Character table; definition and an example

Posted: April 6, 2011 in Characters of Finite Groups, Representations of Finite Groups
Tags: ,

Before giving the definition of the character table of a finite group G we need some preparation. We defined an irreducible character: \chi_{\rho} is called irreducible if \rho is an irreducible representation. We also saw that equivalent representations have equal characters and we mentioned that the converse of this fact is also true, which we will prove it later (see Remark 2 in here). We proved that if G has r conjugacy classes, then the number of non-equivalent irreducible representations of G is exactly r (see the theorem in here). Thus the number of distinct irreducible characters of G is exactly r. Finally, recall that the value of a character is constant on each conjugacy class (see part iii) of Remark 1 in here).

Let \{\mathcal{C}_1, \ldots , \mathcal{C}_r \} be the set of conjugacy classes of G, where we will choose \mathcal{C}_1 to be the conjugacy class of 1, the identity element of G. So \mathcal{C}_1=\{1\}. Let \{\chi_1, \ldots , \chi_r \} be the set of distinct irreducible characters of G. We will choose \chi_1 to be the trivial character of G, i.e. \chi_1 is the character afforded by the trivial representation \rho : G \longrightarrow \mathbb{C}^{\times} defined by \rho(g)=1, for all g \in G. So \chi_1(g)=1, for all g \in G. We are now ready to define the character table of G.

Definition. The character table of G is the r \times r table (or matrix) X(G)=[a_{ij}] defined by a_{ij}=\chi_i(c_j), for all 1 \leq i,j \leq r, where c_j is any element of \mathcal{C}_j.

Remark 1. Each entry in the first row of X(G) is 1 because, since by convention \chi_1 is the trivial character of G, we have a_{1j}=\chi_1(c_j)=1. The i-th entry in the first column of X(G) is \deg \chi_i because a_{i1}=\chi_i(1)= \deg \chi_i, by part i) of Remark 1 in here.

Example. Find the character table of S_3.

Solution. We know that S_3 is isomorphic to the dihedral group of order 6. Thus we can write S_3=\{1,g_1,g_2,g_2^2,g_1g_2,g_1g_2^2\}, where we choose g_1 and g_2 to be cycles of length 2 and 3 respectively. We know from group theory, and it’s easy to see, that S_3 has 3 conjugacy classes:

\mathcal{C}_1 = \{1\}, \ \mathcal{C}_2 = \{g_2,g_2^2\}, \ \mathcal{C}_3 = \{g_1,g_1g_2,g_1g_2^2 \}.

So S_3 has three non-equivalent irreducible representations \rho_1, \rho_2, \rho_3 affording distinct irreducible characters \chi_1, \chi_2, \chi_3, where \chi_1 is the trivial character. So our character table is 3 \times 3. In Example 2 in here we determined \rho_1, \rho_2 and \rho_3. We showed that S_3, in fact S_n in general, has only two representations of degree one: the trivial representation \rho_1 and \rho_2 = \text{sgn}. Let \zeta = \exp(2 \pi i/3). Then, the third representation \rho_3 has degree two and is defined by

\rho_3(g_1^jg_2^k)= \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}^j \begin{pmatrix} \zeta & 0 \\ 0 & \zeta^{-1} \end{pmatrix}^k, \ \ \ \ \ \ \ \ (*)

for all 0 \leq j \leq 1 and 0 \leq k \leq 2. Let’s pick an element of each conjugacy class: 1 \in \mathcal{C}_1, \ g_2 \in \mathcal{C}_2 and g_1 \in \mathcal{C}_3. Of course, you may choose any element you like and that wouldn’t change the character table. Since \chi_1 is trivial, we have \chi_1 = 1, as we also mentioned in the above remark. Let’s now find the values of \chi_2. Well, we have \chi_2(1) = \deg \chi_2 = 1. Now, since g_1 is a cycle of length two, it is odd and hence its signature is -1. Thus \chi_2(g_1)=-1. The cycle g_2 has length three and so it is even. Thus \chi_2(g_2)=\text{sgn}(g_2)=1. Finally, let’s find the values of \chi_3. We have \chi_3(1) = \deg \chi_3 =2. Now, from (*) and the fact that \zeta + \zeta^{-1}=-1, it is clear that \chi_3(g_2)=\text{Tr}(\rho_3(g_2)) = -1 and \chi_3(g_1)=\text{Tr}(\rho_3(g_1))=0. So, the character table of S_3 is

\begin{array}{c | ccc} \ & \mathcal{C}_1 & \mathcal{C}_2 & \mathcal{C}_3 \\ \hline \chi_1 & 1 & 1 & 1 \\ \chi_2 & 1 & 1 & -1 \\ \chi_3 & 2 & -1 & 0 \end{array}.

Remark 2. It is possible for two non-isomorphic groups to have the same character tables; e.g. D_8 and Q_8. We will show this later. So character tables do not determine groups up to isomorphism.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s