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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-03
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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 HH of a cyclic group G=⟨g⟩G=\langle g\rangle is cyclic. If H≠{e}H\ne\{e\}, then the least positive integer dd for which gd∈Hg^d\in H satisfies H=⟨gd⟩H=\langle g^d\rangle.

Facts & Assumptions

Given: A cyclic group G=⟨g⟩G=\langle g\rangle and a subgroup H≤GH\le G.

[L1]

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

[L2]

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

[L3]

Integer powers satisfy gr+s=grgsg^{r+s}=g^rg^s, (gr)−1=g−r(g^r)^{-1}=g^{-r}, and (gd)q=gdq(g^d)^q=g^{dq} (Exponent laws in a group: gm+n=gmgng^{m+n} = g^{m}g^{n} and (gm)n=gmn(g^{m})^{n} = g^{mn} for all m,n∈Zm, n \in \mathbb{Z}, and (gh)n=gnhn(gh)^{n} = g^{n}h^{n} when gg and hh commute).

[L4]

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

[L5]

For d>0d>0 and k∈Zk\in\mathbb Z, there are q,r∈Zq,r\in\mathbb Z with k=qd+rk=qd+r and 0≤r<d0\le r<d (Division with remainder in Z\mathbb{Z}: for a∈Za \in \mathbb{Z} and b>0b > 0 there are unique q,r∈Zq, r \in \mathbb{Z} with a=qb+ra = qb + r and 0≤r<b0 \le 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}H=\{e\}, then H=⟨e⟩H=\langle e\rangle and is cyclic. Assume henceforth that H≠{e}H\ne\{e\}.

L1L2given
1.2

The set S:={n∈N:n≥1 and gn∈H}S:=\{n\in\mathbb N:n\ge1\text{ and }g^n\in H\} is nonempty: choose h∈H∖{e}h\in H\setminus\{e\} and write h=gkh=g^k by [L2]. Since g0=eg^0=e, k≠0k\ne0; by [L6], either k>0k>0 or −k>0-k>0. In the first case use hh, and in the second use h−1=g−kh^{-1}=g^{-k}.

L1L2L3L6given
2.1

Let dd be the least element of SS, supplied by [L4]. Then gd∈Hg^d\in H, so ⟨gd⟩⊆H\langle g^d\rangle\subseteq H by the smallest-subgroup property in [L2].

step 1.2L2L4
3.1

For h∈Hh\in H, write h=gkh=g^k by [L2] and write k=qd+rk=qd+r with 0≤r<d0\le r<d by [L5]. Then gr=h(gd)−q∈Hg^r=h(g^d)^{-q}\in H.

step 2.1L1L2L3L5
4.1

If r>0r>0 in step 3.1, then r∈Sr\in S contradicts the minimality of dd; hence r=0r=0, so h=(gd)q∈⟨gd⟩h=(g^d)^q\in\langle g^d\rangle.

step 3.1L2L4
5.1

Thus H⊆⟨gd⟩H\subseteq\langle g^d\rangle in the nontrivial case, and step 2.1 gives equality; together with step 1.1 this proves every subgroup of GG is cyclic.

step 1.1step 2.1step 4.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 58 results over 20 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.

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