False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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FALSE: every finite abelian group is $Gal (Q (μ_{n}) / Q)$ for some $n$

Statement

False claim. For every finite abelian group $G$ there is an $n \geq 1$ with

$Gal (Q (μ_{n}) / Q) ≅ G .$

The obstruction is visible already at the level of cardinality: the cyclic group $C_{3}$ occurs as a Galois group over $Q$ , but never as the Galois group of a cyclotomic field.

Facts & Assumptions

Given: Cyclotomic Galois groups and the theorem realising finite abelian groups over $Q$ .

[L1]

$Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ , so its order is $φ (n)$ ( $[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ , The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L2]

Every finite abelian group is the Galois group of some finite Galois extension of $Q$ (Every finite abelian group is the Galois group of some finite Galois extension of $Q$ ).

[L3]

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

Refutation

technique · direct

1.1

For $n = 1$ and $n = 2$ , the group $(Z / n)^{\times}$ is trivial, so $φ (1) = φ (2) = 1$ .

L1algebra

1.2

If $n \geq 3$ , then the unit class $[- 1]$ in $(Z / n)^{\times}$ has order $2$ : one has $[- 1]^{2} = [1]$ , and $[- 1] \neq [1]$ because otherwise $n$ would divide $2$ , contrary to $n \geq 3$ . Therefore [L3] makes the group order $φ (n) = ∣ (Z / n)^{\times} ∣$ even.

L1L3algebra

2.1

Steps 1.1 and 1.2 show that $φ (n)$ is never $3$ . So no cyclotomic field $Q (μ_{n})$ has Galois group of order $3$ , and in particular none has Galois group isomorphic to $C_{3}$ .

step 1.1step 1.2L1

3.1

By [L2], however, some finite Galois extension of $Q$ does have Galois group $C_{3}$ . Hence the false claim fails.

step 2.1L2∎

Remarks

What the true theorem says instead. The proved result is that every finite abelian group is the Galois group of a subfield of a cyclotomic field, not of the cyclotomic field itself.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

49 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

P. L. Clark, Field Theory (course notes/monograph), Corollary 9.12 (standard reference, not scraped)
K. Conrad, Cyclotomic Extensions (expository blurb), Section 2 (standard reference, not scraped)

FALSE: every finite abelian group is Gal⁡(Q(μn)/Q)Gal(Q(μn​)/Q) for some nn