Character table for $G:= \langle x,y \mid x^5=y^4=1\text{ and }yx=x^2y\rangle$ Let $G$ be the group of order $20$ defined in terms of generators and relations: $$G:= \langle x,y \mid x^5=y^4=1\text{ and }yx=x^2y\rangle.$$
Can anyone help me to derive the character table? According to gap the conjugacy classes can be represented by $1, x, y, y^2$, and $y^3$. I can't determine the sizes of these classes which is proving to make my task more difficult.
 A: First let's handle the conjugacy classes. By the second relation, any element $g\in G$ will be represented by $x^jy^k$, by the first relation we may choose $0\leq j\leq 4$ and $0\leq k\leq 3$. 

Claim : $x^jy^k$ is conjugate with $x^ly^m$ if and only if $j=l=k=m=0$ or $k=m>0$ or $k=m=0$ and $j,l>0$. 

Just use the fact that $x$ conjugates $x$ to $x$, $y$ to $x^{-2}y$ and $y$ conjugates $x$ to $x^2$ and $y$ to $y$ so that $x^s(x^jy^k)x^{-s}=x^{j-2ks}y^k$ and so on...
Hence we have five conjugacy classes in $G$, the conjugacy class of $1_G$ is of cardinal $1$, the conjugacy class of $x$ is of cardinal $o(x)-1=4$, the conjugacy class of $y^k$ with $1\leq k\leq 3$ is of cardinal $o(x)=5$. 
Finally this leads to exactly $5$ irreducible representations. The abelianization of $G$ is given by the following presentation :
$$G^{Ab}=\langle x,y\mid y^5=x^4=1\text{, } yx=x^2y\text{, } yx=xy\rangle $$
It follows that in $G^{Ab}$, $x$ is trivial and hence :
$$G^{Ab}\text{ is generated by the image of }y\text{ of order } 4 $$
For our problem this gives already four irreducibles representation of dimension $1$ :
$$\phi_k:G\rightarrow \mathbb{C}^* $$
$$x\mapsto 1$$
$$y\mapsto i^k$$
For $k=0,1,2,3$ this gives indeed four different group morphisms. Now let $\chi$ be the last irreducible representation then its dimension $n$ will be $n^2=20-4=16=4^2$ so the last representation we are looking for is of dimension $4$. 
Now just check that the group in $GL(4,\mathbb{C})$ generated by those two matrices is isomorphic to $G$ and conclude ($\xi$ is a primitive $5$-root of the unity) :
$$X:=\begin{pmatrix}\xi&&&\\&\xi^3&&\\&&\xi^4&\\&&&\xi^2\end{pmatrix}$$
$$Y:=\begin{pmatrix}&&&1\\1&&&\\&1&&\\&&1&\end{pmatrix}$$
