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Cyclic groups


\begin{thm}
Let $G$\ be a group, and let $a\in G$\ be an arbitrary element.
The ...
...rs of $a$is a subgroup of $G$. Its order is equal to the order of $a$.
\end{thm}


\begin{proof}
% latex2html id marker 2041For $a^i,a^j\in H$\ we clearly have $...
... $a$\ are $e=a^0$, $a$, \ldots, $a^{n-1}$, so that $\vert H\vert=n$.
\end{proof}


\begin{de}
The subgroup $H$\ from the above theorem is called the
{\em cyclic su...
...$G=H$\ then $G$\ is said to be a {\em cyclic group}
(generated by $a$).
\end{de}


\begin{ex}
% latex2html id marker 2054The additive group $\mathbb{Z}$\ is cycl...
...mathbb{Z}}_n$\ generates
${\mathbb{Z}}_n$\ if and only if $(a,n) = 1$.
\end{ex}


\begin{ex}
Consider the group ${\mathbb{Z}}_7\setminus\{0\}$\ under multiplicati...
... fact in number theory.
It is far from obvious what its generators are.
\end{ex}


\begin{thm}
Every cyclic group is abelian.
\end{thm}


\begin{proof}
Suppose $G = \langle a \rangle$. Then if $x,y \in G$, $x=a^i, y=a^...
...me $i,j$\ so $xy=a^ia^j=a^{i+j} = a^ja^i = yx$\ and $G$\ is abelian.
\end{proof}


\begin{ex}
$\mathbb{Q}$\ is not cyclic. For if $q \in \mathbb{Q}$\ and $H$\ is t...
...H$, $\vert x\vert \geq \vert q\vert$. Thus $\vert q/2\vert \not \in H$.
\end{ex}


\begin{thm}
Every subgroup of a cyclic group is cyclic.
\end{thm}


\begin{proof}
Let $G$\ be a cyclic group generated by $a$,
and let $H\leq G$. If...
...$, it follows that $r=0$, and hence
$a^i=(a^m)^q$, a power of $a^m$.
\end{proof}


\begin{ex}
Let us list all subgroups of the group ${\mathbb{Z}}_{12}$:
\begin{di...
...{ 0,2,4,6,8,10\},\
\{0,3,6,9\},\ \{0,4,8\},\ \{0,6\}.
\end{displaymath}\end{ex}



Edmund F Robertson

11 September 2006