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Thursday, 24 January 2013

Basic properties of nilpotent and solvable groups (unfinished)

Theorem If G is nilpotent or solvable so is any subgroup.
proof: Let G have the series 1=G0G1Gn=G. Suppose HG then 1=G0HG1HGnH=H is a series for H by ABC.
Now suppose G was nilpotent, by the alternative characterization of a central factor we have that [G,Gi]Gi1 and we deduce H is nilpotent from [GH,GiH]Gi1H.
Now suppose G was solvable we deduce that H is too from the fact that if Gi/Gi1 is abelian so is GiH/Gi1HGi1(GiH)/Gi1Gi/Gi1 (by ABC) as it's the subgroup of an abelian group.

Theorem If G is nilpotent or solvable so is any quotient.
proof: Suppose G was solvable, NG and let ν:GG/N be the natural inclusion homomorphism which we know to be surjective. Take Gs composition series as before. Let us call ν(Gi) by ˉHi so by Noether(-1) and the fact it's a composition series ˉHiˉHi+1 (so we have a series for G/N) also it's an abelian series since ˉHi+1/ˉHi is abelian: let ˉx,ˉyˉHi+1 say ˉx=νx, ˉy=νy then since Hi+1/Hi is abelian xy=yxd for some dHi, so taking ν we find ˉxˉy=ˉyˉxˉd for some ˉdˉHi, thus: (ˉxˉHi)(ˉyˉHi)=ˉyˉxˉdˉHi=ˉyˉxˉHi=ˉyˉHiˉxˉHi.

Theorem If H and K are both nilpotent or both solvable so is H×K
proof:

Proposition Nilpotent groups aren't closed under taking extensions. Example S3 is an extension of A3C3 by S3/A3C2 but isn't nilpotent.

Theorem Solvable groups are closed under taking extensions: Suppose KG then if K and G/K are both solvable so is G.
proof:

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