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Proving Jordan-Holder theorem

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose that [itex]N \triangleleft G[/itex]. Show that given normal series [itex]S[/itex] for [itex]N[/itex] and [itex]T[/itex] for [itex]G / N[/itex] one can construct a normal series [itex]U[/itex] for [itex]G[/itex] such that the first part of [itex]U[/itex] is isomorphic to [itex]S[/itex] and the rest is isomorphic to [itex]T[/itex].


    2. Relevant equations

    This is from the last couple of weeks of an undergraduate Abstract Algebra course. The teacher assigned it as homework while discussing a proof of the Jordan-Holder theorem.

    3. The attempt at a solution

    I'd like to simply construct [itex]U[/itex] from [itex]S[/itex] and [itex]T[/itex]. Using [itex]S[/itex] would be straightforward as this is already a normal series from [itex]\left\{ e \right\}[/itex] to [itex]N[/itex]. However, I'd hoped to use correspondence theorem to map the normal series [itex]T[/itex] to a normal series from [itex]N[/itex] to [itex]G[/itex]. I believe, however that there is a problem with the part where it says this part of the series should be isomorphic to [itex]T[/itex].
     
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  3. Apr 14, 2012 #2

    morphism

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    What does it mean to say that two normal series are isomorphic?
     
  4. Apr 14, 2012 #3
    I guess that was the part I was confused about as well. My roommate now informs me that we defined two normal series to be isomorphic as follows:

    Series [itex]S[/itex] and [itex]T[/itex] are isomorphic if there exists a bijection from the factors of [itex]S[/itex] to the factors of [itex]T[/itex] such that the corresponding factors are isomorphic.

    So, that makes a bit more sense to me now.
     
  5. Apr 14, 2012 #4

    morphism

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    Typically that's referred to as "equivalence", but anyway, your idea does work, i.e. it will produce an equivalent normal series.
     
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