# Tour:Equivalence of definitions of cyclic group

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WHAT YOU NEED TO DO:
• Read, and thoroughly understand, the equivalence of the two definitions of cyclic group. (We'll see better formulations of the proof later, after introducing homomorphisms and quotients)

## 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.

## 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 $g^{a+b} = g^ag^b$ by definition. If the sum of $a$ and $b$ as integers is at least $n$, then $\varphi(a + b) = g^{a+b-n} =g^ag^bg^{-n} = g^ag^b$ (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.