ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17

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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.

The $p$ -cores, Fitting subgroup, and Frattini subgroup of $S_{4}$

Example

For $S_{4}$ , one has $O_{2} (S_{4}) = V_{4}$ , $O_{p} (S_{4}) = 1$ for every odd prime $p$ , $F (S_{4}) = V_{4}$ , and $Φ (S_{4}) = 1$ . Hence $F (S_{4} / Φ (S_{4})) = F (S_{4}) / Φ (S_{4})$ . See The $p$ -core $O_{p} (G)$ as the largest normal $p$ -subgroup.

Facts & Assumptions

Given: The hypotheses and objects in the Example.

[L1]

For a finite group $G$ and a prime $p$ , the $p$ -core $O_{p} (G)$ is the subgroup generated by all normal $p$ -subgroups of $G$ . There are finitely many such subgroups, their product is normal, and repeated use of $∣ A B ∣ = ∣ A ∣ ∣ B ∣ / ∣ A \cap B ∣$ shows that the product is again a $p$ -group. It contains every normal $p$ -subgroup, so it is the unique largest normal $p$ -subgroup. (The $p$ -core $O_{p} (G)$ as the largest normal $p$ -subgroup).

[L2]

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).

[L3]

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).

[L4]

For every finite group $G$ , $F (G / Φ (G)) = F (G) / Φ (G) .$ . ( $F (G / Φ (G)) = F (G) / Φ (G)$ for every finite group).

[L5]

Let $n \in N$ , so that $n = {0, 1, \dots, n - 1}$ (def-natural-numbers). The symmetric group on $n$ letters is $S_{n} : = Sym (n) = Sym ({0, 1, \dots, n - 1}),$ the group of all bijections of $n$ under composition (def-symmetric-group), with the composition convention. (The finite symmetric group $S_{n}$ , one-line notation, and cycle notation).

Verification

technique · direct

1.1

The identity and the three double transpositions form a normal Klein four group $V_{4}$ . Since the three Sylow $2$ -subgroups are not normal, no normal $2$ -subgroup can properly contain $V_{4}$ , so $O_{2} (S_{4}) = V_{4}$ . The four Sylow $3$ -subgroups are nonnormal, and no other odd prime divides $24$ , so every odd $p$ -core is trivial and $F (S_{4}) = V_{4}$ .

L1L2L3L4L5givenalgebra

2.1

The four point stabilizers are maximal subgroups isomorphic to $S_{3}$ , and their intersection fixes every point and is therefore $1$ . Thus the intersection of all maximal subgroups is $Φ (S_{4}) = 1$ . Again the quotient formula reduces to the identity $F (S_{4}) = F (S_{4})$ . This proves the stated claim.

step 1.1givenalgebra∎

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The pp-cores, Fitting subgroup, and Frattini subgroup of S4S4​