Theorem 6.10

Return to Theorems, Glossary, Homework Problems.
Statement:

Let $G$ be a cyclic group with generator $a$. If the order of $G$ is infinite,then $G$ is isomorphic to $\langle\mathbb Z,+\rangle$. If $G$ has finite order $n$,then $G$ is isomorphic to $\langle\mathbb Z_n,+_n\rangle$.

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Proof:

Case I
For all positive integers $m$,$a^m\neq e$.In this case we claim that no two distinct exponents $h$ and $k$ can give equal elements $a^h$ and $a^k$ of $G$.
Suppose that $a^h=a^k$ and say $h>k$.Then

(1)

\begin{equation} a^ha^{-k}=a^{h-k}=e, \end{equation}

contrary to our Case I assumption.Hence every element of $G$ can be expressed as $a^m$ for a unique $m \in Z$.The map $\phi:G\rightarrow \mathbb Z$ given by $\phi(a^i)=i$ is thus well defined,one to one,and onto $\mathbb Z$.Also,

(2)

\begin{align} \phi(a^i a^j)=\phi(a^{i+j})=i+j=\phi(a^i)+\phi(a^j), \end{align}

so the homomorphism property is satisfied and $\phi$ is an isomorphism.

Case II
$a^m=e$ for some positive integer m.Let $n$ be the smallest positive integer such that $a^n=e$.If $s\in \mathbb Z$ and $s=nq+r$ for 0$\leqslant r < n$,then $a^s=a^{nq+r}=(a^n)^q a^r=e^q a^r=a^r$.As in Case 1,if 0$<k<h<n$ and $a^h=a^k$,then $a^{h-k}=e$ and 0$<k<h<n$,contradicting our choice of $n$.Thus the elements

(3)

\begin{equation} a^0=e,a,a^2,a^3,…,a^{n-1} \end{equation}

are all distinct and comprise all elements of $G$.The map $\psi:G\rightarrow \mathbb Z_n$ given by $\psi(a^i)=i$ for $i=0,1,2,...,n-1$ is thus well defined,one to one,and onto $\mathbb Z_n$.Because $a^n=e$,we see that $a^i a^j=a^k$ where $k=i+_n j$.Thus

(4)

\begin{align} \psi(a^i a^j)=i+_nj=\psi(a^i)+_n\psi(a^j), \end{align}

so the homomorphism property is satisfied and $\psi$ is an isomorphism.