Cyclic Groups

Subjects: Group Theory
Links: Groups, Integers modulo n, Subgroups, Group Homomorphisms and Isomorphisms, Integer Numbers

Def: We define the powers $x^n$ of $x$ (for $n\in \mathbb{Z}$) as follows:

• $x^0=e$
• $x^{n+1}=x^{n}x$ for $n\in \mathbb{Z}$
• $x^{n-1}=x^n{x}^{-1}$

Prop: Let $G$ be a group and let $x\in G$. Let $m,n\in \mathbb{Z}$. Then:

• $x^m{x}^n=x^{m+n}$
• $(x^n{)}^{-1}=x^{-n}$
• $(x^m{)}^n=x^{mn}$

Def: For $G$ a group and $x\in G$ define the order of $x$ to be the smallest positive integer $n$ such that $x^n=1$, and denote this integer by $o(x)$. In the case $x$ is said to be of order $n$. If no positive power of $x$ is the identity, the order of $x$ is defined to te $o(x)=\mathrm{\infty }$, or infinite order.

Def: For $x\in G$, then we consider the subgroup generated by $x$ $⟨x⟩:=\{x^n\in G\mid n\in \mathbb{Z}\}$. A group $G$ is called cyclic if there is an element $x\in G$ such that $G=⟨x⟩$; then $x$ is then called a generator for $G$.

Th: If $G$ is cyclic, then $G$ is abelian

Prop: If $G$ is cyclic, then every subgroup of $G$ is cyclic

Prop: Let $G$ be a group and $x\in G$.

• $o(x)=o(x^{-1})$
• if $o(x)=n$ and $x^m=e$, then $n\mid m$
• if $o(x)=n$ and $(m,n)=d$, then $o(x^m)=n/d$
• Let $y\in G$, then $o(xy)\mid o(x)o(y)$

Cor: If $G=⟨x⟩$, then $G=⟨x^{-1}⟩$.

Th: Let $G=⟨x⟩$. If $o(x)=\mathrm{\infty }$, then $x^j\ne x^k$, for $j\ne k$, and as a consequence $G$ is infinite. If $o(x)=n$, then $j\equiv k(\mathrm{mod}n)$, and as a consequence the distinct elements of $G$ are $e,x,x^2,\dots ,x^{n-1}$.

Def: The order of a group $G$, defined by $|G|$ is the cardinality of the set

Cor: If $G=⟨x⟩$, then $|G|=o(x)$.

Th: If $G$ is an infinite subgroup generated by $g$, then every power of $g$ is distinct.

Th: If $G$ is a finite group of order $n$, the following statements are equivalent:

• $G$ is cyclic
• For every divisor $d$ of $n$, $G$ has at most one subgroup of order $d$

Th: If $G$ is a finite cyclic group of order $n$, then $G$ is isomorphic to ${\mathbb{Z}}_n$. If $G$ is a cyclic and of infinite order, then $G$ is isomorphic to $\mathbb{Z}$.

Def: Since the cyclic groups are unique up to isomorphism, we can just denoted $C(n)$ or $C_n$ as the cyclic of order $n$.

Th: Every subgroup of $\mathbb{Z}$ are of the form $n\mathbb{Z}$ for $n\in \mathbb{Z}$.