ProofOfCauchysTheorem 2002-02-19 10:03:21 2007-06-04 03:11:23 Proof Cauchy's theorem 0 \usepackage{amsfonts} % The following lines should work as the command % \renewcommand{\bibname}{References} % without creating havoc when rendering an entry in % the page-image mode. \makeatletter \@ifundefined{bibname}{}{\renewcommand{\bibname}{References}} \makeatother \PMlinkescapeword{contains} \PMlinkescapeword{divisible} \PMlinkescapeword{multiple} Let $G$ be a finite group, and suppose $p$ is a prime divisor of $|G|$. Consider the set $X$ of all $p$-tuples $(x_1, \ldots, x_p)$ for which $x_1\cdots x_p = 1$. Note that $|X| = |G|^{p-1}$ is a multiple of $p$. There is a natural group action of the cyclic group $\mathbb{Z}/p\mathbb{Z}$ on $X$ under which $m \in \mathbb{Z}/p\mathbb{Z}$ sends the tuple $(x_1, \ldots, x_p)$ to $(x_{m+1}, \ldots, x_p, x_1, \ldots, x_m)$. By the Orbit-Stabilizer Theorem, each orbit contains exactly $1$ or $p$ tuples. Since $(1,\ldots, 1)$ has an orbit of cardinality $1$, and the orbits partition $X$, the cardinality of which is divisible by $p$, there must exist at least one other tuple $(x_1,\ldots, x_p)$ which is left fixed by every element of $\mathbb{Z}/p\mathbb{Z}$. For this tuple we have $x_1 = \ldots = x_p$, and so $x_1^p=x_1\cdots x_p=1$, and $x_1$ is therefore an element of order $p$. \begin{thebibliography}{9} \bibitem{mckay} James H.~McKay. {\it Another Proof of Cauchy's Group Theorem}, American Math.~Monthly, 66 (1959), p119. \end{thebibliography}