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Let $G$ be a finite group. Prove following statement. $\forall g \in G$ such that $g \neq e$ $ \exists$ a complex irrep $\rho$ such that $\rho(g) \neq E$

I have no idea how to start it. I can prove this statement for cyclic group. I tried for any finite group. Assume that $ \exists g\neq e \in G$ such that $\forall $ complex irrep $\rho$ $\rho(g) = E.$ I am not sure how to continue. Can I show that $\rho(g) $ is trivial?

Thanks a lot in advance for any help! I am sorry for my bad English.

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  • $\begingroup$ Can we use the second orthogonal relation? $\endgroup$ – gaoxinge Jan 4 '14 at 0:55
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This is probably an overkill but by standard character theory this follows easily. To any representation $\rho$ one associates naturally a character $\chi(g) = {\rm Tr}\rho(g)$. One then constructs a character table $\chi(C)$ indexed by inequivalent characters and conjugacy classes in $G$ (since character is constant on a given class).

Then the orthogonality relations (proved using Schur's lemma) state that the matrix given by this table has a full rank. Because we have $\chi(e) = {\rm Tr} E$ for all the characters $\chi$, at every other $g \in G$ there must be at least one $\chi$ so that $\chi(g) \neq {\rm Tr} E$ (otherwise the columns for $e$ and $g$ would be linearly dependent) and therefore the representation $\rho$ it is associated to must be non-trivial.

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Hint: While it is not necessarily irreducible, can you conclude that $\rho(g) \neq E$ when $\rho$ is, for example, the regular representation?

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  • $\begingroup$ I think the OP meant complex representation, @Jim $\endgroup$ – DonAntonio Jan 4 '14 at 0:54
  • $\begingroup$ @DonAntonio: I don't think the choice of field makes a difference in this argument. $\endgroup$ – Jim Jan 4 '14 at 1:05
  • $\begingroup$ I like this argument but not this particular hint. Regular representation has $\rho(g) \neq E$ by the very definition, so there is nothing to conclude. The real work is in reducing into irreducibles. $\endgroup$ – Marek Jan 4 '14 at 1:15
  • $\begingroup$ There's no work there either. Nothing needs to be shown. I'm not sure how to hint at that without giving away the entire answer. $\endgroup$ – Jim Jan 4 '14 at 1:20
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Hint to @Jim's hint: if a matrix is block-diagonalizable, and each block is the identity, what kind of matrix is it?

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