TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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

Statement

For every finite abelian group $G$ there is a finite Galois extension $L / Q$ (Finite Galois extensions and $Gal (K / F)$ ) with

$Gal (L / Q) ≅ G,$

and $L$ may be taken inside a cyclotomic field $Q (μ_{N})$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ).

Facts & Assumptions

Given: A finite abelian group $G$ ; $Q$ is an ordered field (The rationals form a totally ordered field), so $char Q = 0$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise) and divides no positive integer (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L1]

There are positive integers $n, k$ and a surjective group homomorphism $(Z / n)^{k} \to G$ (Every finite abelian group is a quotient of $(Z / n)^{k}$ for some $n$ and $k$ , The external direct product $G \times H$ with componentwise multiplication).

[L2]

For every $n \geq 1$ the set of primes $p$ with $p \equiv 1 (m o d n)$ is not finite (For every $n \geq 1$ there are infinitely many primes $p$ with $p \equiv 1 (m o d n)$ , Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ , Finite, countably infinite, countable, uncountable, Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[L3]

For a finite pairwise-coprime list $n_{0}, \dots, n_{r - 1}$ of positive integers with $N = \prod_{i} n_{i}$ , the map $[x]_{N} \mapsto ([x]_{n_{i}})_{i}$ is a bijection $Z / N \to \prod_{i} Z / n_{i}$ preserving addition, multiplication, $[0]$ and $[1]$ (Chinese remainder theorem for a finite pairwise-coprime list: simultaneous residues determine one class modulo the product, and the resulting bijection preserves addition and multiplication, The congruence class $[a]_{n}$ and the quotient set $Z / n$ , For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold, Coprime integers: $\gcd (a, b) = 1$ , Common divisor, and the greatest common divisor $\gcd (a, b)$ , with the convention $\gcd (0, 0) : = 0$ ).

[L4]

$(Z / p)^{\times}$ is a finite subgroup of the unit group of the field $Z / p$ , hence cyclic (Every finite subgroup of the unit group of an integral domain is cyclic, For every prime $p$ , the two operations on $Z / p$ make it a field, Field), of order $φ (p) = p - 1$ ( $φ (1) = 1$ , and $φ (p) = p - 1$ for every prime $p$ , The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L5]

A cyclic group of finite order $m$ is isomorphic to $(Z / m, +)$ (Every cyclic group is isomorphic to $(Z, +)$ or to $(Z / n, +)$ for its finite order $n \geq 1$ ).

[L6]

$Gal (Q (μ_{N}) / Q) ≅ (Z / N)^{\times}$ ( $[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ ), and this group is abelian (The Galois group of a cyclotomic extension is abelian).

[L7]

For $K / Q$ finite Galois with group $G^{'}$ and $H \leq G^{'}$ , the field $K^{H}$ is an intermediate field (The fundamental theorem of finite Galois theory); it is Galois over $Q$ exactly when $H$ is normal (Normal subgroup: invariance under conjugation), and then $Gal (K^{H} / Q) ≅ G^{'} / H$ (Normal subgroups, conjugate fields, and quotient groups in the Galois correspondence, The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

[L8]

For a homomorphism $f : A \to B$ the rule $a \ker f \mapsto f (a)$ is an isomorphism $A / \ker f \to im f$ (First isomorphism theorem for groups: $G / \ker f ≅ im f$ ).

Proof

technique · direct

1.1L1

Fix $n, k \geq 1$ and a surjection $π ⁣ : (Z / n)^{k} \to G$ by [L1].

2.1step 1.1L2

Choose pairwise distinct primes $p_{1}, \dots, p_{k}$ with $p_{i} \equiv 1 (m o d n)$ : the set of such primes is not finite by [L2], so at each of the $k$ steps one may pick a prime outside the finitely many already chosen. Put $N : = p_{1} \dots p_{k}$ .

3.1step 2.1givenalgebra

Distinct primes are coprime: a positive common divisor of $p_{i}$ and $p_{j}$ is $1$ or $p_{i}$ , and is $1$ or $p_{j}$ , so if it is not $1$ then $p_{i} = p_{j}$ . Hence $p_{1}, \dots, p_{k}$ is a pairwise-coprime list.

3.2step 2.1L4L5

For each $i$ there is a surjective homomorphism $(Z / p_{i})^{\times} \to Z / n$ : by [L4] the group $(Z / p_{i})^{\times}$ is cyclic of order $m_{i} = p_{i} - 1$ , which $n$ divides by step 2.1; [L5] identifies it with $(Z / m_{i}, +)$ , and $[a]_{m_{i}} \mapsto [a]_{n}$ is well defined because $n ∣ m_{i}$ , is a homomorphism, and is onto.

4.1step 3.1L3

By [L3] the map $[x]_{N} \mapsto ([x]_{p_{i}})_{i}$ is a bijection $Z / N \to \prod_{i} Z / p_{i}$ preserving multiplication and $[1]$ , so it carries units to units bijectively and restricts to a group isomorphism $(Z / N)^{\times} \to \prod_{i} (Z / p_{i})^{\times}$ .

5.1step 1.1step 4.1step 3.2L6

Taking the product of the maps of step 3.2 and composing with step 4.1 and with $π$ gives a surjective group homomorphism $Ψ_{0} ⁣ : (Z / N)^{\times} \to G$ ; composing with the isomorphism of [L6] gives a surjective homomorphism $Ψ ⁣ : Gal (Q (μ_{N}) / Q) \to G$ .

6.1step 5.1L6L7

Put $G^{'} : = Gal (Q (μ_{N}) / Q)$ and $H : = \ker Ψ$ . The group $G^{'}$ is abelian by [L6], so every subgroup is normal, $g H g^{- 1} = H$ holding for all $g$ ; hence $L : = Q (μ_{N})^{H}$ is an intermediate field, $L / Q$ is finite Galois, and $Gal (L / Q) ≅ G^{'} / H$ by [L7].

7.1step 5.1step 6.1L8∎

By [L8] applied to $Ψ$ , which is surjective, $G^{'} / H ≅ im Ψ = G$ ; hence $Gal (L / Q) ≅ G$ , with $L$ inside $Q (μ_{N})$ .

Remarks

What is produced is a subfield, not a cyclotomic field. The construction realises $G$ as the Galois group of an intermediate field of $Q (μ_{N}) / Q$ , and it must: the Galois group of $Q (μ_{N})$ itself is $(Z / N)^{\times}$ , whose order $φ (N)$ is even for $N \geq 3$ , so most finite abelian groups are not of that form. The companion page spells out that failure in FALSE: every finite abelian group is $Gal (Q (μ_{n}) / Q)$ for some $n$ ↗.
The distinctness of the primes is needed twice. It makes the list pairwise coprime so that the Chinese remainder theorem applies, and it makes the product $N$ have exactly the intended unit group. Repeating a prime would collapse two factors into one.

Depends on

Used by

FALSE: every finite abelian group is Gal(ℚ(μₙ)/ℚ) for some n False statement

Dependency tree · two levels

130 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 5 (standard reference, not scraped)

Every finite abelian group is the Galois group of some finite Galois extension of QQ