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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17
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The Fitting subgroup is nilpotent and is the largest normal nilpotent subgroup of a finite group

Statement

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

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a finite group G, the Fitting subgroup is F(G):=pGOp(G), the product of its p-cores (def-p-core-of-a-finite-group). The factors are normal, so their finite product is a normal subgroup and does not depend on the order of multiplication. For the trivial group the product is empty and equals 1. (The Fitting subgroup F(G)=pOp(G) of a finite group).

[L2]

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).

[L3]

Let N0,,Nr1G. They form an internal direct product of G if and only if every gG has a unique expression g=n0nr1 with niNi, equivalently the multiplication map i<rNiG is an isomorphism. (Internal direct products are external direct products, equivalently every element has a unique factorisation).

[L4]

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).

[L5]

The p-core Op(G) is the unique largest normal p-subgroup of the finite group G. (The p-core Op(G) as the largest normal p-subgroup).

[L6]

Every p-subgroup of a finite group is contained in a Sylow p-subgroup, and 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]

The order of a subgroup of a finite group divides the order of the group. (Lagrange's theorem: G=[G:H]H for every subgroup H of a finite group G).

Proof

technique · direct
1.1

For distinct primes p and q, normality puts every commutator of an element of Op(G) with an element of Oq(G) in Op(G)Oq(G). By [L7], the order of this intersection divides powers of both p and q, so the intersection is trivial and the two p-cores centralize one another. If a product pxp with xpOp(G) equals 1, then for each p the element xp lies both in the p-group Op(G) and in the commuting product of the other prime-power groups; its order divides two coprime numbers and is therefore 1. Thus product expressions are unique, so [L3] identifies F(G) with the internal direct product of the p-cores. The direct-product order shows that Op(G) is the Sylow p-subgroup of F(G); [L2] makes F(G) nilpotent, and [L1] gives its normality in G.

L1L2L3L5L7givenalgebra
2.1

If NG is nilpotent, [L2] makes each Sylow subgroup normal in N, and conjugacy [L6] makes it unique. Automorphisms preserve orders, so this unique Sylow subgroup is characteristic in N; [L4] makes it normal in G, and the maximality clause [L5] puts it in the corresponding p-core.

L2L4L5L6step 1.1givenalgebra
3.1

By [L2], N is the product of its Sylow subgroups, and step 2.1 puts every factor in the corresponding p-core. Hence NF(G).

L1L2step 2.1given
4.1

For G=1 the family of p-cores is empty and F(G)=1, which is nilpotent and contains every normal nilpotent subgroup. This proves the stated claim.

L1step 1.1step 3.1givenalgebra

Depends on

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Dependency tree · next 3 levels

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