# 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.

• An irreducible representation is completely reducible, so it can not be the "opposite" of it. Commented Apr 7, 2020 at 9:03
• Note that on the other hand "irreducible" always imply "totally reducible". Commented Apr 7, 2020 at 9:03
• @DietrichBurde I mean that if a representation is not irreducible, is it completely reducible then? Commented Apr 7, 2020 at 9:06
• @CaptainLama then my definitions are wrong? Commented Apr 7, 2020 at 9:06
• You could, but that would be a bad idea. Anyway, the answer to your question is yes if the characteristic of $K$ does not divide the order of $G$, and no otherwise. Commented Apr 7, 2020 at 9:11

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.