# 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?

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?

• As a direct sum of simpme G invariant subspaces. 0 is not simple – Amr Oct 18 '16 at 4:06

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 $30 = 5 \times 3 \times 2 \times 1 \times 1 \times \cdots \times 1$.
• @user20327 Writing a representation as a direct sum of 0 and itself does not count as fully reducing it if the representation itself is not already irreducible. "Irreducible" and "fully reducible" do not exhaust all possibilities. For instance, the group algebra ${\bf F}_p[C_p]$ is not irreducible (the kernel of the trace map is a $C_p$-invariant space, which has no $C_p$-invariant complement) nor is it semisimple. – anon Mar 18 '13 at 23:03