irrep of a non unit element in the finite group 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.
 A: Hint: While it is not necessarily irreducible, can you conclude that $\rho(g) \neq E$ when $\rho$ is, for example, the regular representation?
A: 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.
A: Hint to @Jim's hint: if a matrix is block-diagonalizable, and each block is the identity, what kind of matrix is it?
