Let $X$ be a matrix representation. And $N ={\{g \in G: X(g) = I}\}$.

If $X$ has character $\chi$ and degree $d$. Prove that $g \in N$ if and only if $\chi(g) = d$ . Hint: Show that $\chi(g)$ is a sum of roots of unity.

proof: Let $X : G → GL(V)$ be a representation of a group $G$. And $\chi" G → \mathbb{C}$. We also know $\chi(g) = Tr(X(g)$ for all $g \in G$.

Suppose that $X(e) = I_d $, so $\chi(e) = trI_d = d$.

I was thinking in trying to show the eigenvalues of $X(g)$ are roots of unity.

$\chi(g) = \lambda_1 + ....+ \lambda_d = d$, show this implies the sum of the $\lambda_i = 1$ .

Suppose $g \in G$, then $X(g) = I$ so $\chi(g) = d$.

However, I dont really know what to do. Please can someone please help me? Any feedback would be appreciate it. Thanks!

  • $\begingroup$ Is G assumed to be finite? $\endgroup$ – mich95 Feb 25 '16 at 1:40
  • $\begingroup$ I forgot to mention it. It is finite $\endgroup$ – user29442 Feb 25 '16 at 1:42

Let $G$ have order $n$. Notice that any $X(g)$ is diagonalisable, since $X(g)^{n}=Id$ and hence the minimal polynomial of $X(g)$ divides $x^{n}-1$ which splits completely and is seprable in $\mathbb{C}$, and the eigenvalues of $X(g)$ are also roots of $x^{n}-1$, hence $n-th$ oots of unity. Therefore, $\chi((g))$ is the sum of the eigenvalues of $X(g)$. Now, when is the sum of $d$ distinct roots of unity is $d$ ?

  • $\begingroup$ Can you rephrase your question please? $\endgroup$ – user29442 Feb 25 '16 at 1:59

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