# Character table of the non-abelian group of order 21

I'm working my way through the first Chapter of Fulton and Harris' Representation Theory and I'm trying exercise 3.26:

There is a unique nonabelian group $G$ of order 21, which can be realized as the group of affine transformations $T_{\alpha,\beta}(x) = \alpha x + \beta$ of the line over the field with seven elements, with $a$ a cube root of unity in that field. Find the irreducible representations and character table.

To start, I was trying to find the conjugacy classes of $G$, $\alpha = 1,2,4$. Of course $\{T_{1,0}\}$ is the easiest conjugacy class.

Furthermore, I found that $T_{\alpha^{-1},-\alpha^{-1}\beta}$ is the inverse of $T_{\alpha,\beta}$ so using that to find conjugates: $$T_{\alpha,\beta}\cdot T_{\gamma,\delta} \cdot T_{\alpha^{-1},-\alpha^{-1}\beta} = T_{\gamma, -\gamma \beta + \alpha \delta + \beta}$$

We see that conjugacy classes will be of the form $\{T_{\alpha, x}\}$ so we have at least 3 conjugacy classes. How do I proceed from here? Is this even the right way to tackle this exercise?

• Looks like a very good start to me. But why would this force 7 conjugacy classes? $\gamma$ takes only three values! – Jyrki Lahtonen Oct 12 '14 at 19:54
• Whoops! Definitely right. Typo. Fixed. – Krijn Oct 12 '14 at 19:55
• At the end I would also do an independent recount using Sylow theory. The subgroup $\{T_{1,\beta}\mid \beta\in\Bbb{Z}_7\}$ is normal. In addition to the class of the identity it seems to have two conjugacy classes of size three. The remaining 14 elements are the non-trivial elements of the seven cyclic Sylow 3-subgroups. Each of those is self-normalizing, and thus each element of order three generates its own centralizer. Therefore all of the have 7 conjugates, so the 14 elements split into 2 conjugacy classes. – Jyrki Lahtonen Oct 12 '14 at 19:59
• (cont'd) This gives us a total of five conjugacy classes, and suggests that if $\gamma\neq1$ all the elements $T_{\gamma,\beta}$ with fixed $\gamma$ form a conjugacy class of seven. Can you show that using your calculation? – Jyrki Lahtonen Oct 12 '14 at 20:01
• The hint indeed helps: if $\delta = 0, \gamma \neq 1$ then $-\gamma \beta + \beta = \beta ( -\gamma + 1)$ and thus $-\gamma + 1 = \epsilon \neq 0$ so we will take all seven values when multiplying $\epsilon$ with all values of $\mathbb{F_7}$ – Krijn Oct 12 '14 at 20:13

Thanks to Jyrki Lahtonen and his useful comments I have found the conjugacy classes myself:

We of course already had $\{T_{1,0}\}$ and we know that if two elements are conjugated then it would be of the shape $T_{\gamma,x}$ where $x = -\gamma \beta + \alpha \delta + \beta$.

Let us first look at the case where $\gamma = 1$ then $-\gamma \beta + \alpha \delta + \beta = \alpha \delta$ and, because $\alpha = 1,2,4$ this would mean that $T_{1,\delta}$ is conjugated with $T_{1,1\cdot\delta}, T_{1,2\cdot \delta}$ and $T_{1,4 \cdot \delta}$. Thus, two other conjugacy classes are $\{ T_{1,1}, T_{1,2}, T_{1,4} \}$ and $\{ T_{1,3}, T_{1,5}, T_{1,6} \}$.

Now, if $\gamma \neq 1, \delta = 0$ then $T_{\gamma, 0}$ is conjugated to $T_{\gamma, \beta(-\gamma +1)}$ because then $-\gamma \beta + \alpha \delta + \beta = \beta(-\gamma +1)$. But because $\gamma \neq 1 \rightarrow -\gamma + 1 = \epsilon \neq 0$ and thus $\beta \cdot \epsilon$ will be any element of $\mathbb{F_7}$ if we choose $\beta$ right.

This gives us another two conjugacy classes: $\{T_{2, \delta} | \delta \in \mathbb{F_7} \}$ and $\{T_{4, \delta} | \delta \in \mathbb{F_7} \}$

We now list all the conjugacy classes so for, along with the number of elements:

$$|\{T_{1,0}\}| = 1$$ $$|\{ T_{1,1}, T_{1,2}, T_{1,4} \}| = 3$$ $$|\{ T_{1,3}, T_{1,5}, T_{1,6} \}| = 3$$ $$|\{T_{2, \delta} | \delta \in \mathbb{F_7} \}| = 7$$ $$|\{T_{4, \delta} | \delta \in \mathbb{F_7} \}| = 7$$

We conclude that these are all the conjugacy classes, as $1+3+3+7+7=21$.

EDIT: After working on my own for quite a while, I think I have also answered the actual question 3.26, and I would be very grateful if someone could check these results:

We must now find the characters of $G$. We have ofcourse the trivial character, and we'll start using for different characters by looking at the subgroup $H = \{ T_{1,\alpha} | \alpha \in \mathbb{F_7} \}$. At first, we see that $G/H = \{ H, T_{2,0}H, T_{4,0}H \} \cong C_3$, of which we know the character table (with $\omega^3 = 1$):

$$\begin{array}{c|c c c} \text{Class rep.} & T_{1,0}H & T_{2,0}H & T_{4,0}H\\ \hline\hline \chi_{triv} & 1 & 1 & 1 \\ \chi_1 & 1 & \omega & \omega^2 \\ \chi_2 & 1 & \omega^2 & \omega \\ \end{array}$$

Which we can expand into the following irreducible characters of $G$

$$\begin{array}{c|c c c c c} \text{Class rep.} & T_{1,0} & T_{1,1} & T_{1,3} & T_{2,0} & T_{4,0}\\ \hline\hline \chi_{triv} & 1 & 1 & 1 & 1 & 1 \\ \chi_1 & 1 & 1 & 1 & \omega & \omega^2 \\ \chi_2 & 1 & 1 & 1 & \omega^2 & \omega \\ \end{array}$$

We also know that $\sum \dim(\chi)^2 = |G| = 21$ and thus we are looking for two $3$ dimensional representations, as $1^2 + 1^2 + 1^2 + 3^2 + 3^2 = 21$.

From the hint in the book, these representations come from induced representations of $H$, so let us take the easiest non-trivial representation of $H$, where $\alpha^7 = 1$: $$\chi_\alpha : 1, \alpha, \alpha^2, \alpha^3, \alpha^4, \alpha^5, \alpha^6$$

We know that $\chi_{Ind(\alpha)}(g) = 1/|H|\sum_{x \in G} \chi_{\alpha}^{'}(x^{-1}gx)$ So we can easily calculate: $\chi_{Ind(\alpha)}(T_{1,0}) = 1/7\cdot(21) = 3$ as expected. $\chi_{Ind(\alpha)}(T_{1,1}) = 1/7\cdot(7\cdot\alpha + 7\cdot\alpha^2 + 7\cdot\alpha^4) = \alpha + \alpha^2 + \alpha^4$ $\chi_{Ind(\alpha)}(T_{1,3}) = 1/7\cdot(7\cdot\alpha^3 + 7\cdot\alpha^5 + 7\cdot\alpha^6) = \alpha^3 + \alpha^5 + \alpha^6$
$\chi_{Ind(\alpha)}(T_{2,0}) = 0$
$\chi_{Ind(\alpha)}(T_{4,0}) = 0$

So we find $$\chi_3 : 3, \alpha + \alpha^2 + \alpha^4, \alpha^3 + \alpha^5 + \alpha^6, 0, 0$$ Which also gives us $$\chi_4 : 3, \alpha^3 + \alpha^5 + \alpha^6,\alpha + \alpha^2 + \alpha^4, 0, 0$$

Concluding:

$$\begin{array}{c|c c c c c} \text{Class length} & 1 & 3 & 3 & 7 & 7 \\ \hline \text{Class rep.} & T_{1,0} & T_{1,1} & T_{1,3} & T_{2,0} & T_{4,0}\\ \hline\hline \chi_{triv} & 1 & 1 & 1 & 1 & 1 \\ \chi_1 & 1 & 1 & 1 & \omega & \omega^2 \\ \chi_2 & 1 & 1 & 1 & \omega^2 & \omega \\ \chi_3 & 3 & \alpha + \alpha^2 + \alpha^4 & \alpha^3 + \alpha^5 + \alpha^6 & 0 & 0 \\ \chi_4 & 3 & \alpha^3 + \alpha^5 + \alpha^6 & \alpha + \alpha^2 + \alpha^4 & 0 & 0 \\ \end{array}$$

With $\omega$ as a third root of unity.
With $\alpha$ as a seventh root of unity.