TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-02 (claude-opus-5)

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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.

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$a\mathbb{Z} + b\mathbb{Z} = \gcd(a,b)\,\mathbb{Z}$ and $a\mathbb{Z} \cap b\mathbb{Z} = \operatorname{lcm}(a,b)\,\mathbb{Z}$ ; equivalently, in $(\mathbb{Z},+)$ the subgroup generated by $\{a,b\}$ is $\langle \gcd(a,b) \rangle$ and $\langle a \rangle \cap \langle b \rangle = \langle \operatorname{lcm}(a,b) \rangle$

Statement

Let $a, b \in \mathbb{Z}$ , and put

$g\mathbb{Z} := \{\, gk : k \in \mathbb{Z} \,\}, \qquad a\mathbb{Z} + b\mathbb{Z} := \{\, ax + by \;:\; x, y \in \mathbb{Z} \,\} .$

Then $a\mathbb{Z} + b\mathbb{Z}$ and $a\mathbb{Z} \cap b\mathbb{Z}$ are subgroups of $(\mathbb{Z},+)$ (Subgroup), and

$a\mathbb{Z} + b\mathbb{Z} \;=\; \gcd(a,b)\,\mathbb{Z}, \qquad a\mathbb{Z} \cap b\mathbb{Z} \;=\; \operatorname{lcm}(a,b)\,\mathbb{Z}$

(Common divisor, and the greatest common divisor $\gcd(a,b)$ , with the convention $\gcd(0,0) := 0$ , Common multiple, and the least common multiple $\operatorname{lcm}(a,b)$ , taken to be $0$ when $a = 0$ or $b = 0$ ). Equivalently, in the group $(\mathbb{Z},+)$ ,

$\langle \{a,b\} \rangle \;=\; \langle \gcd(a,b) \rangle, \qquad \langle a \rangle \cap \langle b \rangle \;=\; \langle \operatorname{lcm}(a,b) \rangle$

(The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$ , and cyclic groups). Since $\gcd(a,b) \ge 0$ and $\operatorname{lcm}(a,b) \ge 0$ , and every subgroup of $(\mathbb{Z},+)$ has exactly one nonnegative generator (Every subgroup of $(\mathbb{Z}, +)$ is $\langle n \rangle = n\mathbb{Z}$ for exactly one natural number $n$ ), these are identifications of subgroups by their canonical generator, not merely equalities of two sets that happen to coincide.

Facts & Assumptions

Given: Integers $a$ and $b$ ; $d := \gcd(a,b)$ and $\ell := \operatorname{lcm}(a,b)$ .

[L1]

$\mathbb{Z}$ 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; 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]

$(\mathbb{Z},+,0)$ is an abelian group; $g\mathbb{Z} = \langle g \rangle$ is a subgroup for every $g$ ; and every subgroup equals $\iota(n)\mathbb{Z}$ for exactly one $n \in \mathbb{N}$ (Every subgroup of $(\mathbb{Z}, +)$ is $\langle n \rangle = n\mathbb{Z}$ for exactly one natural number $n$ , Group and abelian group).

[L3]

A subgroup contains the identity and is closed under the operation and under inverses; a nonempty $S \subseteq \mathbb{Z}$ with $x - y \in S$ for all $x, y \in S$ is a subgroup of $(\mathbb{Z},+)$ (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 $G$ ).

[L4]

$\langle S \rangle$ is the smallest subgroup containing $S$ : it contains $S$ and is contained in every subgroup containing $S$ (The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$ , and cyclic groups).

[L5]

An intersection of subgroups is a subgroup (The intersection of a nonempty family of subgroups of $G$ is a subgroup of $G$ ).

[L6]

$u \mid v$ means $v = uk$ for some $k \in \mathbb{Z}$ ; equivalently $v \in u\mathbb{Z}$ (Divisibility in $\mathbb{Z}$ : $d \mid a$ when $a = dq$ for some integer $q$ ).

[L7]

$\gcd(a,b) \ge 0$ and is a common divisor of $a$ and $b$ ; $\gcd(0,0) = 0$ (Common divisor, and the greatest common divisor $\gcd(a,b)$ , with the convention $\gcd(0,0) := 0$ ).

[L8]

For $(a,b) \ne (0,0)$ there are $x_0, y_0$ with $a x_0 + b y_0 = \gcd(a,b)$ (Bézout's identity: for integers $a, b$ not both zero, $\gcd(a,b)$ is the least positive element of $\{\, ax + by : x, y \in \mathbb{Z} \,\}$ ; in particular $ax + by = \gcd(a,b)$ has an integer solution).

[L9]

$\operatorname{lcm}(a,b) \ge 0$ and is a common multiple of $a$ and $b$ (Common multiple, and the least common multiple $\operatorname{lcm}(a,b)$ , taken to be $0$ when $a = 0$ or $b = 0$ ).

[L10]

Every common multiple of $a$ and $b$ is a multiple of $\operatorname{lcm}(a,b)$ (Every common multiple of $a$ and $b$ is a multiple of $\operatorname{lcm}(a,b)$ , and $\gcd(a,b) \cdot \operatorname{lcm}(a,b) = |ab|$ ).

[L11]

The image of $\iota : \mathbb{N} \to \mathbb{Z}$ is exactly the set of nonnegative integers, so every $x \ge 0$ is $\iota(n)$ for a unique $n \in \mathbb{N}$ (The naturals embed in the integers).

Proof

technique · direct

1.1

$I := a\mathbb{Z} + b\mathbb{Z}$ is a subgroup of $(\mathbb{Z},+)$ : it contains $0 = a \cdot 0 + b \cdot 0$ , so is nonempty, and for $ax + by$ and $ax' + by'$ in $I$ the difference is $a(x - x') + b(y - y') \in I$ by distributivity, so the one-step test applies.

L1L3

1.2

$I \subseteq d\mathbb{Z}$ . Since $d$ is a common divisor, write $a = d a_1$ and $b = d b_1$ ; then $ax + by = d(a_1 x + b_1 y) \in d\mathbb{Z}$ for all $x, y$ .

L1L6L7

1.3

$d \in I$ . If $(a,b) \ne (0,0)$ this is [L8]. If $(a,b) = (0,0)$ then $d = 0 = a \cdot 0 + b \cdot 0 \in I$ . The two cases are exhaustive.

L1L7L8

1.4

$a\mathbb{Z} \cap b\mathbb{Z}$ is a subgroup by [L2] and [L5], and by [L6] it is exactly the set of common multiples of $a$ and $b$ : $m \in a\mathbb{Z}$ says $a \mid m$ , and $m \in b\mathbb{Z}$ says $b \mid m$ .

L2L5L6

2.1

$I = \langle \{a,b\} \rangle$ . It contains $a = a \cdot 1 + b \cdot 0$ and $b = a \cdot 0 + b \cdot 1$ , so $\langle \{a,b\} \rangle \subseteq I$ by [L4]. Conversely any subgroup $H$ containing $a$ and $b$ contains $\langle a \rangle = a\mathbb{Z}$ and $\langle b \rangle = b\mathbb{Z}$ by [L2] and [L4], hence contains $ax + by$ for all $x, y$ by closure under the operation; taking $H = \langle \{a,b\} \rangle$ gives $I \subseteq \langle \{a,b\} \rangle$ .

step 1.1L1L2L3L4

2.2

Hence $d\mathbb{Z} = \langle d \rangle \subseteq I$ , because $I$ is a subgroup containing $d$ and $\langle d \rangle$ is the smallest such. With step 1.2, $I = d\mathbb{Z}$ .

step 1.1step 1.2step 1.3L2L4

2.3

$\ell\mathbb{Z} \subseteq a\mathbb{Z} \cap b\mathbb{Z}$ : $\ell$ is a common multiple by [L9], so $\ell$ lies in the subgroup $a\mathbb{Z} \cap b\mathbb{Z}$ , and $\ell\mathbb{Z} = \langle \ell \rangle$ is the smallest subgroup containing $\ell$ .

step 1.4L2L4L9

3.1

$a\mathbb{Z} \cap b\mathbb{Z} \subseteq \ell\mathbb{Z}$ : every common multiple $m$ satisfies $\ell \mid m$ by [L10], that is $m \in \ell\mathbb{Z}$ . With step 2.3, $a\mathbb{Z} \cap b\mathbb{Z} = \ell\mathbb{Z}$ .

step 1.4step 2.3L6L10

4.1

Both identities are proved, and in group-theoretic form they read $\langle \{a,b\} \rangle = I = d\mathbb{Z} = \langle d \rangle$ by step 2.1 and step 2.2, and $\langle a \rangle \cap \langle b \rangle = a\mathbb{Z} \cap b\mathbb{Z} = \ell\mathbb{Z} = \langle \ell \rangle$ by step 3.1 and [L2].

step 2.1step 2.2step 3.1L2

5.1

Finally $d \ge 0$ and $\ell \ge 0$ , so each is $\iota(n)$ for a natural $n$ by [L11], and by [L2] a subgroup of $(\mathbb{Z},+)$ has exactly one such generator; hence $d$ and $\ell$ are the canonical generators of the two subgroups, and the identities identify the subgroups and not merely the underlying sets.

step 4.1L2L7L9L11∎

Remarks

This is the seam between the arithmetic and the group theory of this page, and it is the concrete shadow of the statement that $\mathbb{Z}$ is a principal ideal domain. Read from left to right, it says that the set of integer combinations of $a$ and $b$ is exactly the set of multiples of their greatest common divisor — which is Bézout's identity and the divisibility characterisation of $\gcd$ packed into one equation between subgroups.
The classification is used for one thing only. Both inclusions above come from Bézout and divisibility; Every subgroup of $(\mathbb{Z}, +)$ is $\langle n \rangle = n\mathbb{Z}$ for exactly one natural number $n$ is invoked in step 5.1, for the uniqueness of the nonnegative generator, and nowhere else.
Boundary pairs. At $(a,b) = (0,0)$ the theorem reads $\{0\} + \{0\} = 0\mathbb{Z} = \{0\}$ and $\{0\} \cap \{0\} = 0\mathbb{Z}$ , using $\gcd(0,0) = 0$ and $\operatorname{lcm}(0,0) = 0$ ; at $a = 0$ and $b \ne 0$ it reads $\{0\} + b\mathbb{Z} = |b|\mathbb{Z}$ and $\{0\} \cap b\mathbb{Z} = \{0\}$ . Both are true as stated, and neither needed a separate clause.

Depends on

Used by

12ℤ + 18ℤ = 6ℤ and 12ℤ ∩ 18ℤ = 36ℤ, the arithmetic of gcd and lcm read off the subgroups of (ℤ,+) Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 77 results over 25 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Bézout's identity (Wikipedia) (standard reference, not scraped)
Cyclic group (Wikipedia) (standard reference, not scraped)
Least common multiple (Wikipedia) (standard reference, not scraped)