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Suppose an n-dimensional irreducible complex representation is not faithful. Then a non-identity element gets mapped to the identity matrix in $GL_n(\mathbb{C})$ so that the value of its associated character on the conjugacy class of this element is $n$. Thus, $n$ appears at least twice in the corresponding row of the group's character table.

I suspect the converse is true: if the row corresponding to an irreducible $n$-dimensional complex representation contains the dimension of the representation in more than one column, then the representation is not faithful. I have looked in a few of the standard algebra references and have been unable to find a proof. Can anyone point me in the right direction? We proved this for $n=2$, but it seems that it would be difficult and messy to generalize. I wonder if there is a simpler proof.

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I think it would be wise for Vitaly Lorman to choose an answer, as both are correct. This is an excellent question. Also. –  Jon Beardsley Apr 13 '11 at 3:49
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up vote 4 down vote accepted

What you are saying is true and is in fact contained in all the standard references (e.g. Isaacs). The idea is that any character of an $n$-dimensional representation, evaluated on an element of order $d$ is a sum of $n$ $d$-th roots of unity (why?). Your statement now follows easily by the triangle inequality: $|\chi(g)|\leq \chi(1)$ for all $g$ and $\chi(g)=\chi(1)$ iff $g$ is sent to the identity matrix.

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Beat me by 5 minutes. One really has to be quick on this site! –  Gerry Myerson Apr 13 '11 at 0:36
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If $\chi$ is the character, and $\chi(g)=\chi(1)=n$ for some group element $g$, then $\rho(g)$ is an $n\times n$ matrix $A$ whose eigenvalues are all complex numbers of modulus 1 and whose trace is $n$ (here $\rho$ is the representation whose character is $\chi$). Also, some power of $A$ is the identity. Can you see how this forces $A$ to be the identity matrix?

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Sorry, Gerry. Still +1 for you :-) –  Alex B. Apr 13 '11 at 2:09
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