CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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A modulus with primitive roots has exactly $φ (φ (n))$ primitive roots

Statement

If $n \geq 1$ admits a primitive root, then it has exactly $φ (φ (n))$ primitive roots.

Facts & Assumptions

Given: A positive modulus $n$ admitting a primitive root.

[L1]

Primitive roots are exactly generators of the unit group (A unit is a primitive root modulo $n$ if and only if it generates $(Z / n Z)^{\times}$ ).

[L2]

A cyclic group of order $m$ has $φ (m)$ generators (The generators of a cyclic group of order $m$ are the $g^{a}$ with $\gcd (a, m) = 1$ , so there are $φ (m)$ of them).

Proof

technique · direct

1.1

By the Given and [L1], the unit group is cyclic of order $φ (n)$ , and its generators are exactly the primitive roots.

givenL1

2.1

Applying [L2] with $m = φ (n)$ yields $φ (φ (n))$ primitive roots. At $n = 1$ this is $φ (1) = 1$ , counting the unique class.

step 1.1L2∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 45 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

William Stein, Elementary Number Theory, Proposition 2.5.12 (standard reference, not scraped)

A modulus with primitive roots has exactly φ(φ(n))φ(φ(n)) primitive roots