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

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

Statement

For every $k \geq 3$ ,

$(Z / 2^{k})^{\times} ≅ C_{2} \times C_{2^{k - 2}} .$

More precisely, every unit has a unique representation $(- 1)^{ε} 5^{j}$ with $ε \in {0, 1}$ and $j$ modulo $2^{k - 2}$ .

Facts & Assumptions

Given: An integer $k \geq 3$ .

[L1]

The class of $5$ has order $2^{k - 2}$ modulo $2^{k}$ (For $k \geq 3$ , the class of $5$ has order $2^{k - 2}$ modulo $2^{k}$ ).

[L2]

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

[L3]

A direct product of finite groups has the product of their orders (For finite groups $G$ and $H$ , $∣ G \times H ∣ = ∣ G ∣ ∣ H ∣$ ).

[L4]

The units modulo $2^{k}$ are the classes represented by odd integers (For $n \geq 1$ , $[a]_{n}$ is a unit if and only if $\gcd (a, n) = 1$ ).

[L5]

Multiplication modulo $2^{k}$ is commutative and restricts to the unit group (The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L6]

A bijective group homomorphism is a group isomorphism (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

Proof

technique · direct

1.1L1algebra

The class of $- 1$ has order $2$ , and [L1] gives order $2^{k - 2}$ for $5$ .

1.2algebra

Every power of $5$ is $1$ modulo $4$ , whereas $- 1$ is $3$ modulo $4$ ; hence $⟨ - 1 ⟩ \cap ⟨ 5 ⟩ = {1}$ .

2.1step 1.1step 1.2L4L5

The map $C_{2} \times C_{2^{k - 2}} \to (Z / 2^{k})^{\times}$ given by $(ε, j) \mapsto (- 1)^{ε} 5^{j}$ is a homomorphism by [L5], and step 1.2 makes it injective.

3.1step 2.1L2L3

Its domain has $2^{k - 1}$ elements by [L3], equal to the size of the target by [L2]; thus it is bijective.

4.1step 3.1L6∎

By [L6] the map is an isomorphism, and its bijectivity is exactly the asserted unique representation.

Depends on

Used by

Dependency tree · two levels

33 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.8 (standard reference, not scraped)

For k≥3k≥3, (Z/2kZ)×≅C2×C2k−2(Z/2kZ)×≅C2​×C2k−2​, generated uniquely as (−1)ε5j(−1)ε5j