ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck 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.
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$n Z$ is a subgroup of $(Z, +)$ for every $n \in Z$ , and every subgroup of $(Z, +)$ has this form

Example

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

$n Z : = {n k : k \in Z} .$

Then:

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

In particular every subgroup of $(Z, +)$ is cyclic.

Facts & Assumptions

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

[L1]

$Z$ 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]

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

[L3]

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

[L4]

$⟨ g ⟩$ 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 $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups, $⟨ g ⟩ = {g^{n} : n \in Z}$ , 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 $a \in Z$ and $b > 0$ there are $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ (Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ ).

[L6]

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

[L7]

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

[L8]

Induction on $N$ (The principle of mathematical induction).

Verification

technique · direct

1.1

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

L1L3

1.2

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

L1L2L4L8

1.3

Let $H \leq (Z, +)$ with $H \neq {0}$ . Choose $x \in H$ with $x \neq 0$ ; 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 $⟨ n ⟩$ , the set of integer powers of $n$ , is exactly ${n k : k \in Z} = n Z$ . With step 1.1 this is claim 1.

step 1.1step 1.2L4

2.2

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

step 1.3L2L6L7

3.1

$n Z \subseteq H$ : by step 2.1, $n Z = ⟨ n ⟩$ , and $⟨ n ⟩$ is contained in every subgroup containing $n$ , in particular in $H$ .

step 2.1step 2.2L4

4.1

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

step 2.2step 3.1L1L2L3L5L7

5.1

So $H = n Z$ with $n$ the least positive element of $H$ ; and if $H = {0}$ then $H = 0 Z$ , since $0 \cdot k = 0$ for every $k$ . This is claim 2, and with claim 1 it shows every subgroup of $(Z, +)$ is $⟨ n ⟩$ for some $n \geq 0$ , 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 $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ 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: $m Z \subseteq n Z$ holds exactly when $n ∣ m$ . Indeed the inclusion applied to $k = 1$ gives $m \in n Z$ , that is $m = n c$ for some $c$ ; and conversely $m = n c$ gives $m k = n (c k) \in n Z$ for every $k$ . The systematic study of the divisibility relation belongs to a later page.

Depends on

Used by

Nothing in the library uses this result yet.

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

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

nZnZ is a subgroup of (Z,+)(Z,+) for every n∈Zn∈Z, and every subgroup of (Z,+)(Z,+) has this form