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Composition Factors cyclic IFF finite group soluble

  1. Mar 31, 2013 #1
    Hey, just trying to get my head around the logic of this. I can see that if composition factors are cyclic then clearly the group is soluble, since there exists a subnormal series with abelian factors, but I am struggling to see how the converse holds. If a group is soluble, then it has a subnormal series with abelian factors, and if this is already the composition series then given that the factors are both simple and abelian, they are cyclic. But when refining a series and adding in new "terms", why should the new factors be abelian? I know they are simple if they are in the composition series, but just trying to see why it must be that they are abelian.
     
  2. jcsd
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