TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-24

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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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).

[L3]

If $1 = K_{0} \leq \dots \leq K_{c} = G$ is a central series, then $[G, K_{i}] \leq K_{i - 1}$ for each $i \geq 1$ . With the library's convention $[h, z] = h z h^{- 1} z^{- 1}$ , this gives $z h z^{- 1} = [h, z]^{- 1} h$ for $h \in G$ and $z \in K_{i}$ . (Central factors are equivalent to adjacent commutator containments).

Proof

technique · direct

1.1L1given

Let $H < G$ . By [L1], choose a central series $1 = K_{0} \leq \dots \leq K_{c} = G$ . Since $K_{0} = 1 \leq H$ and $K_{c} = G ≰ H$ , there is a least $i \geq 1$ with $K_{i} ≰ H$ . Then $K_{i - 1} \leq H$ ; choose $z \in K_{i} ∖ H$ .

2.1step 1.1L2L3

For every $h \in H$ , [L3] gives $[h, z] \in K_{i - 1} \leq H$ , hence $z h z^{- 1} = [h, z]^{- 1} h \in H$ . Thus $z H z^{- 1} \subseteq H$ . Since $z^{- 1} \in K_{i}$ , the same calculation with $z^{- 1}$ gives $z^{- 1} H z \subseteq H$ ; conjugating this inclusion by $z$ gives $H \subseteq z H z^{- 1}$ . Therefore $z H z^{- 1} = H$ and $z \in N_{G} (H) ∖ H$ .

3.1step 2.1L2∎

Every $h \in H$ normalizes $H$ , so $H \leq N_{G} (H)$ ; the element $z \notin H$ from step 2.1 makes the inclusion strict. If $G = 1$ there is no proper subgroup and the claim is vacuous.

Depends on

Used by

Dependency tree · two levels

11 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

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