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A group is solvable if and only if it has a subnormal series with abelian factors

Statement

A group G is solvable if and only if it has a finite subnormal series G=G0⊵G1⊵⋯⊵Gn=1 whose factors Gi/Gi+1 are abelian. Moreover, for every such series, G(i)≤Gi for 0≤i≤n.

Facts & Assumptions

Given: A group G.

[F1]

A subnormal series has Gi+1⊴Gi at every adjacent pair (Subnormal and normal series, factors, refinements, and equivalence).

[F2]

G is solvable exactly when G(n)=1 for some n (The derived series, solvable groups, and derived length).

[L1]

Derived series terms are functorial under inclusions and quotient maps (Homomorphisms respect commutator subgroups and derived series).

[L2]

If N⊴H, then H/N is abelian if and only if H′≤N (G/N is abelian if and only if [G,G]⊆N).

Proof

technique · direct
1.1

Suppose G is solvable, and choose n with G(n)=1. The derived chain G=G(0)⊵⋯⊵G(n)=1 is subnormal, and [L2] makes every factor G(i)/G(i+1) abelian.

assume-hypF1F2L2
1.2

Conversely, suppose G=G0⊵⋯⊵Gn=1 is subnormal with abelian factors. By [L2], Gi′≤Gi+1 for every i<n.

assume-hypF1L2
2.1

Starting with G(0)=G0, if G(i)≤Gi, then [L1] gives G(i+1)=(G(i))′≤Gi′≤Gi+1; hence G(i)≤Gi for every i≤n.

step 1.2L1F2
3.1

Step 2.1 gives G(n)≤Gn=1, so G is solvable by [F2]. Steps 1.1 and 3.1 prove both directions.

step 1.1step 2.1F2∎

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