ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-24

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Inversion on $C_{3}$ is detected on its Frattini quotient

Example

Inversion on $C_{3}$ is a nontrivial automorphism of order two, and its action on the Frattini quotient is nontrivial. Thus Hall–Burnside detects it; the theorem does not assert that coprime automorphisms are absent.

Facts & Assumptions

Given: The cyclic group $C_{3} = ⟨ g ⟩$ .

[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]

If a $p^{'}$ -subgroup of $Aut (P)$ acts trivially on $P / Φ (P)$ , then it is trivial (Hall–Burnside: coprime automorphisms are detected on the Frattini quotient).

[L3]

For $C_{n} = ⟨ g ⟩$ , every unit class $[a] \in (Z / n)^{\times}$ defines the automorphism $g \mapsto g^{a}$ ( $Aut (C_{n}) ≅ (Z / n Z)^{\times}$ ).

Verification

technique · direct

1.1

By [L1], $Φ (C_{3}) = ⟨ g^{3} ⟩ = 1$ . The unit $[- 1]_{3}$ gives inversion by [L3]; it sends $g$ to $g^{- 1} = g^{2} \neq g$ and its square is the identity, so it is a nonidentity automorphism of order two.

givenL1L3algebra

2.1

Since the Frattini subgroup is trivial, the induced quotient action is the same nontrivial inversion. This is consistent with [L2], which forbids this order-two subgroup from acting trivially because $2$ is coprime to $3$ .

step 1.1L2∎

Depends on

Used by

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Dependency tree · two levels

19 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

K. Conrad, Generating Sets, Example 6.9 (standard reference, not scraped)

Inversion on C3C3​ is detected on its Frattini quotient