TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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The unit group modulo $n$ is the product of its odd-prime cyclic factors and its explicit $2$ -power factor

Statement

Let $n \geq 1$ have prime-power factorisation $n = 2^{a} \prod_{i < r} p_{i}^{k_{i}}$ , where the $p_{i}$ are distinct odd primes and $k_{i} \geq 1$ . Then

$(Z / n)^{\times} ≅ U_{2, a} \times \prod_{i < r} C_{p_{i}^{k_{i} - 1} (p_{i} - 1)},$

where $U_{2, 0}$ and $U_{2, 1}$ are trivial, $U_{2, 2} = C_{2}$ , and

$U_{2, a} = C_{2} \times C_{2^{a - 2}} (a \geq 3) .$

For $n = 1$ the product is empty and hence trivial.

Facts & Assumptions

Given: A positive integer $n$ and its displayed prime-power factorisation.

[L1]

CRT gives an isomorphism from a unit group to the product of the unit groups of pairwise coprime factors (For pairwise coprime positive moduli, the Chinese remainder bijection restricts to an isomorphism of unit groups).

[L2]

For odd $p$ , $(Z / p^{k})^{\times}$ is cyclic of order $p^{k - 1} (p - 1)$ (For every odd prime $p$ and $k \geq 1$ , $(Z / p^{k} Z)^{\times}$ is cyclic of order $p^{k - 1} (p - 1)$ ).

[L3]

For $a \geq 3$ , $(Z / 2^{a})^{\times} ≅ C_{2} \times C_{2^{a - 2}}$ (For $k \geq 3$ , $(Z / 2^{k} Z)^{\times} ≅ C_{2} \times C_{2^{k - 2}}$ , generated uniquely as $(- 1)^{ε} 5^{j}$ ).

[L4]

Given an injective list of primes containing every prime divisor of $n \geq 1$ , one has $n = \prod_{i} p_{i}^{v_{p_{i}} (n)}$ , the exponents being determined by $n$ (For $n \geq 1$ and any injective list $p : r \to Z$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i < r} p_{i}^{v_{p_{i}} (n)}$ ; the exponents are determined by $n$ , and $v_{q} (n) = 0$ for every prime $q$ outside the list). The Given of this theorem supplies exactly such a list, namely the primes of the displayed factorisation.

[L5]

$φ (2) = 1$ and $φ (4) = 2$ by the prime-power formula (For a prime $p$ and $k \geq 1$ , $φ (p^{k}) = p^{k} - p^{k - 1}$ ).

[L7]

The totient is the cardinality of the unit group: $φ (n) = ∣ (Z / n)^{\times} ∣$ (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L6]

Every group of prime order is cyclic (A finite group of prime order is cyclic and every nonidentity element generates it).

Proof

technique · direct

1.1L4L1

By [L4], the displayed factors are pairwise coprime, so [L1] decomposes the unit group into its $2$ -power factor and the odd-prime-power factors.

2.1step 1.1L2L3

Substitute [L2] for every odd factor and [L3] for the $2$ -power factor when $a \geq 3$ .

2.2step 1.1L5L6L7

If $a = 0$ there is no $2$ -factor. If $a = 1$ , then [L5] gives $φ (2) = 1$ and [L7] reads that as $∣ (Z / 2)^{\times} ∣ = 1$ , so the unit group is trivial. If $a = 2$ , then [L5] and [L7] give $∣ (Z / 4)^{\times} ∣ = 2$ , prime order, so [L6] makes it cyclic.

3.1step 2.1step 2.2L1∎

Steps 2.1 and 2.2 give the asserted decomposition. When $n = 1$ , [L1] identifies the empty product with the one-element unit group.

Depends on

Used by

Dependency tree · two levels

59 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

Peter Hackman, Elementary Number Theory, §§C.IV–C.V (standard reference, not scraped)

The unit group modulo nn is the product of its odd-prime cyclic factors and its explicit 22-power factor