# 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 $n$?

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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 $p^{\frac{e}{2}}+1.$
It is an $e \times e$ matrix which has $1$'s on the superdiagonal in rows $1$ through $e-1$ (zeroes elsewhere on those rows)and its entries in row $e$ apart from the diagonal are the coefficients of $p(x)$ each multiplied by signs to make its characteristic polynomial $p(x).$ Eg, the companion matrix of $x^{2} +1$ is $\left( \begin{array}{clcr} 0 & 1\\-1 & 0 \end{array} \right)$. –  Geoff Robinson Jul 22 '12 at 12:28
For the cyclic group, it is the number of irreducible factors of degree $e$ of $x^{n}-1.$ For the dihedral group (apart from $n =2$), I think it is half this number if $n$ does not divide $p^{\frac{e}{2}}+1$ and twice that number if $n$ does divide $p^{\frac{e}{2}}+1,$ but I suggest you check that yourself: it is a question of Clifford theory. –  Geoff Robinson Jul 22 '12 at 13:42