I have a group $G$ with the following presentation $G=\langle x,y \mid x^4=y^3=1,yx=xy^2\rangle$. I know its conjugacy classes and I have to compute the character table :
Now since $y$ and $y^2$ are in the same conjugacy class, I have $\lambda_i(y)=\lambda_i(y)^2$ so $\lambda_i(y)=1$ or $\lambda_i(y)=0$ but since $y^3=1 \lambda_i^3(y)=1$ so we cannot have $\lambda_i(y)=0$. Then I am stuck. I feel I have to use the fact that $x^2$ is in the center.
\begin{array}{rrrrrrrrrrr} & \{1\} & \{x^2\} & \{x,xy,xy^2\} & \{y,y^2\} & \{x^3,x^3y\} & \{x^3,y^3\}\\ \text{Triv}& 1 & 1 & 1 & 1 & 1 & 1\\ \lambda_1&1&&&1&&\\ \lambda_2&1&&&1&&\\ \lambda_3&1&&&1&&\\ \chi_1&2&&&&&\\ \chi_2&2&&&&& \end{array}