counterexample: degree of representation $\leq$ index of normal subgroup if I have a finite group $G$ with an abelian normal subgroup $N$  and an irreducible representation $\pi$ of $G$ over $K$. Then I know, that $deg(\pi) \leq [G:N]$, if $K$ has positive characteristic and is a splitting field for $G$. My professor claimed (without a proof), that this is also true if I take $K=\mathbb{Z}/p\mathbb{Z}$. Why should it work? I mean, $K$ must not be a splitting field here? Or are there any counterexamples?
I tried to get a counterexample for $G=\mathbb{Z}/p'\mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}$ with $gcd(p,p')=1$. You should get here, that the dimension of an irreducible $K(G)$-module is 1 or 2. But I don't get a result.
regards, Khanna
 A: Counterexamples are pretty common. Your idea of taking a dihedral group $G$ is good. For instance, consider a faithful irreducible representation of the dihedral group of order 10 over a field of size 3. Obviously[1] it has a faithful irreducible representation, so we just want to figure out its dimension. 
The dimension is not 1, because $\newcommand{\GL}{\operatorname{GL}}\GL(1,3)$ has order $2 < 10$. The dimension is not even 2 since $\GL(2,3)$ has order $48 \neq 0 \mod 10$, so Lagrange's theorem shows that $G$ has no faithful representation of dimension 2 over a field of size 3. In fact the explicit matrices[2] show that dimension 4 is large enough, and a consideration of eigenvalues shows that dimension 4 is minimal.
[1] Every normal subgroup is the kernel of a representation, and since the field has order coprime to the group, every representation is semi-simple, so only the irreducible representations in the representation matter (not their multiplicities). Since $G$ has a unique minimal normal subgroup, it must have a faithful irreducible representation in order for the trivial subgroup to be a kernel.
[2] Also, I give some explicit matrices in this answer. The question and answer use the Frobenius automorphism and traces to find minimal dimensions of faithful irreducible representations.
Smallest counterexample
Let $G$ be cyclic of order 3, and let $K=\mathbb{Z}/2\mathbb{Z}$ be a field of order 2. Let $N=G$ and check that $N$ is an abelian normal subgroup of index $[G:N]=1$. However, $\GL(1,K)$ has order 1, and so every representation of a group of dimension 1 over $K$ is the trivial representation. However, $G$ has more than just the trivial representation: it has an irreducible representation of dimension 2:
$$ a = \begin{bmatrix} . & 1 \\ 1 & 1 \end{bmatrix} \in \GL(2,K)$$
It is not reducible, since being the direct sum of two one-dimensional representations over $K$ is the same as being the direct sum of two trivial representations over $K$, is the same as every group element mapping to the identity matrix. In particular, here is an irreducible representation $\pi$ such that $\deg(\pi)=2 > [G:N]=1$.
The only smaller groups are $G$ of order 1 and 2, and in those cases all irreducible representations over every field are dimension 1, so there is no problem.
