Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solvable groups

  1. Jul 6, 2008 #1
    I'm supposed to show that a subgroup of a solvable group is solvable.

    (I am using the Fraleigh Abstract Algebra book and the given definition of a solvable group is a group which has a COMPOSITION series in which each of the factor groups is abelian. In other books I have looked at a solvable group is defined as one which has a SUBNORMAL series in which all the factor groups are abelian.)

    Let K be a subgroup of a solvable group G. Let

    {e}=Ho<H1<...<Hn=G be a composition series for G in which all the factor groups are abelian.

    I showed that the distinct groups from among (K & Hi) (&= set intersection)
    form a subnormal series for K.

    Then I used the second isomorphism to show that (K & Hi)/(K & Hi-1) is isomprphic to Hi-1(K & Hi)/Hi-1.

    I then showed that Hi-1(K & Hi) is a subgroup of Hi. Since Hi/Hi-1 is abelian, Hi-1(K & Hi)/Hi-1 is abelian, and so (K & Hi)/(K & Hi-1) is abelain.

    What I can't show is that the distinct groups among (K & Hi) from a composition series for K. That is, I can't show that (K & Hi)/(K & Hi-1) is simple.

    Please help!
     
  2. jcsd
  3. Jul 6, 2008 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    These are the same: factor groups only exist for normal subgroups.

    please could you tex this clearly so people can understand what you mean?


    Surely you mean abelian, not simple?
     
  4. Jul 6, 2008 #3
    Thank you for responding to my question.

    As I said in the preface to my question, I need to find a composition series for K in which each of the factor groups is abelian. So I need each of the factor groups (K & Hi)/(K & Hi-1) to be both abelian and simple.
     
  5. Jul 6, 2008 #4

    gel

    User Avatar

    Why do you need to show that? [itex]K\cap H_{i}/K\cap H_{i-1}[/itex] is isomorphic to a subgroup of [itex]H_i/H_{i-1}[/itex], so it is abelian, and this is enough to show that K is simple.

    Actually, if K is a strict subgroup of G, you will have [itex]H_i\cap K=H_{i-1}\cap K[/itex] for at least one i. So, [itex]K\cap H_i[/itex] will not be a composition series -- you would have to remove some of them from the series.
     
  6. Jul 6, 2008 #5

    gel

    User Avatar

    I've got 2 further comments on your question.

    First, you don't need to use a composition series. Any subnormal series for which the quotients [itex]H_i/H_{i-1}[/itex] are Abelian will do. Then, you can eliminate any [itex]H_i[/itex] for which [itex]H_i=H_{i-1}[/itex]. Then, if any of the quotients are not simple you can enlarge the series, until you reach a maximum one. This will be a composition series and the quotients will still be abelian.

    Second, the only simple abelian groups are the cyclic groups of prime order. Any subgroup will be either the trivial group, or the whole group, so still simple. So, the way you have set up the problem it will be the case that you get a composition series once you eliminate any terms with [itex]K\cap H_i=K\cap H_{i-1}[/itex].
     
  7. Jul 6, 2008 #6
    gel:

    Thank you for your comments. They were very helpful.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Solvable groups
  1. Solvable groups (Replies: 2)

  2. Solvable group (Replies: 2)

  3. Solvable Groups (Replies: 6)

  4. Finite solvable groups (Replies: 2)

Loading...