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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

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Not yet. This theorem we did not have. Maybe that is the reason, why there wasn't a proof in class. I am going to read about it. Thank you – Khanna Aug 1 '12 at 13:56
If what your professor said was correct, then all meta-abelian groups would be supersolvable. :-) C3 (and its incarnation A4) is the smallest counterexample, but my answer goes along with your idea of using dihedral groups. – Jack Schmidt Aug 1 '12 at 14:02
You may hear the term “absolutely irreducible $K$-module” for a module where $K$ is close enough to being a splitting field (it is a splitting field for that module, maybe not for every module). If $\pi$ is absolutely irreducible, then the theorem holds. In my example, the representations are irreducible, but not absolutely irreducible. – Jack Schmidt Aug 1 '12 at 14:17
up vote 3 down vote accepted

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.

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Thank you, it seems like a difficult counterexample, but you explained it really good. In the answer of the other question, it sounds like an interesting example for simple $KG$-modules in positive characteristic. But I just don't see, what the number of all the simple modules is? You wrote, there is just one faithful repr., but are there other irred. repr., which are not faithful? – Khanna Aug 1 '12 at 14:17
@Khanna: yes. A dihedral group of order $10$ over a field of size $q \in \{3,13\}$ has exactly 3 irreducible representations: the trivial (everything goes to [1]), the sign (reflections go to [-1], rotations go to [1]), and the faithful irreducible. – Jack Schmidt Aug 1 '12 at 14:20
And in general (for the dihedral goup): Are there always just the 2 one-dim. irreducible representations and the other are faithful irreducible representations? – Khanna Aug 1 '12 at 14:26
@Khanna: yes exactly. The number of faithful irreducibles is always between $1$ and $(p-1)/2$ (a divisor of $(p-1)/2$ even) and has to do with the multiplicative order of $q$ mod $p$ (or more exactly the size of the subgroup $\langle q, -1 \rangle \leq \mathbb{Z}/p\mathbb{Z}$. I've only thought about this when $p$ is an odd prime, so check it for non-prime and evens, but I don't think the general case is too much different (presumably $p-1$ becomes $\phi(n)$, etc.). – Jack Schmidt Aug 1 '12 at 14:32
It sound like a good topic for a bachelor thesis. Thank you for your inspiration :) – Khanna Aug 1 '12 at 14:39

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