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Here is a question in the book "Representation theory of finite group, an introductory approach" of Benjamin Steinberg. (Question 3.8(2), page 25) that I need some hints from you :

Give an example of a finite group $G$ and a decomposable representation $\phi : G\to GL_4(\mathbb{C})$ such that the $\phi_g$ with $g\in G$ do not have a common eigenvector.

I tried to give some examples with cyclic groups of order 3, or abelian group of order 4, but I did not succeed. Please give me some examples that you know.

Thanks in advance.

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this wants to not be true of abelian groups, because if commuting matrices are diagonalizable, they are simultaneously diagonalizble. Also, I think the eigenspace of is invariant under the other, which should let you find a common eigenvector even if they are not diagonalizable. – mike Apr 21 '12 at 12:34
Slightly generalising mike's comment: you certainly don't want your representation to have one-dimensional direct summands. So your only chance is to construct a direct sum of two irreducible two-dimensional representations. Conversely, if a representation is irreducible and more-than-1-dimensional, then the $\phi_g$ cannot have a common eigenvector, since that would generate a subrepresentation. I am sure you will be able to take it from there. – Alex B. Apr 21 '12 at 13:02
@Alex : Thanks for your comment. Of course that we should eliminate the case of one-dimensional representation. I tried to construct a direct sum of two-dimensional representations. – Victor Apr 21 '12 at 13:50
@mike : thanks for your noting me that I should take a nonabelian group. Got it from example 3.1.14 in that chapter. – Victor Apr 21 '12 at 13:52

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