Cyclic Groups

Cyclic Group

A group $G$ is cyclic if $G=⟨a⟩=\left\{\dots ,a^{-3},a^{-2},a^{-1},e,a,a^2,a^3,\dots \right\}$ that is, every $g\in G$ has the form $g=a^n$ for some $n\in \mathbb{Z}$. We say that $G$ is generated by $a$.

e.g. $\mathbb{Z}$ is cyclic, generated by either 1 or −1.

Prop Let $G=⟨a⟩$ be a cyclic group. Then exactly one of the following holds:

1. $G$ is infinite
2. There exists $n\in \mathbb{N}$ such that $G=\{e,a,a^2,...,a^{n-1}\}$. 3

Proof If $G$ is finite, To prove exists $n\in \mathbb{N}$ such that $G=\{e,a,a^2,...,a^{n-1}\}$ is equivalent to prove that for all $k\in \mathbb{Z}$, exists $n<k$ and $r<n$ such that $a^k=a^r$. By the pigeonhole principle, for any integer $l$ exists $m$ that $m<l$ we have $a^l=a^m$. Hence $a^{l-m}=e$, so $a$ has a finite order $o(a)$. Let $n=o(a)$. Then for every $k\in \mathbb{Z}$, $k=qn+r$ with $q\in \mathbb{Z}$ and $0\le r<n$. It follows that $a^k=a^{nq+r}={\left(a^n\right)}^q{a}^r=e^q{a}^r=a^r$

Proposition

If a group is cyclic, then it is abelian.

Prop Suppose $G$ is a group. If there exists $g\in G$ such that $o(g)=|G|$ then $G$ is cyclic.

Prop Any group with prime order is cyclic. Proof See proof.

Definition

A product of two finite cyclic groups with coprime orders is cyclic. i.e.

The Structure Theorem for Finite Abelian Groups

Every finite abelian group $G$ is isomorphic to a direct product (sum) of cyclic groups: $G\cong C_1\times C_2\times \cdots \times C_n,$such that $|C_i|$ divides $|C_{i+1}|$ for all $i\ge 1$, and $|G|=|C_1|\cdots |C_n|$.

e.g.

Proposition

Let $C_n$ be a cyclic group of order $n$. For each $m\mid n$, there exists a unique subgroup $H$ of $C_n$ of order $m$, and it is cyclic. Furthermore, the subgroups of $C_n$ are precisely the cyclic groups of order dividing $n$.