TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17

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Every proper subgroup of a finite nilpotent group is properly contained in its normalizer

Statement

Every proper subgroup of a finite nilpotent group is properly contained in its normalizer. See Nilpotence via central series, the upper central series, and the lower central series.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a group $G$ and $c \in N$ , the following are equivalent: 1. $G$ has a central series $1 = H_{0} \leq \dots \leq H_{c} = G$ ; 2. $Z_{c} (G) = G$ ; 3. $γ_{c + 1} (G) = 1$ . Hence $G$ is nilpotent exactly when its lower central series reaches $1$ , and the least such $c$ is its nilpotency class. (Nilpotence via central series, the upper central series, and the lower central series).

[L2]

Let $H \leq G$ be a subgroup (def-subgroup). The normalizer of $H$ in $G$ is $N_{G} (H) : = {g \in G : g H g^{- 1} = H} .$ Thus $g \in N_{G} (H)$ exactly when the conjugation automorphism $c_{g}$ preserves $H$ setwise (thm-conjugation-is-an-automorphism). The subgroup property is proved in lem-centralizers-and-normalizers-are-subgroups. (The normalizer $N_{G} (H) = {g \in G : g H g^{- 1} = H}$ of a subgroup).

Proof

technique · direct

1.1

We use a central series and take the first term not contained in the proper subgroup $H$ .

L1L2givenalgebra

2.1

The preceding term lies in $H$ , so an element newly appearing at that stage normalizes $H$ modulo the preceding term but is not in $H$ .

step 1.1givenalgebra

3.1

Both boundary cases are admitted and hold. If $G$ has nilpotency class zero then $G = 1$ , which has no proper subgroup, so the claim is vacuously true and step 1.1 is never entered. If $H = 1$ and $G \neq 1$ , then $N_{G} (H) = G$ , which properly contains $H$ . This proves the stated claim.

step 1.1step 2.1givenalgebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 22 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

Keith Conrad, Consequences of the Sylow Theorems, Sections 1-5 (standard reference, not scraped)