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I came across the definition of a completely reducible representation of a group. The way I understand the definition there is no real difference between a reducible representation and a completely reducible representation. The book I'm currently reading states that "This means that a completely reducible representation can be written, with a suitable choice of basis, as the direct sum of irreducible representations.". But, as far as I know, any representation is either reducible or irreducible so, eventually, any reducible representation can be written as a direct sum of irreducible representations.

What am I missing? What is the difference between reducible and completely reducible?

Thanks.

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    $\begingroup$ In characteristic $p$ not every reducible representation of a group is completely reducible. $\endgroup$ – Lord Shark the Unknown Apr 29 '17 at 16:41
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    $\begingroup$ Or the representation $1 \mapsto \left(\begin{array}{cc}1&1\\0&1\end{array}\right)$ of ${\mathbb Z}$ is reducible but not completely reducible. $\endgroup$ – Derek Holt Apr 29 '17 at 16:43
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For a field of characteristic zero we have the Maschke theorem : if $W \subset V$ is a subrepresentation, there is another subrepresentation $U \subset V$ such that $U \oplus W \cong V$. (Here I am assuming that $G$ is finite, as if $G$ is infinite there are already counter-example as given in the comments).

This become false if the field is not of characteristic zero, and a very cute example is $S_2$ acting on the plane over the field $\mathbb F_2$. There is exactly one line which is invariant (namely $L = \text{span}(e_1 + e_2)$) so our representation is reducible but not completely reducible.

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    $\begingroup$ Also, Maschke works more generally if the characteristic of the field does not divide $|G|$ for $G$ finite. $\endgroup$ – user171326 Apr 29 '17 at 16:51
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Def.: A reducible representation $(\rho,V)$ has a(t least one) sub representation $(\rho_W,W)$, where W a subset of V and W is closed under the action of $\rho$.

Def.: Completely reducible means that there are subsets $W_i$ of V such that each $W_i$ forms together with its $\rho_{W_i}$ a sub representation and $V=\oplus_{i} W_i$.

As @Derek Holt mentioned in the comments a counterexample to your statement that every reducible representation can be written as a direct sum of irreducible is the following: $$x \mapsto \left(\begin{array}{cc}1&x\\0&1\end{array}\right)$$ You can check that $$a\left(\begin{array}{c}1\\0\end{array}\right)$$, where a is a real number is an irreducible sub representation but $\left(\begin{array}{cc}1&x\\0&1\end{array}\right)$ cannot be written as a direct sum of irreducible sub representations because it is not diagonalizable (det=1).

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A reducible representation is one that has a proper invariant subspace; a completely reducible representation is one where every invariant subspace splits (i.e., is a direct summand as representations). Let $W\subset V$ be an irrep over a ground field $k$. If $\operatorname{char} k$ is prime to $\#G < \infty$ (or $\operatorname{char} k = 0$), then the operator \begin{align*} x \to \frac{1}{\#G} \sum_{g\in G} g\,\pi_W(g^{-1}x), \end{align*} where $\pi_W:V \to W$ is the projection map, is manifestly $G$-invariant and is trivial on $W$. Hence $W$ splits.

The canonical example of an instance where it fails is the representation of $\mathbb{Z}_p$ on $\mathbb{Z}_p^2$ that sends a generator to $\pmatrix{1 & 1 \\ 0 & 1}$. The subgroup generated by $(1, 0)$ is obviously irreducible, but it doesn't split.

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