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. 
 A: $\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.
