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

• A Template:Green generated by g is

• $⟨g⟩=\{g^n|n\in \mathbb{Z}\}$

• where $g^n=\{\begin{array}{cc}\underset{n}{\underbrace{g\ast g\cdots \ast g}},& n\in \mathbb{Z},n\ge 0\\ \underset{-n}{\underbrace{g^{-1}\ast g^{-1}\cdots \ast g^{-1}}},& n\in \mathbb{Z},n<0\end{array}$

• 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

$g^i=g^j\leftrightarrow i=j\text{mod}n$

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. $i=j\text{mod}n$

    Template:Hidden end

Proof

0. Define   $\begin{array}{cc}f:C_m& \to \mathbb{Z}/m\\ g^i& \mapsto i\text{mod}m\end{array}$

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