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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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For every odd prime $p$ and $k \geq 1$ , $(Z / p^{k} Z)^{\times}$ is cyclic of order $p^{k - 1} (p - 1)$

Statement

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

$φ (p^{k}) = p^{k - 1} (p - 1) .$

Facts & Assumptions

Given: An odd prime $p$ and an integer $k \geq 1$ .

[L1]

Every prime modulus admits a primitive root (Every prime modulus admits a primitive root), and from a primitive root modulo an odd prime $p$ , at least one of $g$ and $g + p$ is primitive modulo $p^{2}$ (For an odd prime $p$ and a primitive root $g$ modulo $p$ , at least one of $g$ and $g + p$ is primitive modulo $p^{2}$ ).

[L2]

If $p ∤ u$ , the class $1 + p u$ has order $p^{k - 1}$ modulo $p^{k}$ (For odd prime $p$ , $p ∤ u$ , and $k \geq 1$ , the class of $1 + p u$ has order $p^{k - 1}$ modulo $p^{k}$ ).

[L3]

$φ (p^{k}) = p^{k - 1} (p - 1)$ (For a prime $p$ and $k \geq 1$ , $φ (p^{k}) = p^{k} - p^{k - 1}$ ).

[L4]

For an element of finite order $r$ , $x^{t} = 1$ exactly when $r ∣ t$ (If $ord (g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n - 1}$ are distinct, and $⟨ g ⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

[L5]

The order of an element of a finite group divides the group order (The order of every element of a finite group divides the order of the group).

[L6]

If two coprime integers divide an integer, then their product divides it (If $\gcd (a, b) = 1$ and $a ∣ b c$ then $a ∣ c$ ; and if $a ∣ c$ , $b ∣ c$ and $\gcd (a, b) = 1$ then $a b ∣ c$ ).

Proof

technique · direct

1.1L1L4choosealgebra

Choose a primitive root modulo $p$ by [L1], and replace its integer representative by the lift supplied there so that it is primitive modulo $p^{2}$ . Reduction modulo $p$ gives $g^{p - 1} = 1 + p u$ for some integer $u$ , while primitivity modulo $p^{2}$ and [L4] show $p ∤ u$ .

2.1step 1.1L2L4

Reduction modulo $p$ shows that the order of $g$ modulo $p^{k}$ is divisible by $p - 1$ , while [L2] applied to step 1.1 shows that the order of $g^{p - 1}$ is $p^{k - 1}$ .

3.1step 2.1L3L5L6algebra

The cyclic subgroup generated by $g^{p - 1}$ lies in that generated by $g$ , so [L5] and step 2.1 make $p^{k - 1}$ divide the order of $g$ . That order is also divisible by $p - 1$ by step 2.1. These two divisors are coprime, so [L6] makes their product divide the order; conversely [L3] and [L5] make the order divide $p^{k - 1} (p - 1)$ . Therefore it equals that group order.

4.1step 3.1L3∎

Thus $g$ generates the entire unit group, proving cyclicity; when $k = 1$ the same argument reduces to the original primitive root modulo $p$ .

Depends on

Used by

Dependency tree · two levels

52 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, Theorem C.IV.4 (standard reference, not scraped)
William Stein, Elementary Number Theory, Theorem 2.5.11 (standard reference, not scraped)

For every odd prime pp and k≥1k≥1, (Z/pkZ)×(Z/pkZ)× is cyclic of order pk−1(p−1)pk−1(p−1)