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 \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[edit]

Theorem[edit]

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[edit]

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[edit]

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