Faithful irreducible representations of cyclic and dihedral groups over finite fields How to determine all the faithful irreducible representations of $\mathbb Z_n$ and $D_{2n}$ over $GF(p)$, where $p$ is a prime not dividing $n$?
 A: The faithful irreducible representations of $\mathbb{Z}_{n}$ over such a field $F$ have dimension $e$, where $e$ is the smallest positive integer such that $n$ divides $p^{e}-1.$ To find an explicit representation of that dimension is straightforward in theory, but may not be so easy in practice. It is necessary to find an irreducible factor, say $p(x)$  of degree $e$ of $x^{n}-1$ in $F[x].$ Given such a factor $p(x)$, there is an irreducible representation of $\mathbb{Z}_{n}$ which sends a generator of the cyclic group to the companion matrix of $p(x).$
Representing the dihedral group with $2n$ elements is a little more subtle. Let $z$ be a generator of the cyclic subgroup of index $2$. The issue is whether an irreducible representation of degree $e$ of $\langle z \rangle$ extends to the whole dihedral group or not. If it does not extend, then it induces to an irreducible representation of dimension $2e.$ So when does it extend? This depends on whether $z^{-1} = z^{p^{d}}$ for some $d$ with $1 \leq d \leq e.$ If yes, then the representation extends. If no, it does not.But since$e$ is the smallest positive integer such that $n$ divides $p^{e}-1,$  and $z$ has order $n,$ it reduces to checking whether $p^{\frac{e}{2}} +1$ is divisible by $n$ or not
(except in the slightly unusual case $n = 2,$ in which case $e = 1$ and the representation does extend). So ( for $n \neq 2$), if $e$ is odd, the representation does not extend. If $e$ is even then $n \neq 2 $ and the representation extends if and only if $n$ divides $p^{\frac{e}{2}}+1.$ 
