I want to understand if there is a group $G$ with $|G|=8$ and a character $\chi$ of $G$ assuming values $-2,-1,0,0,0,0,1,2$.
Let $\rho: G \rightarrow GL(V)$. Because $\chi(1_G)=\dim V$, we must have $\dim V$ being equal to $1$ or $2$. Because $\rho(g)$ must be invertible, we musn't have dimension $1$, because in that case, we would have $\rho(g)=0$...
By Schur's Lemma, if $g_o$ is in the center, $\rho(g_o)= \lambda I$. Because $\rho(g_o^{|g_o|})=I$, this $\lambda$ must be a root of unity and we need $|\lambda|=1$. Therefore $\chi(g_o)=2\lambda$ and $p-$groups have no trivial center so we need $2$, $4$ or $8$ elements with $|\chi (g)|=2$.
We do have $2$ such elements with $|\chi(g)|=2$, so $|Z(G)|=2$. Because $G/Z(G)$ is abelian we need $G' \subset Z(G)$ and because $G$ is not abelian $G'=Z(G)$, so there should be $4$ irreducible characters of dimension $1$.
I cannot seem to get any contradiction, which makes me think I should start looking for a group with these properties... Is there a general strategy for these questions?
I shouldn't have asked this question: turns out it is a straightforward computation if we decompose $\rho$ in terms of irreducible characters $\rho=\oplus_i n_i \rho_i $:
$$\sum_i n_i^2=\left\langle \sum_i n_i\chi_i ,\sum_j n_j\chi_j \right\rangle=\langle \chi ,\chi \rangle=\frac{10}{8}\in \mathbb{Z}$$
This is absurd.
If someone knows of other strategies to tackle such problems in more generality, I would very much like to see them.