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

• 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 C_m 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

C_m 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 $C_{m}$ have m elements and f is injective