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A group is solvable if and only if it has a subnormal series with abelian factors
Statement
A group is solvable if and only if it has a finite subnormal series whose factors are abelian. Moreover, for every such series, for .
Facts & Assumptions
Given: A group .
A subnormal series has at every adjacent pair (Subnormal and normal series, factors, refinements, and equivalence).
is solvable exactly when for some (The derived series, solvable groups, and derived length).
Derived series terms are functorial under inclusions and quotient maps (Homomorphisms respect commutator subgroups and derived series).
If , then is abelian if and only if ( is abelian if and only if ).
Proof
Suppose is solvable, and choose with . The derived chain is subnormal, and [L2] makes every factor abelian.
Conversely, suppose is subnormal with abelian factors. By [L2], for every .
Starting with , if , then [L1] gives ; hence for every .
Step 2.1 gives , so is solvable by [F2]. Steps 1.1 and 3.1 prove both directions.
Depends on
Used by
Dependency tree · two levels
14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- J. S. Milne, Group Theory, Chapter 6 (standard reference, not scraped)
- K. Conrad, Subgroup Series I (standard reference, not scraped)
- K. Igusa, Notes on Jordan-Hölder, section 5 (standard reference, not scraped)