TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03

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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.
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Every subgroup of a cyclic group is cyclic; the least positive exponent in a nontrivial subgroup supplies a generator

Statement

Every subgroup $H$ of a cyclic group $G = ⟨ g ⟩$ is cyclic. If $H \neq {e}$ , then the least positive integer $d$ for which $g^{d} \in H$ satisfies $H = ⟨ g^{d} ⟩$ .

Facts & Assumptions

Given: A cyclic group $G = ⟨ g ⟩$ and a subgroup $H \leq G$ .

[L1]

A subgroup contains the identity and is closed under products and inverses (Subgroup).

[L2]

The elements of $⟨ g ⟩$ are exactly the integer powers of $g$ , and a subgroup containing one element contains the subgroup generated by it ( $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian, The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups).

[L3]

Integer powers satisfy $g^{r + s} = g^{r} g^{s}$ , $(g^{r})^{- 1} = g^{- r}$ , and $(g^{d})^{q} = g^{d q}$ (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).

[L4]

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

[L5]

For $d > 0$ and $k \in Z$ , there are $q, r \in Z$ with $k = q d + r$ and $0 \leq r < d$ (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]

A nonzero integer is positive or has positive negative (The integers form a totally ordered ring).

Proof

technique · direct

1.1

If $H = {e}$ , then $H = ⟨ e ⟩$ and is cyclic. Assume henceforth that $H \neq {e}$ .

L1L2given

1.2

The set $S : = {n \in N : n \geq 1 and g^{n} \in H}$ is nonempty: choose $h \in H ∖ {e}$ and write $h = g^{k}$ by [L2]. Since $g^{0} = e$ , $k \neq 0$ ; by [L6], either $k > 0$ or $- k > 0$ . In the first case use $h$ , and in the second use $h^{- 1} = g^{- k}$ .

L1L2L3L6given

2.1

Let $d$ be the least element of $S$ , supplied by [L4]. Then $g^{d} \in H$ , so $⟨ g^{d} ⟩ \subseteq H$ by the smallest-subgroup property in [L2].

step 1.2L2L4

3.1

For $h \in H$ , write $h = g^{k}$ by [L2] and write $k = q d + r$ with $0 \leq r < d$ by [L5]. Then $g^{r} = h (g^{d})^{- q} \in H$ .

step 2.1L1L2L3L5

4.1

If $r > 0$ in step 3.1, then $r \in S$ contradicts the minimality of $d$ ; hence $r = 0$ , so $h = (g^{d})^{q} \in ⟨ g^{d} ⟩$ .

step 3.1L2L4

5.1

Thus $H \subseteq ⟨ g^{d} ⟩$ in the nontrivial case, and step 2.1 gives equality; together with step 1.1 this proves every subgroup of $G$ is cyclic.

step 1.1step 2.1step 4.1∎

Depends on

Used by

Dependency tree · two levels

40 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

Romyar Sharifi, Abstract Algebra (standard reference, not scraped)