Proof: The orbit of any element of a group is a subgroupProofThatGInGImpliesThatLangleGRangleLeG2003-03-12 02:59:362003-06-04 23:56:22Proofcyclic group0orbitgroupsubgroup\usepackage{graphicx}
%\usepackage{xypic}
\usepackage{bbm}
\newcommand{\Z}{\mathbbmss{Z}}
\newcommand{\C}{\mathbbmss{C}}
\newcommand{\R}{\mathbbmss{R}}
\newcommand{\Q}{\mathbbmss{Q}}
\newcommand{\mathbb}[1]{\mathbbmss{#1}}
\newcommand{\figura}[1]{\begin{center}\includegraphics{#1}\end{center}}
\newcommand{\figuraex}[2]{\begin{center}\includegraphics[#2]{#1}\end{center}}Following is a proof that, if $G$ is a group and $g \in G$, then $\langle g \rangle \le G$. Here $\langle g \rangle$ is the orbit of $g$ and is defined as
$$\langle g \rangle=\{g^n : n\in\Z\}$$
Since $g \in \langle g \rangle$, then $\langle g \rangle$ is nonempty.
Let $a,b \in \langle g \rangle$. Then there exist $x,y \in {\mathbb Z}$ such that $a=g^x$ and $b=g^y$. Since $ab^{-1}=g^x(g^y)^{-1}=g^xg^{-y}=g^{x-y} \in \langle g \rangle$, it follows that $\langle g \rangle \le G$.