LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Every subgroup of $(Z, +)$ is $⟨ n ⟩ = n Z$ for exactly one natural number $n$

Statement

Write $ι : N \to Z$ for the embedding of The naturals embed in the integers, and for $g \in Z$ put

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

Then $(Z, +, 0)$ is an abelian group (Group and abelian group), $g Z$ is the cyclic subgroup $⟨ g ⟩$ generated by $g$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups) for every $g$ , and:

every subgroup $H \leq (Z, +)$ (Subgroup) equals $ι (n) Z = ⟨ ι (n) ⟩$ for exactly one natural number $n$ . When $H = {0}$ that natural number is $0$ ; otherwise it is the natural number whose image is the least positive element of $H$ .

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

Facts & Assumptions

Given: The set $Z$ with the operations of Arithmetic on the integers, and $ι (k) = [(k, 0)]$ (The naturals embed in the integers).

[L1]

$Z$ is a commutative ring: addition and multiplication are associative and commutative, $x + 0 = x$ , $x \cdot 1 = x$ , $x \cdot 0 = 0$ , multiplication distributes over addition, and every $x$ has an additive inverse $- x$ , with $- (- x) = x$ and $(- x) y = - (x y)$ ; we write $u - v$ for $u + (- v)$ (The integers form a commutative ring, Arithmetic on the integers, The integers as equivalence classes of pairs of naturals).

[L2]

The order on $Z$ is total, antisymmetric and transitive, is compatible with addition, and positives are closed under multiplication; $x < y$ means $x \leq y$ together with $x \neq y$ (The integers form a totally ordered ring, Order on the integers).

[L3]

$ι$ 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).

[L4]

On $N$ : $m < n$ if and only if $σ (m) \leq n$ (Discreteness: $σ (n)$ is the immediate successor); $1 = σ (0)$ (The natural numbers $N$ (von Neumann)); $0 \leq k$ for every $k$ (Order on the natural numbers); and every nonempty subset has a least element (The well-ordering principle).

[L5]

A group is a monoid in which every element is invertible; a subgroup is a subset containing the identity and closed under the operation and under inverses, and is itself a group under the restricted operation (Group and abelian group, 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 $G$ ).

[L6]

For $g$ in a group, $⟨ g ⟩$ is the smallest subgroup containing $g$ , and $⟨ g ⟩ = {g^{x} : x \in Z}$ (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).

[L7]

Powers are defined by $g^{0} = e$ and $g^{σ (k)} = g^{k} \cdot g$ for $k \in N$ , this being the unique function on $N$ with those two properties, and $g^{x} = (g^{k})^{- 1}$ when $x < 0$ and $- x = ι (k)$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , The recursion theorem).

[L8]

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 $0 \leq r < b$ ).

Proof

technique · direct

1.1

$(Z, +, 0)$ is an abelian group: addition is an associative and commutative binary operation, $0$ is a two-sided identity, and every $x$ has the two-sided inverse $- x$ .

L1L5

1.2

In this group the power $g^{x}$ is the ring product $g x$ . For $x = ι (k)$ with $k \in N$ , the function $k \mapsto g ι (k)$ satisfies $g ι (0) = g \cdot 0 = 0$ and $g ι (σ (k)) = g (ι (k) + 1) = g ι (k) + g$ , which are exactly the two defining equations of $k \mapsto g^{k}$ written additively; by the uniqueness in [L7] the two functions agree. For $x < 0$ , writing $- x = ι (k)$ , we get $g^{x} = - (g^{k}) = - (g ι (k)) = g (- ι (k)) = g x$ .

L1L3L7

1.3

Discreteness: if $0 < x$ in $Z$ then $1 \leq x$ . Indeed $x \geq 0$ , so $x = ι (k)$ with $k \neq 0$ , hence $0 < k$ and $1 = σ (0) \leq k$ in $N$ ; applying $ι$ gives $1 \leq x$ .

L2L3L4

1.4

Existence, main case. Suppose $H \neq {0}$ and pick $h \in H$ with $h \neq 0$ . Then $- h \in H$ as well, and by totality one of $h$ , $- h$ is positive; so the set $P$ of positive elements of $H$ is nonempty.

L1L2L5

2.1

Hence $⟨ g ⟩ = {g^{x} : x \in Z} = {g k : k \in Z} = g Z$ for every $g \in Z$ ; in particular $g Z$ is a subgroup.

step 1.2L6

2.2

Every element of $P$ is nonnegative, hence of the form $ι (k)$ ; so $T : = {k \in N : ι (k) \in P}$ is a nonempty subset of $N$ . Let $k_{0}$ be its least element and put $n : = ι (k_{0}) \in P$ .

step 1.4L3L4choose

2.3

For $g > 0$ , $g$ is the least positive element of $g Z$ . Indeed $g = g \cdot 1 \in g Z$ and $g > 0$ ; and if $g k > 0$ then $k > 0$ , since $k = 0$ gives $g k = 0$ and $k < 0$ gives $0 < - k$ , hence $0 < g (- k) = - (g k)$ and so $g k < 0$ . Then $1 \leq k$ by step 1.3, so $0 \leq k - 1$ and $0 \leq g (k - 1) = g k - g$ , that is $g \leq g k$ .

step 1.3L1L2

3.1

Existence, trivial case. If $H = {0}$ then $H = 0 Z = ⟨ 0 ⟩ = ⟨ ι (0) ⟩$ , since $0 \cdot k = 0$ for every $k$ .

step 2.1L1L3

3.2

$n$ is the least element of $P$ : for $p \in P$ write $p = ι (k)$ with $k \in T$ ; then $k_{0} \leq k$ , and applying $ι$ , which preserves the order, gives $n \leq p$ .

step 2.2L3

3.3

$⟨ n ⟩ \subseteq H$ , since $n \in H$ and $⟨ n ⟩$ is the smallest subgroup containing $n$ .

step 2.2L6

3.4

Uniqueness. Suppose $ι (m) Z = ι (n) Z = H$ with $m, n \in N$ . If $ι (n) = 0$ then $H = {0}$ , so $ι (m) \in H$ forces $ι (m) = 0$ and hence $m = 0 = n$ by injectivity. Otherwise $ι (n) > 0$ , so $H \neq {0}$ and likewise $ι (m) > 0$ ; by step 2.3 both $ι (m)$ and $ι (n)$ are the least positive element of $H$ , hence equal, and $m = n$ by injectivity.

step 2.3L2L3

4.1

$H \subseteq n Z$ . Let $h \in H$ . Since $n > 0$ , [L8] gives $h = q n + r$ with $0 \leq r < n$ . Now $q n \in n Z = ⟨ n ⟩ \subseteq H$ by step 2.1 and step 3.3, so $r = h - q n \in H$ because $H$ is closed under inverses and under addition. If $r$ were positive it would lie in $P$ and satisfy $r < n$ , contradicting step 3.2 together with antisymmetry; so $r = 0$ and $h = q n \in n Z$ .

step 2.1step 3.2step 3.3L1L2L5L8

5.1

Combining, $H = n Z = ⟨ n ⟩$ with $n = ι (k_{0})$ , which with step 3.1 proves existence for every subgroup.

step 2.1step 3.1step 3.3step 4.1

6.1

Every subgroup of $(Z, +)$ is therefore $⟨ ι (n) ⟩ = ι (n) Z$ for exactly one $n \in N$ , and in particular is cyclic.

step 5.1step 3.4∎

Remarks

Why this is proved here rather than cited. The published example $n Z$ is a subgroup of $(Z, +)$ for every $n \in Z$ , and every subgroup of $(Z, +)$ has this form gives the same classification, but it lives on an examples page, and examples pages are leaves in the library's reading order: no later page may depend on an item homed there. The classification is therefore re-established on a spine, so that this page and later ones have a citable home for it. Neither is used in proving the other.

The two are not quite the same statement, and the difference is the whole reason this one is worded as it is. The example asserts that every subgroup is $n Z$ for some $n \geq 0$ , and adds that $n$ may be taken to be $0$ or the least positive element of $H$ ; it does not assert that no other $n \geq 0$ names the same subgroup. The statement above asserts exactly one, which is strictly stronger. The extra content is small — if $a Z = b Z$ with $a, b \geq 0$ then each of $a, b$ divides the other, so $a = b$ — but it is what the uses below actually need, and it is proved here rather than assumed to come with the example.

What it is used for below. Exactly one thing: the uniqueness of the nonnegative generator, which is what lets $a Z + b Z = \gcd (a, b) Z$ be read as an identification of subgroups rather than as an accident about two sets ( $a Z + b Z = \gcd (a, b) Z$ and $a Z \cap b Z = lcm (a, b) Z$ ; equivalently, in $(Z, +)$ the subgroup generated by ${a, b}$ is $⟨ \gcd (a, b) ⟩$ and $⟨ a ⟩ \cap ⟨ b ⟩ = ⟨ lcm (a, b) ⟩$ ).
The generator is a natural number, hence a set. $n Z$ above always means $ι (n) Z$ ; the natural number $n$ is not itself an element of $Z$ , and the embedding is written out wherever the distinction could matter.

Depends on

Used by

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)

Every subgroup of (Z,+)(Z,+) is ⟨n⟩=nZ⟨n⟩=nZ for exactly one natural number nn