TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17

How statement and proof provenance work

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Nilpotence lifts over the Frattini subgroup of a finite group

Statement

Let $G$ be finite and let $Φ (G) \leq N ⊴ G$ . Then $N$ is nilpotent if and only if $N / Φ (G)$ is nilpotent. In particular, $G$ is nilpotent if and only if $G / Φ (G)$ is nilpotent. See The Frattini subgroup $Φ (G)$ as the intersection of the maximal subgroups of a finite group.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a finite group $G$ , the Frattini subgroup is $Φ (G) : = ⋂ {M \leq G : M is maximal proper} .$ If $G = 1$ , the family is empty and its intersection inside $G$ is $G$ itself. Thus $Φ (1) = 1$ . (The Frattini subgroup $Φ (G)$ as the intersection of the maximal subgroups of a finite group).

[L2]

If $N ⊴ G$ is finite and $P$ is a Sylow $p$ -subgroup of $N$ , then $G = N N_{G} (P) .$ . (Frattini argument: if $N ⊴ G$ and $P$ is Sylow in $N$ , then $G = N N_{G} (P)$ ).

[L3]

For a finite group $G$ , the following are equivalent: $G$ is nilpotent; every Sylow subgroup is normal; and $G$ is the internal direct product of its Sylow subgroups. (A finite group is nilpotent if and only if all Sylow subgroups are normal, if and only if it is their internal direct product).

[L4]

For $K ⊴ G$ , subgroups of $G / K$ correspond to subgroups of $G$ containing $K$ , and the correspondence preserves normality. (Correspondence theorem: subgroups of $G / N$ correspond to subgroups of $G$ containing $N$ , with normality preserved).

[L5]

Every subgroup and every quotient of a nilpotent group is nilpotent, and every finite direct product of nilpotent groups is nilpotent. (Subgroups, quotients, and finite direct products of nilpotent groups are nilpotent).

[L6]

Let $G$ be finite, let $P$ be a Sylow $p$ -subgroup, and let $H \leq G$ be a $p$ -subgroup. There is $g \in G$ with $H \leq g P g^{- 1}$ ; in particular all Sylow $p$ -subgroups are conjugate. (Sylow II: in a finite group every $p$ -subgroup lies in a conjugate of any Sylow $p$ -subgroup, and the Sylow $p$ -subgroups form a single conjugacy class).

[L7]

If $K$ is characteristic in $N$ and $N ⊴ G$ , then $K ⊴ G$ . (If $K$ is characteristic in $N$ and $N$ is normal in $G$ , then $K$ is normal in $G$ ).

Proof

technique · direct

1.1L1L5given

If $N$ is nilpotent, then its quotient $N / Φ (G)$ is nilpotent by [L5].

1.2L1L3L4L6givenalgebra

Conversely, assume that $N / Φ (G)$ is nilpotent, and let $P$ be a Sylow $p$ -subgroup of $N$ . Every automorphism of $G$ permutes its maximal proper subgroups, so [L1] makes $Φ (G)$ characteristic and hence normal in $G$ . If $S$ is Sylow in $Φ (G)$ , then [L6] gives $n \in N$ with $S \leq n P n^{- 1}$ ; normality of $Φ (G)$ gives $n^{- 1} S n \leq P \cap Φ (G)$ . Thus $P \cap Φ (G)$ contains a Sylow $p$ -subgroup of $Φ (G)$ and, being a $p$ -subgroup, is itself Sylow there. Consequently $P Φ (G) / Φ (G)$ is a Sylow $p$ -subgroup of $N / Φ (G)$ .

2.1step 1.2L3L4L6L7givenalgebra

By nilpotence and [L3], this quotient Sylow subgroup is normal; conjugacy [L6] makes it unique, hence characteristic in $N / Φ (G)$ . Since $N / Φ (G) ⊴ G / Φ (G)$ , [L7] and [L4] give $P Φ (G) ⊴ G$ .

3.1L1L2step 1.2step 2.1givenalgebra

Because $P \cap Φ (G)$ is Sylow in $Φ (G)$ , the product-order formula shows that $P$ is Sylow in $P Φ (G)$ . Apply the Frattini argument [L2] to the normal subgroup $P Φ (G)$ : $G = P Φ (G) N_{G} (P) = Φ (G) N_{G} (P) .$ If $N_{G} (P)$ were proper, finiteness would place it in a maximal subgroup $M$ ; [L1] gives $Φ (G) \leq M$ , contradicting the displayed equality. Hence $N_{G} (P) = G$ .

4.1L3step 1.1step 3.1given

Thus every Sylow subgroup of $N$ is normal, so [L3] makes $N$ nilpotent. Together with step 1.1 this proves both directions.

5.1L1step 4.1given∎

Taking $N = G$ gives the asserted special case. When $G = 1$ , one has $N = Φ (G) = 1$ , and both groups in the equivalence are trivial and nilpotent.

Depends on

Used by

Dependency tree · two levels

31 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

David A. Craven, Finite Group Theory, Sections 1.4 and 2.3 (standard reference, not scraped)