I'm supposed to show that a subgroup of a solvable group is solvable.(adsbygoogle = window.adsbygoogle || []).push({});

(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!

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# Solvable groups

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