# 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.

• You should at least be able to find the linear characters. You can also find one of degree $2$ by inducing up a character from the cyclic subgroup of order $10$. – Derek Holt Jan 1 at 16:03
• Derek says that restricting to $a = \pm 1$ yields an abelian (cyclic) subgroup $A$ of index $2$. So for a character $\psi : A \to \mathbb{C}^\times$ then $Ind_A^G \psi$ is a $2$ dimensional representation of $G$ and iff it is not irreducible then $Ind_A^G$ is the sum of two abelian representations thus abelian iff its kernel contains $[G,G] = ...$ – reuns Jan 1 at 16:50
• You asked for a hint, not a detailed answer. – Derek Holt Jan 6 at 18:23

## 2 Answers

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.

$$\begin{array}{c|rrrrrrrr} \rm class&\rm I&\rm -I&\rm4A&\rm4B&\rm5A&\rm5B&\rm10A&\rm10B\cr \rm size&1&1&5&5&2&2&2&2\cr \hline \chi_{1}&1&1&1&1&1&1&1&1\cr \chi_{2}&1&1&-1&-1&1&1&1&1\cr \chi_{3}&1&-1&-i&i&1&1&-1&-1\cr \chi_{4}&1&-1&i&-i&1&1&-1&-1\cr \chi_{5}&2&2&0&0&\frac{-1+\sqrt{5}}{2}&\frac{-1-\sqrt{5}}{2}&\frac{-1-\sqrt{5}}{2}&\frac{-1+\sqrt{5}}{2}\cr \chi_{6}&2&2&0&0&\frac{-1-\sqrt{5}}{2}&\frac{-1+\sqrt{5}}{2}&\frac{-1+\sqrt{5}}{2}&\frac{-1-\sqrt{5}}{2}\cr \chi_{7}&2&-2&0&0&\frac{-1+\sqrt{5}}{2}&\frac{-1-\sqrt{5}}{2}&\frac{1+\sqrt{5}}{2}&\frac{1-\sqrt{5}}{2}\cr \chi_{8}&2&-2&0&0&\frac{-1-\sqrt{5}}{2}&\frac{-1+\sqrt{5}}{2}&\frac{1-\sqrt{5}}{2}&\frac{1+\sqrt{5}}{2}\cr \end{array}$$

• +1 from me! I returned to the question today and see that you already have done a great job! – Nicky Hekster Jan 8 at 18:14

(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) $$k(G)=8$$.

• It is worth noting also that you can get all of the characters of degree $2$ by starting with one of them and multiplying that by the four linear characters. – Derek Holt Jan 2 at 10:10
• Yes Derek, agree ($Lin(G)$ always acts on the non-linear). And this would be a way to start calculating the $8 \times 8$ character table. – Nicky Hekster Jan 2 at 10:41