Find all irreducible characters of a matrix group on finite field $\mathbb F_5$ Find all irreducible characters of matrix group $G =\left\{ \left( \begin{array}{cc}
a & b \\0 & a^{-1}\end{array} \right)|\,\,\, a,b \in\mathbb F_5, a\not=0 \right\}$.
The former question is to find irreducible characters of subgroup of $G$
$H =\left\{ \left( \begin{array}{cc}
a & 0 \\0 & a^{-1}\end{array} \right)|\, a\in\mathbb F_5^{\times}  \right\}$ and 
$N =\left\{ \left( \begin{array}{cc}
1 & b \\0 & 1 \end{array} \right)|\, b\in\mathbb F_5 \right\}$ and I can work it out.
But situation of $G$ is much complicated than I thought.
 A: (Partial answer / hint, starting with character degrees) I am assuming that you are taking about representations over $\mathbb{C}$. As indicated $G=HN$, $N \lhd G$, $H \cap N=1$ with $H \cong C_4$ and $N \cong C_5$. So $|G|=20$. Note that $G/N \cong C_4$, whence abelian, so $G' \subseteq N$. Since $G$ is not abelian (it is easy to find a non-commuting pair of matrices here) and $|N|=5$, $G'=N$. This means that there are $|G:G'|=4$ linear characters.
A simple computation shows that $Z(G)=\{\pm I\}$, which implies that $A=NZ(G)$ is an abelian (normal) subgroup of order $10$. If $\chi \in Irr(G)$ is non-linear, then $\chi(1) \leq |G:A|=2$. Hence, all non-linear characters must have degree $2$. Let $k$ be the number of non-linear characters, then $20=1^2+1^2+1^2+1^2+k \cdot 2^2$, which implies $k=4$ and the number of characters (=the number of conjugacy classes) $k(G)=8$.
A: From Nicky's answer it is fairly straight forward to obtain the full table.  Assuming the reader has calculated the conjugacy classes.  We know that $C_4$ is a quotient so we can lift the nontrivial characters from it.  We should determine which classes are mapped where.  Since we know that the copy of $C_{10}$ in the group contains the center, elements of order 10 will be mapped to the involution in $C_4$.  Elements of order 4 in $G$ will be mapped to elements of order 4 in the quotient and involutions to the involution.  This combined with the character table of $C_4$ (below) gives us $\chi_1,\dots\chi_4$.
$$
\begin{array}{c|rrrr}
  \rm class&\rm1&\rm2&\rm4A&\rm4B\cr
  \rm size&1&1&1&1\cr
\hline
  \rho_{1}&1&1&1&1\cr
  \rho_{2}&1&1&-1&-1\cr
  \rho_{3}&1&-1&-i&i\cr
  \rho_{4}&1&-1&i&-i\cr
\end{array}
$$
All that remains now is to find a 2-dimensional representation.  We can do this by calculating the quotient group $G/Z(G)$.  It turns out this is isomorphic to $D_5$ (exercise - you also need to write down where the conjugacy classes go).  Now, we can lift a two dimensional character from $D_5$ (below), say $\eta_3$ and then multiply it by the three linear characters we have already calculated.  This will give us $\chi_4,\dots,\chi_8$.
$$
\begin{array}{c|rrrr}
  \rm class&\rm1&\rm2&\rm5A&\rm5B\cr
  \rm size&1&5&2&2\cr
\hline
  \eta_{1}&1&1&1&1\cr
  \eta_{2}&1&-1&1&1\cr
  \eta_{3}&2&0&\frac{-1-\sqrt{5}}{2}&\frac{-1+\sqrt{5}}{2}\cr
  \eta_{4}&2&0&\frac{-1+\sqrt{5}}{2}&\frac{-1-\sqrt{5}}{2}\cr
\end{array}
$$
The full table is given here, however, it is left to the reader to calculate the conjugacy classes explicitly.  The number in the conjugacy class name tells you the order of the elements in the class, the $A$ and $B$ is to distinguish classes with elements of the same order.
$$
\begin{array}{c|rrrrrrrr}
  \rm class&\rm I&\rm -I&\rm4A&\rm4B&\rm5A&\rm5B&\rm10A&\rm10B\cr
  \rm size&1&1&5&5&2&2&2&2\cr
\hline
  \chi_{1}&1&1&1&1&1&1&1&1\cr
  \chi_{2}&1&1&-1&-1&1&1&1&1\cr
  \chi_{3}&1&-1&-i&i&1&1&-1&-1\cr
  \chi_{4}&1&-1&i&-i&1&1&-1&-1\cr
  \chi_{5}&2&2&0&0&\frac{-1+\sqrt{5}}{2}&\frac{-1-\sqrt{5}}{2}&\frac{-1-\sqrt{5}}{2}&\frac{-1+\sqrt{5}}{2}\cr
  \chi_{6}&2&2&0&0&\frac{-1-\sqrt{5}}{2}&\frac{-1+\sqrt{5}}{2}&\frac{-1+\sqrt{5}}{2}&\frac{-1-\sqrt{5}}{2}\cr
  \chi_{7}&2&-2&0&0&\frac{-1+\sqrt{5}}{2}&\frac{-1-\sqrt{5}}{2}&\frac{1+\sqrt{5}}{2}&\frac{1-\sqrt{5}}{2}\cr
  \chi_{8}&2&-2&0&0&\frac{-1-\sqrt{5}}{2}&\frac{-1+\sqrt{5}}{2}&\frac{1-\sqrt{5}}{2}&\frac{1+\sqrt{5}}{2}\cr
\end{array}
$$
