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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11
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Every finite subgroup of the unit group of an integral domain is cyclic

Statement

Let D be an integral domain. Every finite subgroup G≤D× of the unit group of D is cyclic.

Facts & Assumptions

Given: An integral domain D and a finite subgroup G≤D×.

[L1]

An integral domain is a commutative ring with no zero divisors and with 0≠1 (Zero divisor, and integral domain: a commutative ring with 1≠0 and no zero divisors).

[L3]

A subgroup contains the identity and is closed under multiplication and inverses (Subgroup).

[L4]

A group is cyclic when it equals ⟨g⟩ for some element g (The subgroup ⟨S⟩ generated by a subset, the cyclic subgroup ⟨g⟩, and cyclic groups).

[L5]

A finite set has a natural-number cardinality ∣G∣ (The cardinality ∣A∣ of a finite set).

[L6]

The exponent e of a finite group is the least positive integer such that ge=1 for every group element g (The exponent of a finite group).

[L7]

If G≅Cn1×⋯×Cnr with 1<n1∣⋯∣nr, then ∣G∣=n1⋯nr and exp⁡(G)=nr (Invariant factors determine the order and exponent of a finite abelian group).

[L8]

A nontrivial finite abelian group is cyclic exactly when its invariant-factor list has one entry (A nontrivial finite abelian group is cyclic if and only if it has one invariant factor).

[L9]

A monic polynomial has leading coefficient 1, and a nonzero polynomial has a natural-number degree (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L10]

Evaluation substitutes a ring element into a formal polynomial, and a root is an element with value zero (Evaluation and roots of a polynomial in a commutative target ring).

[L11]

A nonzero polynomial of degree n over an integral domain has at most n distinct roots (A nonzero polynomial of degree n over an integral domain has at most n distinct roots).

[L12]

Every finite abelian group has a unique invariant-factor list 1<n1∣⋯∣nr, with the trivial group corresponding to the empty list (Fundamental theorem of finite abelian groups: invariant-factor form).

Proof

technique · direct
1.1

If G={1}, then G=⟨1⟩ and [L4] makes it cyclic.

givenL2L3L4
1.2

Suppose G is nontrivial. Since D is commutative by [L1], [L2] and [L3] make G a finite abelian group; let e≥1 be its exponent from [L6].

givenL1L2L3L5L6
2.1

Every g∈G satisfies ge=1, so all ∣G∣ distinct elements of G are roots, in the sense of [L10], of the nonzero monic polynomial Te−1, whose degree is e by [L9]; [L11] gives ∣G∣≤e.

step 1.2L5L6L9L10L11
3.1

By [L12], write the invariant-factor list of G as 1<n1∣⋯∣nr. By [L7], ∣G∣=n1⋯nr and e=nr, so e≤∣G∣; step 2.1 gives equality. Hence n1⋯nr=nr, and because every ni>1, this forces r=1. Fact [L8] makes G cyclic, while step 1.1 covers the trivial case.

step 1.1step 2.1L7L8L12∎

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Sources