TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck 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 $F (G) = \prod_{p} O_{p} (G)$ of a finite group.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a finite group $G$ , the Fitting subgroup is $F (G) : = \prod_{p ∣ ∣ G ∣} O_{p} (G),$ 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 $F (G) = \prod_{p} O_{p} (G)$ 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 $N_{0}, \dots, N_{r - 1} ⊴ G$ . They form an internal direct product of $G$ if and only if every $g \in G$ has a unique expression $g = n_{0} \dots n_{r - 1}$ with $n_{i} \in N_{i}$ , equivalently the multiplication map $\prod_{i < r} N_{i} \to G$ 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 $N ⊴ G$ , then $K ⊴ G$ . (If $K$ is characteristic in $N$ and $N$ is normal in $G$ , then $K$ is normal in $G$ ).

[L5]

The $p$ -core $O_{p} (G)$ is the unique largest normal $p$ -subgroup of the finite group $G$ . (The $p$ -core $O_{p} (G)$ 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

1.1L1L2L3L5L7givenalgebra

For distinct primes $p$ and $q$ , normality puts every commutator of an element of $O_{p} (G)$ with an element of $O_{q} (G)$ in $O_{p} (G) \cap O_{q} (G)$ . 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 $\prod_{p} x_{p}$ with $x_{p} \in O_{p} (G)$ equals $1$ , then for each $p$ the element $x_{p}$ lies both in the $p$ -group $O_{p} (G)$ 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 $O_{p} (G)$ is the Sylow $p$ -subgroup of $F (G)$ ; [L2] makes $F (G)$ nilpotent, and [L1] gives its normality in $G$ .

2.1L2L4L5L6step 1.1givenalgebra

If $N ⊴ G$ 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.

3.1L1L2step 2.1given

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

4.1L1step 1.1step 3.1givenalgebra∎

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.

Depends on

Used by

The Frattini subgroup is contained in the Fitting subgroup Corollary
The Fitting subgroup of A₅ does not contain its centralizer Counterexample
The Fitting and Frattini subgroups of S₃ Example
F(G/Φ(G))=F(G)/Φ(G) for every finite group Theorem
Philip Hall: in a finite solvable group the Fitting subgroup contains its own centralizer Theorem
The generalized Fitting subgroup contains its centralizer Theorem

Dependency tree · two levels

35 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)