CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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A nonsurjective homomorphism need not carry the Frattini subgroup into the target Frattini subgroup

Statement refuted

For every group homomorphism $f : G \to H$ , one has $f (Φ (G)) \leq Φ (H)$ . This fails for the embedding $f : C_{4} ↪ S_{5}$ below: $f (Φ (C_{4})) ≰ Φ (S_{5})$ .

Facts & Assumptions

Given: The cyclic group $C_{4} = ⟨ g ⟩$ , the symmetric group $S_{5}$ with the cycle convention of The finite symmetric group $S_{n}$ , one-line notation, and cycle notation, and the homomorphism $f : C_{4} \to S_{5}$ defined by $f (g) = (1 2 3 4)$ (Monoid homomorphism and group homomorphism).

[L1]

If $P = ⟨ g ⟩$ has order $p^{n}$ with $n \geq 1$ , then $Φ (P) = ⟨ g^{p} ⟩$ and $d (P) = 1$ (The Frattini subgroup of a nontrivial cyclic $p$ -group).

[L2]

The Frattini subgroup of every finite group is nilpotent (The Frattini subgroup of a finite group is nilpotent).

[L3]

For $n \geq 5$ , the normal subgroups of $S_{n}$ are $1, A_{n}, S_{n}$ (For $n \geq 5$ , the only proper nontrivial normal subgroup of $S_{n}$ is $A_{n}$ ).

[L4]

The groups $A_{5}$ and $S_{5}$ are not solvable ( $A_{5}$ and $S_{n}$ for $n \geq 5$ are not solvable).

[L5]

Every nilpotent group is solvable (Nilpotent groups, and in particular finite $p$ -groups, are solvable).

[L6]

The Frattini subgroup is characteristic and hence normal (The Frattini subgroup of a finite group is characteristic, The Frattini subgroup $Φ (G)$ as the intersection of the maximal subgroups of a finite group).

Counterexample

technique · direct

1.1

The $4$ -cycle has order four, so $f$ is an embedding. By [L1], $Φ (C_{4}) = ⟨ g^{2} ⟩$ , and $f (g^{2}) = (1 3) (2 4) \neq 1$ .

givenL1algebra

1.2

By [L2] and [L6], $Φ (S_{5})$ is nilpotent and normal. The list [L3] leaves only $1, A_{5}, S_{5}$ ; [L4] and [L5] exclude the latter two, so $Φ (S_{5}) = 1$ .

givenL2L3L4L5L6algebra

2.1

Step 1.1 exhibits a nonidentity element of $f (Φ (C_{4}))$ , while step 1.2 makes the target Frattini subgroup trivial. Hence $f (Φ (C_{4})) ≰ Φ (S_{5})$ .

step 1.1step 1.2∎

Depends on

Used by

FALSE: every homomorphism carries the Frattini subgroup into the target Frattini subgroup False statement

Dependency tree · two levels

36 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

D. A. Craven, The Theory of p-Groups, §2.2 (standard reference, not scraped)
J. S. Milne, Group Theory (standard reference, not scraped)