# 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.

• Can we use the second orthogonal relation? – gaoxinge Jan 4 '14 at 0:55

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.
Hint: While it is not necessarily irreducible, can you conclude that $\rho(g) \neq E$ when $\rho$ is, for example, the regular representation?
• 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. – Marek Jan 4 '14 at 1:15