ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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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.

$n\mathbb{Z}$ is a subgroup of $(\mathbb{Z}, +)$ for every $n \in \mathbb{Z}$ , and every subgroup of $(\mathbb{Z}, +)$ has this form

Example

Work in the abelian group $(\mathbb{Z}, +, 0)$ . For $n \in \mathbb{Z}$ put

$n\mathbb{Z} \;:=\; \{\, nk \;:\; k \in \mathbb{Z} \,\}.$

Then:

$n\mathbb{Z}$ is a subgroup of $(\mathbb{Z},+)$ for every $n \in \mathbb{Z}$ (Subgroup), and $n\mathbb{Z} = \langle n \rangle$ (The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$ , and cyclic groups);
conversely, every subgroup $H \le (\mathbb{Z},+)$ equals $n\mathbb{Z}$ for some $n \ge 0$ ; and $n$ may be taken to be $0$ if $H = \{0\}$ and otherwise the least positive element of $H$ .

In particular every subgroup of $(\mathbb{Z},+)$ is cyclic.

Facts & Assumptions

Given: The abelian group $(\mathbb{Z},+,0)$ (The integers as equivalence classes of pairs of naturals, Arithmetic on the integers, Group and abelian group), and the embedding $\iota : \mathbb{N} \to \mathbb{Z}$ of The naturals embed in the integers.

[L1]

$\mathbb{Z}$ is a commutative ring, with $u - v := u + (-v)$ (The integers form a commutative ring, Arithmetic on the integers); its order is total, antisymmetric, transitive and compatible with addition, and positives are closed under multiplication (The integers form a totally ordered ring, Order on the integers).

[L2]

$\iota$ is injective, preserves addition, multiplication and order, and its image is exactly the nonnegative integers; $\iota(0) = 0$ , $\iota(1) = 1$ (The naturals embed in the integers).

[L3]

One-step test for subgroups, written additively: a nonempty $H \subseteq \mathbb{Z}$ with $x - y \in H$ for all $x, y \in H$ is a subgroup (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $gh^{-1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ , Subgroup).

[L4]

$\langle g \rangle$ is the smallest subgroup containing $g$ , and equals the set of integer powers of $g$ , which in additive notation are the multiples of $g$ (The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$ , and cyclic groups, $\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$ , and every cyclic group is abelian, Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L5]

Division with remainder: for $a \in \mathbb{Z}$ and $b > 0$ there are $q, r \in \mathbb{Z}$ with $a = qb + r$ and $0 \le r < b$ (Division with remainder in $\mathbb{Z}$ : for $a \in \mathbb{Z}$ and $b > 0$ there are unique $q, r \in \mathbb{Z}$ with $a = qb + r$ and $0 \le r < b$ ).

[L6]

Every nonempty subset of $\mathbb{N}$ has a least element (The well-ordering principle).

[L7]

On $\mathbb{N}$ : exactly one of $s < t$ , $s = t$ , $t < s$ holds (Trichotomy of the order on $\mathbb{N}$ ); every $k \ne 0$ is a successor $\sigma(j) = 1 + j$ , so $k \ne 0$ gives $1 \le k$ (Every nonzero natural number is a successor, Addition is commutative, Order on the natural numbers, The natural numbers $\mathbb{N}$ (von Neumann)).

[L8]

Induction on $\mathbb{N}$ (The principle of mathematical induction).

Verification

technique · direct

1.1

$n\mathbb{Z}$ is nonempty, containing $0 = n \cdot 0$ ; and for $x = nk$ , $y = nk'$ in it, $x - y = nk - nk' = n(k - k') \in n\mathbb{Z}$ by distributivity. By the one-step test $n\mathbb{Z}$ is a subgroup of $(\mathbb{Z},+)$ .

L1L3

1.2

In the additive group $(\mathbb{Z},+)$ the $k$ -th power of $n$ is $nk$ . For $k \in \mathbb{N}$ this holds by induction: the $0$ -th power is the identity $0 = n \cdot 0$ , and the power at $\sigma(k)$ is the power at $k$ plus $n$ , which is $n\iota(k) + n = n(\iota(k)+1) = n\iota(\sigma(k))$ . For a negative integer $x$ with $-x = \iota(k)$ , the $x$ -th power is the additive inverse of $n\iota(k)$ , namely $n(-\iota(k)) = nx$ .

L1L2L4L8

1.3

Let $H \le (\mathbb{Z},+)$ with $H \ne \{0\}$ . Choose $x \in H$ with $x \ne 0$ ; then $-x \in H$ as well, and by totality one of $x$ , $-x$ is positive, so $H$ contains a positive integer.

L1L3choose

2.1

Hence $\langle n \rangle$ , the set of integer powers of $n$ , is exactly $\{\, nk : k \in \mathbb{Z} \,\} = n\mathbb{Z}$ . With step 1.1 this is claim 1.

step 1.1step 1.2L4

2.2

Every positive integer is $\iota(k)$ for a unique $k \in \mathbb{N}$ with $k \ne 0$ , hence with $k \ge 1$ . So $S := \{\, k \in \mathbb{N} : k \ge 1 \text{ and } \iota(k) \in H \,\}$ is nonempty by step 1.3; let $m$ be its least element and put $n := \iota(m)$ , a positive element of $H$ .

step 1.3L2L6L7

3.1

$n\mathbb{Z} \subseteq H$ : by step 2.1, $n\mathbb{Z} = \langle n \rangle$ , and $\langle n \rangle$ is contained in every subgroup containing $n$ , in particular in $H$ .

step 2.1step 2.2L4

4.1

$H \subseteq n\mathbb{Z}$ : let $h \in H$ and divide $h = qn + r$ with $0 \le r < n$ , legitimate since $n > 0$ . Then $qn \in n\mathbb{Z} \subseteq H$ , so $r = h - qn \in H$ because $H$ is closed under subtraction. If $r > 0$ then $r = \iota(t)$ with $t \ge 1$ , and $\iota(t) < \iota(m)$ gives $t < m$ , since otherwise $m \le t$ would give $\iota(m) \le \iota(t)$ , contradicting antisymmetry; that puts $t$ in $S$ below its least element, which is impossible. Hence $r = 0$ and $h = qn \in n\mathbb{Z}$ .

step 2.2step 3.1L1L2L3L5L7

5.1

So $H = n\mathbb{Z}$ with $n$ the least positive element of $H$ ; and if $H = \{0\}$ then $H = 0\mathbb{Z}$ , since $0 \cdot k = 0$ for every $k$ . This is claim 2, and with claim 1 it shows every subgroup of $(\mathbb{Z},+)$ is $\langle n \rangle$ for some $n \ge 0$ , hence cyclic.

step 2.1step 2.2step 3.1step 4.1L1L4∎

Remarks

Both halves use the division algorithm, but only the second one visibly. The first half is pure closure arithmetic; the second is the standard argument that a subgroup containing a least positive element $n$ can contain nothing strictly between the multiples of $n$ , and it is exactly Division with remainder in $\mathbb{Z}$ : for $a \in \mathbb{Z}$ and $b > 0$ there are unique $q, r \in \mathbb{Z}$ with $a = qb + r$ and $0 \le r < b$ that produces the offending remainder.
$n$ and $-n$ generate the same subgroup, since $(-n)k = n(-k)$ , so the $n$ in claim 2 is unique only after normalising it to be nonnegative. The normalisation is what the phrase "the least positive element" achieves.
Inclusion among these subgroups is divisibility: $m\mathbb{Z} \subseteq n\mathbb{Z}$ holds exactly when $n \mid m$ . Indeed the inclusion applied to $k = 1$ gives $m \in n\mathbb{Z}$ , that is $m = nc$ for some $c$ ; and conversely $m = nc$ gives $mk = n(ck) \in n\mathbb{Z}$ for every $k$ . The systematic study of the divisibility relation belongs to a later page.

Depends on

Used by

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Sources

Cyclic group (Wikipedia) (standard reference, not scraped)
Subgroup (Wikipedia) (standard reference, not scraped)

nZn\mathbb{Z}nZ is a subgroup of (Z,+)(\mathbb{Z}, +)(Z,+) for every n∈Zn \in \mathbb{Z}n∈Z, and every subgroup of (Z,+)(\mathbb{Z}, +)(Z,+) has this form