Is a reducible representation of a finite group always completely reducible? Let $G$ be a finite group and let $V$ be a finite-dimensional vector space over the field $K$. Let $\rho : G \rightarrow GL(V)$ be a representation of $G$. Let $W\subset V$ be a proper nontrivial subspace of $V$ which is stable under the action of $G$, i.e.
$$ \rho(g)w \in W$$
for all $w \in W$ and for all $g \in G$. Then the representation is said to be reducible. If this is not the case, the representation is said to be irreducible. On the other hand, we say a  representation is completely reducible if the vector space on which the group $G$ is represented can be written as a direct sum of $G$-invariant subspaces, i.e. $V= \bigoplus_i W_i$, with each $W_i$ being $G$-invariant as defined above. 
EDIT: I assume here that the sum has more than one element, so that an irreducible representation is never completely reducible at the same time. 
My question: does reducible imply completely reducible? Where can I find a proof? I remember a theorem saying that for any finite group, if a representation is reducible then it is completely reducible, but I can't find it right now. In terms of matrix representations, the question is whether we can ever have
$$\left(\begin{array}{@{}c|c@{}}
  \rho^{(1)}(g) & b(g) \\
\hline
  0 & \rho^{(2)} (g)
\end{array}\right) \, ,$$
with $b(g) \neq 0$. If so, why?
EDIT #2: I think Maschke's theorem is what I am looking for: if the characteristic of the field is zero, then what I said above is true.
 A: Basically your question is "is any finite-dimensional representation of a finite group totally reducible?". Because as I mentioned in my comment an irreducible representation is trivially totally reducible, so restricting the question to reducible representations is not really relevant.
And the answer is: yes if the characteristic of $K$ does not divide the order of $G$. You can find that in literally any book that mentions group representations. Basically the trick is that since you can divide by $|G|$, you can use averages to define a $G$-invariant projection $V\to W$, the kernel of which gives a $G$-stable supplement of $W$.
On the other hand, the answer is (in general) no when the characteristic of $K$ divides $|G|$. The simplest example is to take $K=\mathbb{F}_2$ (the field with two elements), $G=\mathbb{Z}/2\mathbb{Z}$, $V=K^2$, and the non-trivial element of $G$ acting by
$$\begin{pmatrix} 0 & 1 \\ 1 & 0  \end{pmatrix}.$$
Then the line $W$ generated by $\begin{pmatrix} 1 \\ 1\end{pmatrix}$ is stable but does not have any stable supplement.
