LemmaStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedaudited 2026-09-24

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A nonabelian supersolvable group has a noncentral normal abelian subgroup

Statement

Every nonabelian finite supersolvable group has an abelian normal subgroup that is not central.

Facts & Assumptions

[F1]

A finite supersolvable group has a normal series with prime-order factors (Supersolvable groups and monomial characters).

[F2]

A group of prime order is cyclic (A finite group of prime order is cyclic and every nonidentity element generates it).

[F3]

Every term of a normal series is normal in the whole group (Subnormal and normal series, factors, refinements, and equivalence).

Proof

Given: $1 = G_{0} ◃ \dots ◃ G_{r} = G$ is a supersolvable series.

1.1F1F2given

Let $i$ be maximal with $G_{i}$ abelian; such an $i$ exists since $G_{0} = 1$ . If $i = r$ , then $G$ is abelian, contrary to the hypothesis, so $i < r$ . Suppose $G_{i} \leq Z (G)$ . By [F1] and [F2], $G_{i + 1} / G_{i}$ is cyclic of prime order; choose $x \in G_{i + 1}$ whose coset generates it. Then every element of $G_{i + 1}$ has the form $a x^{k}$ with $a \in G_{i}$ . All such elements commute because $G_{i} \leq Z (G)$ and powers of $x$ commute, so $G_{i + 1}$ is abelian, contradicting maximality of $i$ . Thus $G_{i} ≰ Z (G)$ , and there are $a \in G_{i}$ and $g \in G$ with $a g \neq g a$ .

2.1F3step 1.1∎

By [F3], $G_{i} ⊴ G$ . It is abelian by construction and not central by step 1.1, so it is the required abelian normal subgroup.

Depends on

Used by

Finite supersolvable groups are monomial Theorem

Dependency tree · two levels

20 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

Tammo tom Dieck, Representation Theory, Lemma 4.3.3 (standard reference, not scraped)