# Prove that $GL_n(\mathbb{F}_p)$ contains an element of order $p^n-1$

If $p$ is prime and $n$ is a positive integer, prove that GL$_n(\mathbb{F}_p)$ contains an element of order $p^n-1$.

I am a bit stuck on how to get started on this problem. I found it in the Galois theory section on a list of past qualifying exams, so I'm guessing some kind of Galois theory should be involved, but I can't seem to get anywhere with it. I know that $\mathbb{F}_{p^n}$ and $\mathbb{F}_p^n$ are isomorphic as $\mathbb{F}_p$ vector spaces, so $Gal(\mathbb{F}_{p^n}/\mathbb{F}_p)$ is isomorphic to a cyclic subgroup of GL$_n(\mathbb{F}_p)$ of order $n$, though this doesn't seem too useful (but maybe I'm wrong!). My only other idea at this point is to try to find some $\mathbb{F}_p$-linear transformation $T$ of $\mathbb{F}_{p^n}$ with characteristic polynomial $x^{p^n-1}-1$ and then use Cayley-Hamilton to get that $T$ has order dividing $p^n-1$, and hopefully the definition of $T$ (whatever it may be) will show it's order can't be lower than $p^n-1$. Can anyone see a more clever way to attack this problem?

• These elements are called Singer cycles. Fun fact about them: no other element of $\mathrm{GL}(n,p)$ (or more generally $\mathrm{GL}(n,q)$ for arbitrary prime powers $q$) has as big an order as they do. Commented Jun 22, 2018 at 23:08

Hint: If $\alpha$ is a generator of the cyclic group $\mathbb{F}_{p^n}^*$, then what is the order of $\alpha \cdot : \mathbb{F}_{p^n} \to \mathbb{F}_{p^n}$?
• It's also interesting to note that with a bit more work, you can extend this to show: if $f$ is an irreducible polynomial over $\mathbb{F}_p$ of degree $n$, then the corresponding cyclic matrix (the one with 1s just below the diagonal and then negatives of the coefficients in the last column) has degree $p^n-1$. Commented Jun 22, 2018 at 23:39