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Every cyclic group is isomorphic to $(\mathbb Z,+)$ or to $(\mathbb Z/n,+)$ for its finite order $n\ge1$

Statement

If $G=\langle g\rangle$ is cyclic, then exactly one of the following applies:

if $g$ has infinite order, $G\cong(\mathbb Z,+)$ ;
if $g$ has finite order $n$ , necessarily $n\ge1$ , then $G\cong(\mathbb Z/n,+)$ .

Facts & Assumptions

Given: A group $G$ and an element $g\in G$ with $G=\langle g\rangle$ .

[L1]

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

[L2]

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

[L3]

Integer powers satisfy $g^{r+s}=g^rg^s$ 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 $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$ , and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute, Group isomorphisms, automorphisms and the set $\operatorname{Aut}(G)$ ).

[L4]

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

Proof

technique · direct

1.1

Suppose first that $g$ has infinite order. The map $\phi:\mathbb Z\to G$ , $r\mapsto g^r$ , 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].

L1L2L3given

1.2

Suppose instead that $g$ has finite order $n$ . Then $n\ge1$ by the definition of element order, and $[r]\mapsto g^r$ defines a map $\bar\phi:\mathbb Z/n\to G$ because $[r]=[s]$ means $n\mid(r-s)$ , so [L2] gives $g^{r-s}=e$ and [L3] gives $g^r=g^{r-s}g^s=g^s$ .

L2L3L4given

2.1

Hence in the infinite-order case $\phi$ is an isomorphism $\mathbb Z\cong G$ .

step 1.1L3

2.2

The map $\bar\phi$ 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 $n\mid(r-s)$ , and is surjective by [L1].

step 1.2L1L2L3

3.1

Thus $\bar\phi$ is an isomorphism $(\mathbb Z/n,+)\cong G$ in the finite-order case.

step 2.2L3L4

4.1

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

step 2.1step 3.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 84 results over 25 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

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

Every cyclic group is isomorphic to (Z,+)(\mathbb Z,+)(Z,+) or to (Z/n,+)(\mathbb Z/n,+)(Z/n,+) for its finite order n≥1n\ge1n≥1