TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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Subgroups, quotients, and finite direct products of nilpotent groups are nilpotent

Statement

Every subgroup and every quotient of a nilpotent group is nilpotent. Every finite direct product of nilpotent groups is nilpotent; the class of a subgroup or quotient is at most the class of the original group, and the class of a nonempty finite product is at most the maximum of the factor classes. The empty product is the trivial group of class zero.

Facts & Assumptions

Given: A nilpotent group $G$ , a subgroup $H \leq G$ , a normal subgroup $N ⊴ G$ , and nilpotent groups $G_{1}, \dots, G_{t}$ .

[F1]

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

[L1]

For every group $K$ and natural number $c$ , the conditions that $K$ has a central series of length $c$ , that $Z_{c} (K) = K$ , and that $γ_{c + 1} (K) = 1$ are equivalent; the least such $c$ is the nilpotency class (Nilpotence via central series, the upper central series, and the lower central series).

[L2]

Direct products have coordinatewise multiplication and inverses ( $G \times H$ is a group with identity $(e_{G}, e_{H})$ , coordinatewise inverses, and homomorphic coordinate projections).

Proof

technique · direct

1.1F1algebra

Induction on $r$ gives $γ_{r} (H) \leq γ_{r} (G)$ : it is clear at $r = 1$ , and subgroup commutators preserve an inclusion at the next term.

1.2F1algebra

For the quotient map $q : G \to G / N$ , induction on $r$ gives $q (γ_{r} (G)) = γ_{r} (G / N)$ , because $q$ is surjective and sends commutators onto commutators.

1.3F1L2algebra

Coordinatewise commutators from [L2] give $γ_{r} (K \times L) = γ_{r} (K) \times γ_{r} (L)$ for every $r$ , by induction.

2.1step 1.1step 1.2F1L1

If $G$ has class $e \leq c$ , then [L1] gives $γ_{e + 1} (G) = 1$ , and [F1] keeps every later lower-central term trivial, so $γ_{c + 1} (G) = 1$ . Steps 1.1 and 1.2 make $γ_{c + 1} (H)$ and $γ_{c + 1} (G / N)$ trivial; [L1] then makes both nilpotent of class at most $c$ .

2.2step 1.3L1choose

For a nonempty finite product, choose the maximum $c$ of the finitely many factor classes. Repeated use of step 1.3 makes its $(c + 1)$ -st lower-central term trivial, so [L1] gives class at most $c$ ; the empty product is the trivial group of class zero.

3.1step 2.1step 2.2∎

These arguments establish all three closure assertions and their stated class bounds.

Subgroups, quotients, and finite direct products of nilpotent groups are nilpotent

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources