proof: Let G have the series 1=G0⊲G1⊲⋯⊲Gn=G. Suppose H≤G then 1=G0∩H⊲G1∩H⊲⋯⊲Gn∩H=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]≤Gi−1 and we deduce H is nilpotent from [G∩H,Gi∩H]≤Gi−1∩H.
Now suppose G was solvable we deduce that H is too from the fact that if Gi/Gi−1 is abelian so is Gi∩H/Gi−1∩H≃Gi−1(Gi∩H)/Gi−1≤Gi/Gi−1 (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, N⊴G and let ν:G→G/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 ˉHi⊲ˉHi+1 (so we have a series for G/N) also it's an abelian series since ˉHi+1/ˉHi is abelian: let ˉx,ˉy∈ˉHi+1 say ˉx=νx, ˉy=νy then since Hi+1/Hi is abelian xy=yxd for some d∈Hi, so taking ν we find ˉxˉy=ˉyˉxˉd for some ˉd∈ˉHi, 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 A3≃C3 by S3/A3≃C2 but isn't nilpotent.
Theorem Solvable groups are closed under taking extensions: Suppose K⊴G then if K and G/K are both solvable so is G.
proof:
- crazyproject: subgroups-and-quotient-groups-of-solvable-groups-are-solvable
- groupprops/Solvable_group#Metaproperties
- Schaum's Outline of Group Theory - B. Baumslag (Author), B. Chandler (Author)
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