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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.

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$\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.

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  • $\begingroup$ Thanks you very much, @Andreas. Thank you for the help! $\endgroup$ – Sibilla Feb 16 '17 at 7:37

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