LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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For a prime $p$ and $k \geq 1$ , multiplication by $p$ bijects the standard representatives modulo $p^{k - 1}$ with the representatives modulo $p^{k}$ divisible by $p$

Statement

Let $p$ be prime and let $k \in N$ with $k \geq 1$ . Write $k - 1$ for the unique natural $j$ with $j + 1 = k$ . Multiplication by $p$ gives a bijection

${r \in Z : 0 \leq r < p^{k - 1}} ⟶ {s \in Z : 0 \leq s < p^{k}, p ∣ s}, r ⟼ p r .$

Thus the standard representatives modulo $p^{k}$ divisible by $p$ are exactly $0, p, 2 p, \dots, (p^{k - 1} - 1) p$ , and there are $p^{k - 1}$ of them.

Facts & Assumptions

Given: A prime $p$ and a natural $k \geq 1$ .

[F1]

Natural powers satisfy $p^{0} = 1$ and $p^{k} = p^{k - 1} p$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

[F2]

For a positive modulus $m$ , the standard representatives are the integers $r$ with $0 \leq r < m$ , and there are $m$ of them (For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ , The cardinality $∣ A ∣$ of a finite set).

[F3]

A map is bijective when it is injective and surjective (Injection, surjection, bijection).

[L1]

If $p x = p y$ and $p \neq 0$ , then $x = y$ (The integers have no zero divisors; multiplicative cancellation).

[F4]

A prime satisfies $p > 1$ , hence $p > 0$ and $p \neq 0$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

Proof

technique · direct

1.1

If $0 \leq r < p^{k - 1}$ , then $0 \leq p r < p \cdot p^{k - 1} = p^{k}$ , and $p ∣ p r$ . Thus the displayed rule has values in the stated codomain.

F1F4algebra

1.2

The rule is injective: $p r = p r^{'}$ implies $r = r^{'}$ because $p \neq 0$ .

L1F4

1.3

It is surjective: if $0 \leq s < p^{k}$ and $p ∣ s$ , write $s = p r$ . Since $p > 0$ , the inequalities give $0 \leq r < p^{k - 1}$ after using $p^{k} = p p^{k - 1}$ .

F1F4algebrachoose

2.1

Steps 1.1, 1.2 and 1.3 give a bijection, and [F2] transports the domain cardinality $p^{k - 1}$ to the codomain.

step 1.1step 1.2step 1.3F2F3∎

Depends on

Used by

For a prime p and k≥1, φ(pᵏ)=pᵏ-pᵏ⁻¹ Theorem

Dependency tree · two levels

40 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

Euler's totient function (Wikipedia) (standard reference, not scraped)
Mathematics LibreTexts, Euler's phi Function (standard reference, not scraped)
K. Conrad, The Chinese Remainder Theorem (standard reference, not scraped)
Carnegie Mellon, number-theory lecture notes (standard reference, not scraped)

For a prime pp and k≥1k≥1, multiplication by pp bijects the standard representatives modulo pk−1pk−1 with the representatives modulo pkpk divisible by pp