Equivalence of definitions of cyclic group

This article gives a proof/explanation of the equivalence of multiple definitions for the term cyclic group
View a complete list of pages giving proofs of equivalence of definitions

The definitions that we have to prove as equivalent

Definition in terms of modular arithmetic

A group is said to be cyclic (sometimes, monogenic or monogenous) if it is either isomorphic to the group of integers or to the group of integers modulo n for some positive integer $n$.

Definition in terms of generating sets

A group is termed cyclic (sometimes, monogenic or monogenous) if it has a generating set of size 1.

Definition as a quotient

A group is termed cyclic if it is a quotient of the group $\mathbb {Z}$, in other words, there exists a surjective homomorphism from $\mathbb {Z}$ to the group.

Proof of equivalence

From modular arithmetic to generating sets

This is direct: $\mathbb {Z}$ is generated by the element $1\in \mathbb {Z}$, and $\mathbb {Z} /n\mathbb {Z}$ is generated by the element 1.

From generating sets to modular arithmetic

Given: A group $G$ with a generating set $\{g\}$

To prove: $G$ is isomorphic either to $\mathbb {Z}$ (the group of integers) or to $\mathbb {Z} /n\mathbb {Z}$ (the group of integers modulo n)

Proof: We consider two cases.

Case 1: $g$ has finite order. Thus, there exists a minimal positive integer $n$ such that $g^{n}$ is the identity element. Consider now the map $\varphi :\mathbb {Z} /n\mathbb {Z} \to G$ that sends $a$ to the element $g^{a}$. We want to prove that $\varphi$ is an isomorphism.

We first show that $\varphi (a+b)=\varphi (a)\varphi (b)$. For this, observe that if $a$ and $b$ add up to less than $n$ as integers, then $g^{a+b}=g^{a}g^{b}$ by definition. If the sum of $a$ and $b$ as integers is at least $n$, then $\varphi (a+b)=g^{a+b-n}=g^{a}g^{b}g^{-n}=g^{a}g^{b}$ (since $g^{-n}$ is the identity element).

Similarly, $\varphi (0)=g^{0}=e$ by definition, and $\varphi (-a)=\varphi (a)^{-1}$, again because $g^{n}=e$.

Surjectivity: Since $g$ generates $G$, every element of $G$ can be written as a power of $g$, say $g^{m}$ for some integer $m$. Writing $m=nq+r$ where $q,r$ are integers and $r\in \{0,1,2,\dots ,n-1\}$, we get that $g^{m}=g^{r}=\varphi (r)$. Thus, $\varphi$ is surjective.

Injectivity: Finally, if $\varphi (a)=\varphi (b)$ with $a<b$ both in $\{0,1,2,\dots ,n-1\}$, then $g^{b-a}=e$, contradicting the assumption that $g$ has order $n$.

Thus, $\varphi$ is an isomorphism of groups.

Case 2: $g$ does not have finite order. In that case, consider the map $\varphi :\mathbb {Z} \to G$ that sends $n$ to $g^{n}$.

Clearly, by definition, $\varphi (a+b)=\varphi (a)\varphi (b)$, $\varphi (-a)=\varphi (a)^{-1}$, and $\varphi (0)=e$.

Surjectivity: Since $g$ generates $G$, every element of $G$ can be written as $g^{n}$ for some integer $n$.

Injectivity: If $\varphi (a)=\varphi (b)$ for $a<b$, then $g^{b-a}=e$, contradicting the assumption that $g$ does not have finite order.