# Abstract Algebra/Group Theory/Group/a Cyclic Group of Order n is Isomorphic to Integer Moduluo n with Addition

## Theorem

Let Cm be a cyclic group of order m generated by g with ${\displaystyle \ast }$

Let ${\displaystyle (\mathbb {Z} /m,+)}$ be the group of integers modulo m with addition

Cm is isomorphic to ${\displaystyle (\mathbb {Z} /m,+)}$

## Lemma

Let n be the minimal positive integer such that gn = e

${\displaystyle g^{i}=g^{j}\leftrightarrow i=j~{\text{mod}}~n}$
Proof of Lemma
Let i > j. Let i - j = sn + r where 0 ≤ r < n and s,r,n are all integers.
 1. ${\displaystyle g^{i}=g^{j}\;}$ 2. ${\displaystyle e=g^{i-j}=g^{sn+r}=[g^{n}]^{s}\ast g^{r}=[e]^{s}\ast g^{r}=g^{r}}$ as i - j = sn + r, and gn = e 3. ${\displaystyle g^{r}=e}$ 4. ${\displaystyle r=0}$ as n is the minimal positive integer such that gn = e and 0 ≤ r < n 5. ${\displaystyle i-j=sn}$ 0. and 7. 6. ${\displaystyle i=j~{\text{mod}}~n}$

## Proof

0. Define   {\displaystyle {\begin{aligned}f\colon C_{m}&\to \mathbb {Z} /m\\g^{i}&\mapsto i~{\text{mod}}~m\end{aligned}}}
Lemma shows f is well defined (only has one output for each input).
f is homomorphism:
${\displaystyle f(g^{i})+f(g^{j})=i+j~{\text{mod}}~m=f(g^{i+j})=f(g^{i}\ast g^{j})}$
f is injective by lemma
f is surjective as both ${\displaystyle \mathbb {Z} /m}$ and ${\displaystyle C_{m}}$ have m elements and f is injective