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**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 $\langle\mathbb Z_n,+_n\rangle$.

**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

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)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

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