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 $\phi :\mathbb{Z}/n\mathbb{Z}\to G$ that sends $a$ to the element $g^a$. We want to prove that $\phi$ is an isomorphism.

We first show that $\phi (a+b)=\phi (a)\phi (b)$. 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 $\phi (a+b)=g^{a+b-n}=g^a{g}^b{g}^{-n}=g^a{g}^b$ (since $g^{-n}$ is the identity element).

Similarly, $\phi (0)=g^0=e$ by definition, and $\phi (-a)=\phi (a{)}^{-1}$, 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 $g^m=g^r=\phi (r)$. Thus, $\phi$ is surjective.

Injectivity: Finally, if $\phi (a)=\phi (b)$ 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, $\phi$ is an isomorphism of groups.

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

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

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

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