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

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Every cyclic group is isomorphic to $(Z, +)$ or to $(Z / n, +)$ for its finite order $n \geq 1$

Statement

If $G = ⟨ g ⟩$ is cyclic, then exactly one of the following applies:

if $g$ has infinite order, $G ≅ (Z, +)$ ;
if $g$ has finite order $n$ , necessarily $n \geq 1$ , then $G ≅ (Z / n, +)$ .

Facts & Assumptions

Given: A group $G$ and an element $g \in G$ with $G = ⟨ g ⟩$ .

[L1]

A cyclic subgroup is precisely the set of all integer powers of its generator ( $⟨ g ⟩ = {g^{n} : n \in Z}$ , and every cyclic group is abelian).

[L2]

For finite order $n$ , $g^{k} = e$ exactly when $n ∣ k$ , while for infinite order no nonzero integer power of $g$ is the identity (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 $j = k$ ).

[L3]

Integer powers satisfy $g^{r + s} = g^{r} g^{s}$ and $g^{- s} = (g^{s})^{- 1}$ ; a bijective group homomorphism is a group isomorphism (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, Group isomorphisms, automorphisms and the set $Aut (G)$ ).

[L4]

The additive quotient group $(Z, +) / n Z$ is the group $(Z / n, +)$ (For every $n \in N$ , the congruence-class group $(Z / n, +)$ is the quotient group $(Z, +) / n Z$ ).

Proof

technique · direct

1.1L1L2L3given

Suppose first that $g$ has infinite order. The map $ϕ : Z \to G$ , $r \mapsto g^{r}$ , is a homomorphism by the power law in [L3]. If $g^{r} = g^{s}$ , then $g^{r - s} = g^{r} (g^{s})^{- 1} = e$ by [L3], so [L2] forces $r = s$ ; it is surjective by [L1].

1.2L2L3L4given

Suppose instead that $g$ has finite order $n$ . Then $n \geq 1$ by the definition of element order, and $[r] \mapsto g^{r}$ defines a map $\overset{ˉ}{ϕ} : Z / n \to G$ because $[r] = [s]$ means $n ∣ (r - s)$ , so [L2] gives $g^{r - s} = e$ and [L3] gives $g^{r} = g^{r - s} g^{s} = g^{s}$ .

2.1step 1.1L3

Hence in the infinite-order case $ϕ$ is an isomorphism $Z ≅ G$ .

2.2step 1.2L1L2L3

The map $\overset{ˉ}{ϕ}$ is a homomorphism by the power law, is injective because $g^{r} = g^{s}$ gives $g^{r - s} = g^{r} (g^{s})^{- 1} = e$ by [L3] and then [L2] gives $n ∣ (r - s)$ , and is surjective by [L1].

3.1step 2.2L3L4

Thus $\overset{ˉ}{ϕ}$ is an isomorphism $(Z / n, +) ≅ G$ in the finite-order case.

4.1step 2.1step 3.1∎

Steps 2.1 and 3.1 give the asserted classification according as the order of $g$ is infinite or finite.

Depends on

Dependency tree · 2 levels

Sources

Romyar Sharifi, Abstract Algebra

Every cyclic group is isomorphic to (Z,+)(Z,+) or to (Z/n,+)(Z/n,+) for its finite order n≥1n≥1