CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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Nilpotent groups, and in particular finite $p$ -groups, are solvable

Statement

Every nilpotent group is solvable. Consequently every finite $p$ -group is solvable.

Facts & Assumptions

Given: A nilpotent group $G$ .

[F1]

$G^{(0)} = G$ and $G^{(r + 1)} = [G^{(r)}, G^{(r)}]$ (The derived series, solvable groups, and derived length).

[F2]

$γ_{1} (G) = G$ and $γ_{r + 1} (G) = [G, γ_{r} (G)]$ (Subgroup commutators and the lower central series).

[L1]

Nilpotence is equivalent to $γ_{c + 1} (G) = 1$ for some $c$ (Nilpotence via central series, the upper central series, and the lower central series).

[L2]

Every finite $p$ -group is nilpotent (Every finite $p$ -group is nilpotent).

Proof

technique · direct

1.1

For every subgroup $H \leq G$ , one has $[H, H] \leq [G, H]$ .

F2algebra

2.1

Induction on $r$ gives $G^{(r)} \leq γ_{r + 1} (G)$ : equality holds at $r = 0$ , and step 1.1 sends the inclusion at $r$ to $G^{(r + 1)} \leq [G, γ_{r + 1} (G)] = γ_{r + 2} (G)$ .

step 1.1F1F2

3.1

Choose $c$ with $γ_{c + 1} (G) = 1$ using [L1]. Step 2.1 gives $G^{(c)} = 1$ , so $G$ is solvable by [F1].

step 2.1L1F1choose

4.1

A finite $p$ -group is nilpotent by [L2], so step 3.1 applies.

step 3.1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 57 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)

Nilpotent groups, and in particular finite pp-groups, are solvable