TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$Aut (C_{n}) ≅ (Z / n Z)^{\times}$

Statement

For every $n \geq 1$ ,

$Aut (C_{n}) ≅ (Z / n)^{\times} .$

If $C_{n} = ⟨ g ⟩$ , the unit class $[a]$ corresponds to the automorphism $g \mapsto g^{a}$ .

Facts & Assumptions

Given: An integer $n \geq 1$ and a cyclic group $C_{n} = ⟨ g ⟩$ .

[L1]

A cyclic group whose generator has finite order $n$ is isomorphic to $(Z / n, +)$ (Every cyclic group is isomorphic to $(Z, +)$ or to $(Z / n, +)$ for its finite order $n \geq 1$ ).

[L2]

A residue class modulo $n$ is a unit exactly when its representative is coprime to $n$ (For $n \geq 1$ , $[a]_{n}$ is a unit if and only if $\gcd (a, n) = 1$ ).

[L3]

If $g$ has order $n$ , then $g^{k} = 1$ exactly when $n ∣ k$ (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 $j = k$ ).

[L4]

An automorphism is an isomorphism from a group to itself (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

[L5]

If $\gcd (a, n) = 1$ , there are integers $u, v$ with $a u + n v = 1$ (Bézout's identity: for integers $a, b$ not both zero, $\gcd (a, b)$ is the least positive element of ${a x + b y : x, y \in Z}$ ; in particular $a x + b y = \gcd (a, b)$ has an integer solution).

[L6]

The cyclic subgroup generated by $g$ is exactly the set of integer powers of $g$ ( $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian).

Proof

technique · direct

1.1L1L3L6algebra

By [L6] every element of $C_{n}$ is a power $g^{k}$ , and a homomorphism $f$ satisfies $f (g^{k}) = f (g)^{k}$ , so $f$ is determined by $f (g)$ ; writing $f (g) = g^{a}$ , every endomorphism has the form $f_{a} (g^{k}) = g^{a k}$ . By [L3], $f_{a} = f_{b}$ exactly when $a \equiv b (m o d n)$ , so the endomorphisms are indexed by the residue classes of $Z / n$ , which [L1] identifies with $C_{n}$ as an additive group.

1.2L2L3L4L5algebra

The element $g^{a}$ generates $C_{n}$ exactly when $\gcd (a, n) = 1$ : [L5] gives $a u + n v = 1$ , hence $g = (g^{a})^{u}$ , in one direction, while a common divisor greater than one makes every power of $g^{a}$ have exponent divisible by that divisor in the other. Hence $f_{a}$ is an automorphism exactly when $[a]$ is a unit by [L2] and [L4].

2.1step 1.1step 1.2L4∎

Since $f_{a} \circ f_{b} = f_{a b}$ , the correspondence $[a] \mapsto f_{a}$ is a group homomorphism. Steps 1.1 and 1.2 make it bijective, so it is the claimed isomorphism. For $n = 1$ , both groups are trivial.

Depends on

Used by

Dependency tree · two levels

42 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

J. S. Milne, Group Theory (standard reference, not scraped)

Aut⁡(Cn)≅(Z/nZ)×Aut(Cn​)≅(Z/nZ)×