TITLE: Representation of Cyclic Group over Finite Field
QUESTION [8 upvotes]: The post Irreducible representations of a cyclic group over a field of prime order discusses the irreducible representations of a cyclic group of order $N$ over a finite field $\mathbb{F}_p$ where $N$ does not divide $p$.
Where can I find information about the irreducible representations in the case where $p$ does divide $N$?
(I'm interested in this because I'm wondering if, given a finite dimensional vector space $\mathbb{F}_{p}^{M}$ if there are examples of particularly simple invertible linear operators $T$ such that $T$ does not preserve a subspace. Since the vector space contains finitely many elements it seems that this is the same thing as an irreducible representation of a cyclic group. 
However, based on the post I read above, where p does not divide N, it seems like T is just multiplication by a primitive element that generates the extension $\mathbb{F}_{q^{M}}$.)

REPLY [2 votes]: In general, if a finite group $G$ has a normal subgroup $U$ whose order is  power of a prime $p,$ and $G$ acts irreducibly on  finite dimensional vector space $V$ over a field $F$ of characteristic $p,$ then $U$ has a non-zero fixed vector on $V,$ while on the other hand, the fixed-point subspace $V^{U}$ of $U$ on $V$ is easily checked to be $G$-invariant, so we must have $V^{U} = V$ by irreducibility as $V^{U}$ is non-zero. This is a well-known fact, and a more general version of the remark at the end of Qiaochu's answer. The argument when $V$ has finite cardinality is easier considering orbit lengths on non-zero vectors. However, in general, to prove that $V^{U} \neq \{0 \}$ (without further assumption) we can proceed by induction on $|U|.$ The case that $U$ has order $p$ is just linear algebra, as a generator of $U$ has the eigenvalue $1$. If $|U| >p,$ let $Z$ be a subgroup of $Z(U)$ of order $p,$ which is possible as $Z(U) \neq 1$ (since $U$ is  $p$-group). If $Z \neq U,$ then we still have $Z \lhd U$ as $Z$ is characteristic in $U.$ By induction $V^{Z} = V,$ and then we can replace $G$ by $G/U$ and $U$ by $U/Z$ to obtain the desired inclusion by induction.