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What kind of examples of groups and representations should I keep in mind, which do not uniquely decompose into irreducible representations? I am mostly interested in characteristic zero representation, but I am also curiuous about positive characteristic.

What about the traceability of such representation? What algebraic structure give these issues? For which groups, we do not have such issues?

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    Think about how the $p$la$n$e can be s$p$ecified by axes. Even if one demands a right angle ... With re$p$resentations of finite grou$p$s, for exam$p$le, the unique decomposition into subspaces is not into irrecdivcible2011-07-27

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You have to be careful about the difference between irreducibles and indecomposibles. The represenation $\mathbb{Z}\to GL_2$ defined by $1 \mapsto \left( \begin{array}{cc} 1& 1\\ 0&1\end{array}\right)$ is the basic example of a represenation that is reducible but doesn't decompose into irreducibles.

On the other hand if you're asking when things split up into indecomposibles, there is the Krull-Schmidt theorem which says that if a module satisfies both the ascending chain condition and the descending chain condition, then it is a direct sum of indecomposibles. So in particular this would include any finite dimensional representation $G\to GL_n(\mathbb C)$

For counterexamples, maybe you could look at this book, I just now found it by google searching 'krull-schmidt counterexample'... the first example they give is that for $R= \mathbb{Z}[\sqrt{-5}]$ you have $\langle 3,2+\sqrt{-5}\rangle \oplus \langle 3,2-\sqrt{-5}\rangle \cong R\oplus \langle 3\rangle$, so the decompositions aren't unique in this case in the sense that the summands on the LHS and RHS aren't isomorphic as $R$-modules. But this is a statement about rings rather than group representations, so it might not be quite what you're after...

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    It is sort of an example of group representations. You run into something similar looking at integral (as in Z) representations of a cyclic group of order 5. While you mostly stick with finitely generated free abelian groups to build your modules over, as representations they are only finitely generated torsion-free. In particular, they can be projective without being free (as expected), but over rings like *R* you don't have Krull-Schmidt even for projectives. The failure of this is measured by the **ideal class group** and is an old part of algebraic number theory.2011-07-28