LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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$⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian

Statement

Let $G$ be a group and $g \in G$ , with integer powers as in Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ . Then

$⟨ g ⟩ = {g^{n} : n \in Z},$

the cyclic subgroup generated by $g$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups) being exactly the set of integer powers of $g$ . Consequently every cyclic group is abelian, and so is every cyclic subgroup of any group.

Facts & Assumptions

Given: A group $G$ with identity $e$ , an element $g \in G$ , and the set $P : = {g^{n} : n \in Z}$ of its integer powers (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L1]

$⟨ g ⟩$ is the smallest subgroup of $G$ containing $g$ : it is a subgroup, it contains $g$ , and it is contained in every subgroup containing $g$ (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L2]

Exponent laws: $g^{m + n} = g^{m} g^{n}$ , $g^{- m} = (g^{m})^{- 1}$ , and $g^{m} g^{n} = g^{n} g^{m}$ , for all $m, n \in Z$ (Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

[L3]

$g^{0} = e$ , $g^{1} = g$ , and $g^{σ (k)} = g^{k} g$ for $k \in N$ ; a natural number $k$ in an exponent means the integer $ι (k)$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , The naturals embed in the integers).

[L4]

One-step test: a nonempty $S \subseteq G$ with $x y^{- 1} \in S$ for all $x, y \in S$ 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).

[L5]

$Z$ is a commutative ring, so $m + (- n)$ is again an integer and addition is commutative (The integers form a commutative ring, The integers as equivalence classes of pairs of naturals); its order is total, and every $x \geq 0$ is $ι (k)$ for a unique natural $k$ while every $x < 0$ has $- x = ι (k)$ for a unique natural $k$ (The integers form a totally ordered ring, Order on the integers, The naturals embed in the integers).

[L6]

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

Proof

technique · direct

1.1L3given

$P$ is nonempty, since $g^{0} = e \in P$ , and $P \subseteq G$ by construction.

1.2L2L5

For $x = g^{m}$ and $y = g^{n}$ in $P$ : $x y^{- 1} = g^{m} (g^{n})^{- 1} = g^{m} g^{- n} = g^{m + (- n)} \in P$ , since $m + (- n)$ is an integer.

1.3L3

$g = g^{1} \in P$ .

1.4L3L4L6given

Let $H$ be any subgroup of $G$ with $g \in H$ . Then $g^{k} \in H$ for every natural $k$ : the set of such $k$ contains $0$ , because $g^{0} = e \in H$ , and it is closed under $σ$ , because $g^{σ (k)} = g^{k} g$ is a product of two elements of $H$ ; induction finishes it.

2.1step 1.1step 1.2step 1.3L1L4

By steps 1.1 and 1.2 and the one-step test, $P$ is a subgroup of $G$ , and by step 1.3 it contains $g$ ; hence $⟨ g ⟩ \subseteq P$ , since $⟨ g ⟩$ is contained in every subgroup containing $g$ .

2.2step 1.4L3L4L5

Let $H$ be any subgroup of $G$ with $g \in H$ , and let $x \in Z$ . If $x \geq 0$ then $x = ι (k)$ and $g^{x} = g^{k} \in H$ by step 1.4. If $x < 0$ then $- x = ι (k)$ , so $g^{x} = (g^{k})^{- 1}$ , which lies in $H$ because $g^{k} \in H$ by step 1.4 and $H$ is closed under inverses. So $P \subseteq H$ .

3.1step 2.1step 2.2L1

Taking $H = ⟨ g ⟩$ in step 2.2, which is legitimate because $⟨ g ⟩$ is a subgroup containing $g$ , gives $P \subseteq ⟨ g ⟩$ ; with step 2.1 this proves $⟨ g ⟩ = P$ .

4.1step 3.1L2

Any two elements of $P$ commute: $g^{m} g^{n} = g^{n} g^{m}$ for all integers $m, n$ . Hence $⟨ g ⟩$ is abelian, by step 3.1.

5.1step 3.1step 4.1L1∎

If $G$ is cyclic, say $G = ⟨ g ⟩$ , then $G$ is abelian by step 4.1; and every cyclic subgroup of any group is abelian for the same reason.

Remarks

The two inclusions are proved by different means, and both are needed: one shows the set of powers is a subgroup containing $g$ , the other shows every subgroup containing $g$ swallows all the powers. Only together do they identify the smallest one.
Step 1.4 is where induction does the work, and it is stated for natural exponents only; negative exponents are then handled in step 2.2 by closure under inverses, using $g^{x} = (g^{k})^{- 1}$ from the definition of integer powers rather than a second induction.
Nothing here says $⟨ g ⟩$ is finite, or that distinct exponents give distinct powers. Both questions are settled by 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 $j = k$ , which shows the powers repeat with period $ord (g)$ when that is finite and are pairwise distinct when it is not.

$⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

⟨g⟩={ gn:n∈Z }⟨g⟩={gn:n∈Z}, and every cyclic group is abelian

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

$⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian