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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17
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Nilpotence lifts over the Frattini subgroup of a finite group

Statement

Let G be finite and let Φ(G)NG. 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):={MG: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 NG is finite and P is a Sylow p-subgroup of N, then G=NNG(P).. (Frattini argument: if NG and P is Sylow in N, then G=NNG(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 KG, 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 HG be a p-subgroup. There is gG with HgPg1; 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 NG, then KG. (If K is characteristic in N and N is normal in G, then K is normal in G).

Proof

technique · direct
1.1

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

L1L5given
1.2

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 nN with SnPn1; normality of Φ(G) gives n1SnPΦ(G). Thus PΦ(G) contains a Sylow p-subgroup of Φ(G) and, being a p-subgroup, is itself Sylow there. Consequently PΦ(G)/Φ(G) is a Sylow p-subgroup of N/Φ(G).

L1L3L4L6givenalgebra
2.1

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 PΦ(G)G.

step 1.2L3L4L6L7givenalgebra
3.1

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

L1L2step 1.2step 2.1givenalgebra
4.1

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

L3step 1.1step 3.1given
5.1

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.

L1step 4.1given

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 73 results over 14 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.

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