# Exercise 4.7.8 Dixon-Mortimer “Permutation groups”.

Suppose that $m$>$1$ is an integer, and $p$ and $r$ are primes such that $r$ divides $p^{m}-1$ but $r$ does not divide $p^k-1$ for $1$$\leq k <m. Show that GL_m$$($$p$$)$ has an irreducible cyclic subgroup of order $r$. (A theorem of K. Zsigmondy shows that a prime $r$ satisfying these conditions exists for all $p$ and $m$ except for $p$ = $3$ and $m$ = $2$). I don't have any idea how to solve this exercise. Any kind of suggestion is appreciated. Thanks to everyone for the help.

$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$$\DeclareMathOperator{\GL}{GL}Let E be the field with p^{m} elements. This can be regarded as an m-dimensional vector space over the field F with p elements. The multiplicative group of E is cyclic, of order p^{m} - 1. Since r \mid p^{m} - 1, there is an element \alpha of multiplicative order r in E. This induces by multiplication on E an element of \GL(m, p) of order r. Note that \alpha sends 0 to 0, and induces cycles of length r on the non-zero elements of E, as it acts by multiplication. To show that \langle \alpha \rangle is irreducible, assume by way of contradiction that there is a subspace U \subseteq E of dimension k < m on which \alpha acts. Then \alpha induces an element of order r of \GL(k, p). But$$ \Size{\GL(k, p) } = (p^{k} - 1) (p^{k} - p) \cdots (p^{k} - p^{k-1}) = p^{\text{something}} (p^{k} - 1) (p^{k-1} - 1) \cdots (p - 1).$$This is impossible, as by assumption$r\$ does not divide any of these factors.