Abstract Nonsense

The Character Table of S_3xZ_3

Point of post: In this post we use our recently developed theory on the irreducible representations (characters) of the products of groups to find the character table of a more formidable group, namely $S_3\times\mathbb{Z}_3$.

Motivation

Let’s put our newly developed theory to work.

Character Table of $S_3\times\mathbb{Z}_3$

We begin by constructing the character tables for $S_3$ and $\mathbb{Z}_3$. The first of these we have already done and is given by

$\begin{array}{c|ccc} S_3 & e & (1,2) & (1,2,3)\\ \hline \chi^{\text{triv}} & 1 & 1 & 1\\ \chi^{(\alpha_1)} & 1 & -1 & 1\\ \chi^{(\alpha_2)} & 2 & 0 & -1\end{array}$

and the second one is fairly easy. Namely, we know that the number of non-equivalent irreducible characters of $\mathbb{Z}_3$ is $3$. And it’s easily verifiable that the three characters $\chi^{\text{triv}}$ and  $\chi^{(\beta_k)}:\mathbb{Z}_3\to\mathbb{C}:m\mapsto \zeta^{mk}$ where $k=1,2$ and $\zeta=e^{\frac{\pi i}{3}}$. Thus, the character table is given by

$\begin{array}{c|ccc} \mathbb{Z}_3 & 0 & 1 & 2\\ \hline \chi^{\text{triv}} & 1 & 1 & 1\\ \chi^{(\beta_1)} & 1 & \zeta & \zeta^2\\ \chi^{(\beta_2)} & 1 & \zeta^2 & \zeta\end{array}$

From our previous theorem we may conclude that the representatives from the equivalency classes of $S_3\times\mathbb{Z}_3$ are

$\left\{(e,0),(e,1),(e,2),\left((1,2),0\right),\left((1,2),1\right),\left((1,2),2\right),\left((1,2,3),0\right),\left((1,2,3),1\right),\left((1,2,3),2\right)\right\}$

and the irreducible characters are

$\left\{\chi^{\text{triv}\boxtimes\text{triv}},\chi^{\text{triv}\boxtimes(\beta_1)},\chi^{\text{triv}\boxtimes(\beta_2)},\chi^{(\alpha_1)\boxtimes\text{triv}},\chi^{(\alpha_1)\boxtimes(\beta_1)},\chi^{(\alpha_1)\boxtimes(\beta_2)},\chi^{(\alpha_2)\boxtimes\text{triv}},\chi^{(\alpha_2)\boxtimes(\beta_1)},\chi^{(\alpha_2)\boxtimes(\beta_2)}\right\}$

and thus the character table for $S_3\times\mathbb{Z}_3$, using our characterization of character tables of product groups

$\begin{array}{c|ccccccccc} S_3\times\mathbb{Z}_3 & (e,0) & (e,1) & (e,2) & \left((1,2),0\right) & \left((1,2),1\right) & \left((1,2),2\right) & \left((1,2,3),0\right) & \left((1,2,3),1\right) & \left((1,2,3),2\right)\\ \hline \chi^{\text{triv}\boxtimes\text{triv}} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \chi^{\text{triv}\boxtimes(\beta_1)} & 1 & \zeta & \zeta^2 & 1 & \zeta & \zeta^2 & 1 & \zeta & \zeta^2\\ \chi^{\text{triv}\boxtimes(\beta_2)} & 1 & \zeta^2 & \zeta & 1 & \zeta^2 & \zeta & 1 & \zeta^2 & \zeta\\ \chi^{(\alpha_1)\boxtimes\text{triv}} & 1 & 1 & 1 & -1 & -1 & -1 & 1 & 1 & 1\\ \chi^{(\alpha_1)\boxtimes(\beta_1)} & 1 & \zeta & \zeta^2 & -1 & -\zeta & -\zeta^2 & 1 & \zeta & \zeta^2\\ \chi^{(\alpha_1)\boxtimes(\beta_2)} & 1 & \zeta^2 & \zeta & -1 & -\zeta^2 & -\zeta & 1 & \zeta^2 & \zeta\\ \chi^{(\alpha_2)\boxtimes\text{triv}} & 2 & 2 & 2 & 0 & 0 & 0 & -1 & -1 & -1\\ \chi^{(\alpha_2)\boxtimes(\beta_1)} & 2 & 2\zeta & 2\zeta^2 & 0 & 0 & 0 & -1 & -\zeta & -\zeta^2\\ \chi^{(\alpha_2)\boxtimes(\beta_2)} & 2 & 2\zeta^2 & 2\zeta & 0 & 0 & 0 & -1 & -\zeta^2 & -\zeta\end{array}$

We therefore ascertain using our results about character tables that $\mathcal{Z}\left(S_3\times\mathbb{Z}_3\right)=\{e\}\times\mathbb{Z}_3$. Moreover, the normal subgroups of $S_3\times\mathbb{Z}_3$ are $S_3\times\mathbb{Z}_3,$ $S_3\times\{0\},$ $\{(e,0)\},$ $\{e,(1,2,3),)(1,3,2)\}\times\mathbb{Z}_3$, $\{e,(1,2,3),(1,3,2)\}\times\{0\}$ and $\{e\}\times\mathbb{Z}_3$.