# Abstract Algebra/Group Theory/Cyclic groups/Definition of a Cyclic Group

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• A cyclic group generated by g is

• $\langle g\rangle =\lbrace g^{n}\;|\;n\in \mathbb {Z} \rbrace$ • where $g^{n}={\begin{cases}\underbrace {g\ast g\cdots \ast g} _{n},&n\in \mathbb {Z} ,n\geq 0\\\underbrace {g^{-1}\ast g^{-1}\cdots \ast g^{-1}} _{-n},&n\in \mathbb {Z} ,n<0\end{cases}}$ • Induction shows: $g^{m+n}=g^{m}\ast g^{n}{\text{ and }}g^{mn}=[g^{m}]^{n}$ ## Theorem

Let Cm be a cyclic group of order m generated by g with $\ast$ Let $(\mathbb {Z} /m,+)$ be the group of integers modulo m with addition

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

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

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

0. Define   {\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:
$f(g^{i})+f(g^{j})=i+j~{\text{mod}}~m=f(g^{i+j})=f(g^{i}\ast g^{j})$ f is injective by lemma
f is surjective as both $\mathbb {Z} /m$ and $C_{m}$ has m elements and m is injective