Question

Let G be the group of order 21 generated by two elements a, b with the relations a^7 = 1, b^3 = 1, and bab^−1 = a^2. Construct the irreducible complex representations of G, and compute its character table.

Answer 1

In preparation for the computation of the character table of $G$, we first note that $G$ has five conjugacy classes, represented by $1, a, a^3, b, b^2$. (This is an easy group-theoretic computation, which we omit.) Thus, we expect to have five irreducible complex representations. Obviously, $[G, G] = <a>$, so $G / [G, G] \sim <b>$. This shows that there are three 1-dimensional representations $\chi_i: G \to \mathbf{C}^*$, which are trivial on $<a>$, with $\chi_1(b) = 1$, $\chi_2(b) = \omega$, and $\chi_3(b) = \omega^2$, where $\omega$ is a primitive cubic root of unity. Next, we construct a 3-dimensional C-representation $D: G \to \mathrm{GL}_3(\mathbf{C})$ by taking

$$D (a) = \left( \begin{array}{c c} \zeta & \\ & \zeta^ {2} \\ & \zeta^ {4} \end{array} \right), \text { and } D (b) = \left( \begin{array}{c c c} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right),$$

where $\zeta$ is a primitive 7th root of unity. (It is straightforward to check that the relations between $a$ and $b$ are respected by $D$.) If $D$ is a reducible representation, it would have to "contain" a 1-dimensional representation. This is easily checked to be not the case. Thus, $D$ is irreducible, and we get another irreducible 3-dimensional C-representation $D'$ by taking

$$D ^ {\prime} (a) = \overline {{D (a)}} = \left( \begin{array}{c c} \zeta^ {6} & \\ & \zeta^ {5} \\ & \zeta^ {3} \end{array} \right), \text { and } D ^ {\prime} (b) = \overline {{D (b)}} = \left( \begin{array}{c c c} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right).$$

Note that we have $D \not\cong D'$, since they have different characters, say $\chi_4$ and $\chi_5$. We have now computed all complex irreducible representations of $G$, arriving at the following character table:

(where $\alpha = \zeta +\zeta^2 +\zeta^4$)

From the first column of this character table, we see that the Wedderburn decomposition of $CG$ is: $CG \sim C \times C \times C \times M_3(C) \times M_3(C)$.