Characters of irreducible representations Suppose $G$ is a finite group of odd order and $\chi$ is the character corresponding to some 2-dimensional representation of $G$. Must $\chi(x)\neq 0$ for every $x\in G$?
 A: If your question is about complex representations, then the answer is yes-
Note that if $x\in G$ is such that $\chi(x)=0$ then $x$ has even order. To show this, let $\rho:G\to GL_2(\mathbb C)$ be the representation map, and let $x$ have $\chi(x)=0$. Since we are talking about complex representations $\rho(x)$ must have two distinct eigenvalues, namely $\lambda, -\lambda$. Furthermore, since $\rho(x)$ has finite order (as $x$ has), it is forced that $|\lambda|=1$ (since some for some $n\in\mathbb N$, $\lambda^n=(-\lambda)^n=1$). But if $x$ had odd order, then for some odd $n$ we would have
$$1=(-\lambda)^n=-(\lambda)^n=-1$$
which is a contradiction.
Now- consider the group $\rho(G)\subseteq GL_2(\mathbb C)$. Since $\rho$ is onto this group, we have that 
$$|\rho(G)|=\frac{|G|}{|\ker\rho|}$$
and in particular $|\rho(G)|$ bust be odd. Thus, $2$ does not divide the order of $\rho(G)$, and consequently $\rho(G)$ does not contain any elements of even order. So that $\chi(x)\neq 0$ for any $x\in G$.
If we are talking about representations over arbitrary fields this may happen. For example, take $$G=\lbrace \left(\begin{matrix}1&t\\0&1\end{matrix}\right)\mid t\in \mathbb{F}_q\rbrace\subseteq GL_2(\mathbb{F}_q)$$
for $q=p^m$, $p>2$ a prime number. Let $\rho:G\to GL_2(\mathbb{F}_2)$ be the standard representation- that is
$$\rho(\left(\begin{matrix}1&t\\0&1\end{matrix}\right))\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}a+tb\\b\end{matrix}\right)$$
In this case $\chi(x)=0$ for all $x\in G$.
A: Here is a different way to see this for complex characters, which uses a bit more theory:
Since the degree of the character does not divide the order of the group, it cannot be irreducible, so it must in fact be the sum of two linear characters. Call these $\chi_1$ and $\chi_2$. If $(\chi_1+\chi_2)(x) = 0$ then $\chi_1(x) = -\chi_2(x)$. But clearly both $\chi_1(x)$ and $\chi_2(x)$ have odd order, since the group does. On the other hand, if some complex number $z$ has odd order, then $-z$ has even order, so it is not possible to have $(\chi_1+\chi_2)(x) = 0$.
