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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17
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  • 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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F(G/Φ(G))=F(G)/Φ(G) for every finite group

Statement

For every finite group G, F(G/Φ(G))=F(G)/Φ(G). See The Fitting subgroup is nilpotent and is the largest normal nilpotent subgroup of a finite group.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For every finite group G, F(G) is nilpotent and normal, and every normal nilpotent subgroup of G is contained in F(G). (The Fitting subgroup is nilpotent and is the largest normal nilpotent subgroup of a finite group).

[L2]

For every finite group G, Φ(G)≤F(G). (The Frattini subgroup is contained in the Fitting subgroup).

[L3]

Let G be finite and let Φ(G)≤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. (Nilpotence lifts over the Frattini subgroup of a finite group).

[L4]

Correspondence theorem: subgroups of G/N correspond to subgroups of G containing N, with normality preserved. For N⊴G, the maps H↦H/N and K↦π−1(K) are inverse inclusion-preserving bijections between subgroups H with N≤H≤G and subgroups K≤G/N; they preserve normality. (Correspondence theorem: subgroups of G/N correspond to subgroups of G containing N, with normality preserved).

Proof

technique · direct
1.1

The image F(G)/Φ(G) is normal and nilpotent, giving one inclusion.

L1L2L3L4givenalgebra
2.1

For the reverse inclusion, pull F(G/Φ(G)) back to a normal subgroup N of G; the lifting theorem makes N nilpotent, so N≤F(G).

step 1.1givenalgebra
3.1

If G/Φ(G) is trivial, then Φ(G)=G, and Φ(G)≤F(G)≤G from [L2] forces F(G)=G; both sides of the identity are then the trivial group. Together with the two inclusions of steps 1.1 and 2.1 this gives equality in every case. This proves the stated claim.

step 1.1step 2.1L2givenalgebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 60 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.

Sources