# 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 $g^{n} = \begin{cases} \underbrace{g \ast g \cdots \ast g}_{n} , & n \in \mathbb{Z}, n \ge 0\\ \underbrace{g^{-1} \ast g^{-1} \cdots \ast g^{-1}}_{-n}, & n \in \mathbb{Z}, n < 0 \end{cases}$

• Induction shows: $g^{m + n} = g^{m} \ast g^{n} \text{ and } g^{mn} = [g^{m}]^{n}$

## a Cyclic Group of Order n is Isomorphic to the Integers Moduluo n with Addition

### 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{align} f\colon C_m &\to \mathbb{Z}/m \\ g^{i} &\mapsto i ~\text{mod}~m \end{align}

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