Definition completely reducible group representation This should be a simple question, but i'm having a rather hard time though finding a explicit definition of a completely reducible group representation. 
Is it right to say that a presentation 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? 
Because if the answer is "yes", can't I say that every presentation is completely reducible, because it is the direct sum of the full vectorspace with the nullspace which are both $G$-invariant?
I wonder about this, because Mashke's theorem is formulated in my book as follows: 

Every finite-dimensional representation of a finite group $G$ is completely reducible as the direct sum of irreducible representations. 

But in the proof it's mentioned that $V$ itself can be irreducible (but they make no problem out of it), but by the theorem $V$ would then be irreducible and completely reducible at the same time. And if $V$ is completely irreducible, isn't any representation completely reducible then? 
 A: Completely reducible is usually called semisimple, which is to say that it can be written as a direct sum of simples.  What is a simple module $V$?  One that has only the trivial module $\{0\}$ and itself $V$ as submodules.  (It has no proper, non-trivial submodules).
This is analogous to how we exclude the integer 1 from the list of prime numbers, so that we don't write decompositions like $30 = 5 \times 3 \times 2 \times 1 \times 1 \times \cdots \times 1$.
A: A reducible group representation
$\Phi: G \rightarrow GL(n,V)$
is a representation that contains invariant subspaces different from the zero vector space and the complete vector space $V$.
Now, a representation is irreducible if it the only invariant subspaces that it contains are $0$ and $V$
So, a completely irreducible representation is a a representation that can be expressed as the direct sum of irreducible representations.
If you'd like to see a more visual mathematical way of seeing this:
$\Phi$ is completely irreducible $\leftrightarrow \Phi = \oplus \Phi_{\lambda}$ with $\Phi_{\lambda}$ irreducible
I hope this was clear.
