Abstract Nonsense

Crushing one theorem at a time

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.

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April 11, 2011 - Posted by | Algebra, Representation Theory | , , , , , ,

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