If $kG$ were completely decomposable, then there would exist an ideal $I$ for which $kG=kv\oplus I$.
I claim this would imply $$I=\left\{\sum_{g\in G} c_gg:\sum c_g=0\right\}.$$ First, we show that everything in $I$ is of this form. Suppose $x=\sum_{g\in G} c_gg\in I$. This implies that $\sum_{h\in G} h\cdot x\in I$ as well, so that (using standard sum rearrangment and reindexing tricks)
$$
\sum_{h\in G} h\cdot\sum_{g\in G} c_gg=\sum_{g\in G} c_g\sum_{h\in G}hg=\sum_{g\in G} c_g\sum_{h\in G} h=\left(\sum_{g\in G} c_g\right)v\in I
$$
But $\left(\sum_{g\in G} c_g\right)v$ is in $kv$ as well. Since $kv\cap I=\{0\}$, this means that $\sum c_g=0$.
This proves $I\subseteq \left\{\sum c_gg:\sum c_g=0\right\}.$ The reverse inclusion must hold as well, since the both of these are vector subspaces with a dimension of $|G|-1$.
However, it cannot be the case that $kG=kv\oplus I$, since $kv\subset I$! Specifically, the sum of the coefficients of $v=\sum g_i$ is $\sum_{i=1}^{|G|}1=|G|=0$, since $p$ divides $|G|$. Thus, $kG$ cannot be completely decomposable.
Yes. You're right that this is Maschke's Theorem. No. You're not quite right that we can take $V_2$ itself to be invariant.
Consider $G = \mathbb{Z} / 2$ acting on $k^2$ by reflection, for instance. Then obviously the axis of reflection is a sub-$kG$-module, and any other (one dimensional) subspace is a complement of vector spaces. However, it's clear that the only choice of complement that is itself a sub-$kG$-module is the line orthogonal to the axis of reflection (do you see why?).
What Maschke's Theorem buys us is the existence of a complement that's also a $kG$-module, but it might not be the one we started with. When we perform the "averaging" maneuver, we will almost certainly get back a different complement.
So, to answer your original question: How do we get the basis of a $kG$-complement? It may be easy to guess one, but it would be nice to know how to do it in general.
Let $U \subseteq V$ be a sub-$kG$-module. All the usual caveats apply ($G$ should be finite and its order should be coprime to the characteristic of $k$). Then if $\pi : V \to U$ is any old projection of vector spaces, we can average it to get
$$\pi_G = \frac{1}{|G|} \sum_{g \in G} g \pi g^{-1}$$
Then (for a certain definition of "easy to see") it's easy to see that $\pi_G$ is still a $k$-linear map from $V$ onto $U$ (indeed, it still splits the inclusion map), but now it has the bonus property of being a $kG$-linear map (do you see why?). So $\text{ker}(\pi_G)$ (which is itself a $kG$-module) is the desired complement.
Now we're in the home stretch, though, because we can regard $\pi_G$ as an ordinary $k$-linear map. So finding a basis for its kernel is "just linear algebra" in the sense that any of us could do it (or at least, we should be able to... I, for one, am probably rusty). But a computer algebra system like sage can absolutely slaughter this problem. As a (fun?) exercise, you might write up this algorithm in sage. It's not so bad, and it'll convince you that you understand it.
Of course, for many problems there's an "obvious" complement to your subspace. In practice, the guess-and-check method that you employed with the obvious complement is a good way to go. If worse comes to worse, though, your only recourse might be some honest computation:
- Fix any old $k$-linear projection $\pi : V \to U$
- Average it to $\pi_G : V \to U$
- Compute a basis for $\text{ker}(\pi_G)$
I hope this helps ^_^
Best Answer
Actually, I think that approach can be made pretty well-motivated. Here is how I like to see it: we have a $FG$-module $V$ and a submodule $H$, and we would like to know that there is a $G$-stable complement $H'$ (under certain circumstances, e.g., always if the characteristic of $F$ is prime to $|G|$). If such a complement exists, then $V=H \oplus H'$ and the projection onto $H$ with kernel $H'$ is a $G$-module map $V \rightarrow H$ left inverse to the $G$-module map including $H \hookrightarrow V$. Conversely, given a $G$-module map left inverse to this inclusion, its kernel will be a complement.
Now we are left with the problem of how to find such a homomorphism. Here is the idea: consider the vector space $\mathrm{Hom}_F(V,H)$. Inside this vector space live many projections $\pi$ onto $H$. We'd like to find one that is a $G$-module map, that is, such that $g \pi g^{-1}=\pi$ for all $g \in G$. If you stare at that equation for a second, it becomes clear that what you are really looking for is a $G$-fixed point in $\mathrm{Hom}_F(V,H)$, where $G$ acts by conjugation. Now the time-honored way of finding a fixed point for a group action is by averaging---start with any projection whatsoever and average it over all its images by elements of $G$. The condition on the characteristic is precisely what allows you to do this!
A nice thing about this point of view is that it tells you something even when the characteristic is not prime to the order of the group: it says you should really care about the functor of $G$-fixed points. This is a good way to motivate group cohomology to first year grad students, in my experience.