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

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$\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$ , 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

$\langle g \rangle \;=\; \{\, g^{n} \;:\; n \in \mathbb{Z} \,\} ,$

the cyclic subgroup generated by $g$ (The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$ , 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 $P := \{\, g^{n} : n \in \mathbb{Z} \,\}$ of its integer powers (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L1]

$\langle g \rangle$ 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 $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$ , 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 $m, n \in \mathbb{Z}$ (Exponent laws in a group: $g^{m+n} = g^{m}g^{n}$ and $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$ , and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute).

[L3]

$g^{0} = e$ , $g^{1} = g$ , and $g^{\sigma(k)} = g^{k} g$ for $k \in \mathbb{N}$ ; a natural number $k$ in an exponent means the integer $\iota(k)$ (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 $gh^{-1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ , Subgroup).

[L5]

$\mathbb{Z}$ 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 $x \ge 0$ is $\iota(k)$ for a unique natural $k$ while every $x < 0$ has $-x = \iota(k)$ 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 $\mathbb{N}$ (The principle of mathematical induction).

Proof

technique · direct

1.1

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

L3given

1.2

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.

L2L5

1.3

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

1.4

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 $\sigma$ , because $g^{\sigma(k)} = g^{k} g$ is a product of two elements of $H$ ; induction finishes it.

L3L4L6given

2.1

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 $\langle g \rangle \subseteq P$ , since $\langle g \rangle$ is contained in every subgroup containing $g$ .

step 1.1step 1.2step 1.3L1L4

2.2

Let $H$ be any subgroup of $G$ with $g \in H$ , and let $x \in \mathbb{Z}$ . If $x \ge 0$ then $x = \iota(k)$ and $g^{x} = g^{k} \in H$ by step 1.4. If $x < 0$ then $-x = \iota(k)$ , 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 $P \subseteq H$ .

step 1.4L3L4L5

3.1

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

step 2.1step 2.2L1

4.1

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

step 3.1L2

5.1

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

step 3.1step 4.1L1∎

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 $\langle g \rangle$ is finite, or that distinct exponents give distinct powers. Both questions are settled by If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ 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 $\operatorname{ord}(g)$ when that is finite and are pairwise distinct when it is not.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 58 results over 18 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

Cyclic group (Wikipedia) (standard reference, not scraped)
Generating set of a group (Wikipedia) (standard reference, not scraped)

⟨g⟩={ gn:n∈Z }\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}⟨g⟩={gn:n∈Z}, and every cyclic group is abelian