Existence of projection in proof of Maschke's theorem

Question

In a proof of Maschke's theorem, my lecturer writes "If $N$ is a $\mathbb C G$-module and $N\leq M $ is a submodule, let $\pi: M \to M$ be a projection (i.e. a map with $\pi^2=\pi$) with image $N$." I am wondering how we know that such a projection exists.

In one of our problem sheets it is highlighted that the theorem sometimes does and sometimes does not apply for fields other than $\mathbb C$. In the first part of the question I have showed that there does not exist a complement for the $\mathbb F_2 C_2$-submodule $\mathbb F_2 <e+g>$ (where $C_2 = <g>$, the cyclic group of order 2.) In the second part of the question, I am to show that there does exist a complement of the $\mathbb F_3 C_2$-submodule $\mathbb F_3 <e+g>.$

I know that I could tackle the above problem just by trying to spot a complement but I feel like that's cheating somewhat, and I'm also aiming to understand the proof of Maschke's theorem, so I wanted to do the question by forming a projection $\pi : \mathbb F_3C_2 \to \mathbb F_3C_2$ with image $\mathbb F_3 <e+g>.$ I may be missing something silly, but I can't spot one. My first try $\pi(e)=e+g=\pi (f)$ gives $\pi^2(e)=2(e+g)$, so this isn't a projection.

In short I have two questions:

Why does the above projection in the proof of Maschke always exist? And what is a projection that I could use for my exercise, (if it exists) for this $\mathbb F_3 <e+g>$?

Answer 1

The existence of that projection is the point of Maschke's theorem, and it doesn't exist in general; it exists here becaue the group is finite and we're considering representations over a field of characteristic $0$. (Those conditions can be relaxed; Maschke's theorem is usually presented in the context of finite groups and representations over some field $k$ of characteristic prime to $\#G$.) Take the projection $\pi:M \to N$ given by $$\pi(x) = \frac{1}{\#G} \sum_{g\in G} g\,\pi_0(g^{-1}x),$$ where $\pi_0:M\to N$ is a projection of the underlying $\mathbb{C}$-vector spaces. Then $$\pi(\gamma x) = \frac{1}{\#G} \sum_{g\in G} g\,\pi_0(g^{-1} \gamma x) = \frac{1}{\# G}\sum_{g\in G}\gamma g\pi_0(g^{-1} x) = \gamma\pi(x) $$ and \begin{align*} \pi(\pi x) &= \frac{1}{\# G}\sum_{g\in G}g\pi_0(g^{-1}\pi x) = \frac{1}{\# G}\sum_{g\in G}g(g^{-1}\pi x) = \pi x. \end{align*} The problem with the example you give is that $\mathbb{F}_2$ has characteristic $2$, which is not prime to $\#C_2 = 2$. To see what goes wrong in that case, consider the representation $\rho:C_2 \to \mathbb{F}_2^2$ that sends the generator $g$ to $$\rho(g) = \begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}.$$ The matrix $\rho(g)$ has one eigenvalue, $1$, and it has a $1$-dimensional eigenspace. The point of Maschke's theorem is that under its hypotheses, the image of any element is diagonalizable, and its eigenspaces span the representation. That's not the case with an arbitrary representation, and it fails in the case of this representation $\rho: \mathbb{F}_2[C_2]\to\mathbb{F}_2^2$. (The case of representations of a (say, finite) group $G$ over a ground field $k$ with characteristic dividing $\#G$ is called modular representation theory, and it's a very different beast from the alternative case.)

Answer 2

The projection always exists as a $k$-linear map. The hypothesis in Maschke's theorem allow you to average it out and obtain a $kG$-linear map.