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\begin{document}

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\begin{minipage}{118mm}{\small
{\bf R\'esum\'e.} Dans cet article, nous \'etudions des racines primitives.\\
{\bf Abstract.} In this article, we study primitive roots.\\
{\bf Keywords.} Your keywords come here.\\
{\bf Mathematics Subject Classification (2010).} Your MSC numbers come here.
}\end{minipage}

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\section{Introduction}

In this paper, we prove that, for any prime $p\in\mathbb{N}$, the group of units in the quotient ring $\mathbb{Z}/(p)$ is cyclic. This is important in many aspects of number theory.

\section{Primitive roots}

We start with a definition.
\begin{definition}
Let $n\in\mathbb{N}$ be any non-zero number. We say that a number $a\in\mathbb{Z}$ is a {\em primitive root modulo $n$} when $a+(n)\in\mathbb{Z}/(n)$ is a generator for the group of units in the quotient ring $\mathbb{Z}/(n)$.
\end{definition}
Our aim now is to show that, for any prime $p\in\mathbb{N}$, there exists a primitive root modulo $p$. We use a lemma, in which $\varphi$ denotes the {\em Euler phi function}. 
\begin{lemma}
Let $G$ be a finite commutative group (written multiplicatively). Then $G$ is cyclic if and only if, for each divisor $d$ of $|G|$, 
$$|\{g\in G\mid g\mbox{ is of order }d\}|\leq\varphi(d).$$
\end{lemma}
\begin{proof}
Write $m=|G|$ and, whenever $d$ divides $m$, $G_d\subseteq G$ for the set of elements of order $d$. By Lagrange's Theorem these form a partition of $G$:
$$G=\bigcup_{d|m}G_d.$$
Counting on both sides shows that $m=\sum_{d|m}|G_d|$. If $|G_d|\leq\varphi(d)$ for each divisor $d$ of $m$, then 
$$m=\sum_{d|m}|G_d|\leq\sum_{d|m}\varphi(d)=m$$
by a well-known property of $\varphi$. Hence necessarily $|G_d|=\varphi(d)$ for each divisor $d$ of $m$, thus in particular $|G_m|=\varphi(m)\neq 0$, saying precisely that $G$ is cyclic.

The converse is left to the reader.
\end{proof}
Without proof we mention the following.
\begin{proposition}
If $G$ is a finite commutative group, and for every divisor $d$ of $|G|$ we have that
$$|\{g\in G\mid g^d=1\}|\leq d,$$
then $G$ is cyclic.
\end{proposition}
We can now state our main result.
\begin{theorem}
For every prime $p\in\mathbb{N}$, the group $(\mathbb{Z}/(p))^{\times}$ is cyclic.
\end{theorem}
\begin{proof}
The number of roots of the polynomial $X^d-1$, viewed as polynomial with coefficients in the field $\mathbb{Z}/(p)$, is less than or equal to $d$. The previous proposition applies to the group of units of this field.
\end{proof}
\begin{example}
It is easy to verify that $2$ is a primitive root modulo $11$.
\end{example}
\begin{remark}
For more information on the lay-out of this article for the {\em Cahiers}, the official {\em Guide to Authors} can be found at the following address: 
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{\tt http://ehres.pagesperso-orange.fr/Cahiers/Ctgdc.htm}.
\end{center}
\end{remark}




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\bibitem{gauss}[C. F. Gauss, 1801] Disquisitiones Arithmeticae. 

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\noindent
Carl Friedrich Gauss \\
Mathematisches Institut \\
University of G\"ottingen \\
Bunsenstrasse 3-5  \\
37073 G\"ottingen (Germany) \\
my.email@server.univ.xy

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