must be a simple group. I do not understand how to get a group of length m from a simple group.

    I think I'm getting somewhere with this:

    What I was thinking was have G = the cyclic group of order p. Then G would have length 1

    Then I thought to Let H be a subgroup of G such that |G:H|=p^m, so if G=<a>, st |a|=p, then H=<a^p^(m+1)>

    But this doesn't really work because the subgroup would have higher order than G right? Am I missing something, like modular arithmetic?

    Any light you guys could shed would be appreciated
    Thanks
     
  2. jcsd
  3. Mar 28, 2012 #2

    morphism

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    Fix m, and let G be an abelian group of length m. Can you put G inside a simple group? (What simple groups do you know?)
     
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