## Character table; definition and an example

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

## Character of a representation; two remarks

Posted: March 24, 2011 in Characters of Finite Groups, Representations of Finite Groups
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Throughout $G$ is a finite group. Recall that we defined the degree of a character $\chi$ to be the degree of the representation which affords $\chi.$

Remark 1. Let $\rho : G \longrightarrow \text{GL}(V)$ be a representation of $G$ and let $\chi : G \longrightarrow \mathbb{C}$ be its character. Then

i) $\chi(1)=\deg \chi.$

ii) $\chi(g_1g_2)=\chi(g_2g_1),$ for all $g_1,g_2 \in G.$

iii) $\chi(x)=\chi(gxg^{-1}),$ for all $g,x \in G.$

Proof. i) Let $\deg \chi = \deg \rho=\dim V = n.$ By definition, $\chi(1) = \text{Tr}(\rho(1))=\text{Tr}(\text{id}_V).$ But the matrix of $\text{id}_V$ is the  $n \times n$ identity matrix and so its trace is $n.$ Thus $\chi(1)=n=\deg \chi.$

ii) Recall from linear algebra that for any $n \times n$ matrices $A$ and $B$ we have $\text{Tr}(AB)=\text{Tr}(BA).$ Thus

$\text{Tr}(\rho(g_1g_2))=\text{Tr}(\rho(g_1) \rho(g_2))=\text{Tr}(\rho(g_2) \rho(g_1))=\text{Tr}(\rho(g_2g_1)).$

Hence $\chi(g_1g_2)=\chi(g_2g_1).$

iii) By ii), $\chi(gxg^{-1})=\chi(g^{-1}gx)=\chi(x). \ \Box$

Remark 2. Let $\rho_i: G \longrightarrow \text{GL}(V_i), \ i=1,2,$ be two representations of $G.$ If $\rho_1$ and $\rho_2$ are equivalent, then $\chi_{\rho_1}=\chi_{\rho_2}.$

Proof. By $(*)$ in here, there exists a $\mathbb{C}[G]$-module isomorphism $\varphi : V_1 \longrightarrow V_2$ such that $\rho_2(g)=\varphi \rho_1(g) \varphi^{-1},$ for all $g \in G.$ Choose some $\mathbb{C}$-basis for $V_1$ and $V_2$ and look at $\rho_1(g), \rho_2(g)$ and $\varphi$ as matrices. Then, since $\text{Tr}(AB)=\text{Tr}(BA)$ for any $n \times n$ matrices $A$ and $B,$ we have

$\text{Tr}(\rho_2(g))=\text{Tr}(\varphi \rho_1(g) \varphi^{-1})=\text{Tr}(\varphi^{-1} \varphi \rho_1(g))=\text{Tr}(\rho_1(g)).$

Therefore $\chi_{\rho_2}(g)=\chi_{\rho_1}(g),$ for all $g \in G$ and so $\chi_{\rho_2}=\chi_{\rho_1}. \ \Box$

The converse of the statement in Remark 2 is also true, i.e. if $\chi_{\rho_1}=\chi_{\rho_2},$ then $\rho_1$ and $\rho_2$ are equivalent. We will prove this later.

## Character of a representation; basic definitions and remarks

Posted: February 23, 2011 in Characters of Finite Groups, Representations of Finite Groups
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Throughout $G$ is a finite group.

Definition. Let $\rho : G \longrightarrow \text{GL}(V)$ be a representation of $G.$ The map $\chi : G \longrightarrow \mathbb{C}$ defined by $\chi(g) = \text{Tr}(\rho(g)), \ g \in G,$ is called the character of $\rho$ and is denoted by $\chi_{\rho}.$ We say that $\chi_{\rho}$ is an irreducible character if $\rho$ is irreducible. We also define the degree of $\chi_{\rho}$ to be $\deg \rho.$

Note 1. If $\rho$ is understood, we will just write $\chi$ instead of $\chi_{\rho}.$

Note 2. The term $"\rho$ (or $V$) affords $\chi"$ is also used.

Remark 1. In the above definition, $\text{Tr}(\rho(g))$ menas the trace of the linear transformation $\rho(g): V \longrightarrow V$ with respect to a basis of $V.$ To be precise, $\chi(g)$ is the trace of the matrix of $\rho(g)$ with respect to a basis of $V.$ Note that, since the trace of similar matrices are equal, $\chi(g)$ does not depend on the basis we choose for $V$ and so our definition makes sense.

Remark 2. If $\deg \rho = 1,$ then clearly $\rho(g)=\chi_{\rho}(g)$ and so $\chi_{\rho}(g) \neq 0,$ for all $g \in G.$

Remark 3. Recall from linear algebra that the trace of a linear transformation is the sum of its eigenvalues. So $\chi_{\rho}(g)$ is the sum of eigenvalues of $\rho(g).$

Theorem. If $|G|=n, \ \deg \rho = m$ and $g \in G,$ then $\chi_{\rho}(g)=\sum_{i=1}^m \lambda_i,$ where $\lambda_i$ are (not necessarily distinct) $n$-th roots of unity.

Proof. By Remark 3, we only need to show that every eigenvalue of $\rho(g)$ is an $n$-th root of unity. Since $\deg \rho = m,$ the degree of the characteristic polynomial of $\rho(g)$ is $m$ and thus this polynomial has $m$ roots, which are not necessarily distinct. Now if $\lambda$ is an eigenvalue of $\rho(g),$ then $\rho(g)(v)=\lambda v,$ for some $0 \neq v \in V.$ Since $|G|=n,$ we have $g^n=1$ and hence $v = \rho(g^n)(v)=(\rho(g))^n(v)=\lambda^n v.$  Thus $(\lambda^n - 1)v=0$ and hence $\lambda^n=1$ because $v \neq 0.$ So each eigenvalue of $\rho(g)$ is an $n$-th root of unity. $\Box$

Corollary 1. $\chi(g)$ is an algebraic integer for all $g \in G.$

Proof. By the theorem, $\chi(g)$ is a sum of some roots of unity. Obviously any root of unity is an algebraic integer and we know that the set of algebraic integers is a ring and so it is closed under addition. $\Box$

Corollary 2. $\chi(g^{-1})=\overline{\chi(g)},$ where $\overline{\chi(g)}$ is the complex conjugate of $\chi(g).$

Proof.  Let $\rho$ be a representation which affords $\chi.$ Suppose that $\deg \rho = n.$ By the theorem, $\chi(g)=\sum_{i=1}^n \lambda_i,$ where $\lambda_i$ are the eigenvalues of $\rho(g)$ and they are all roots of unity. Since $\rho(g^{-1})=(\rho(g))^{-1},$ the eigenvalues of $\rho(g^{-1})$ are $\lambda_i^{-1}.$ Also, $\lambda_i^{-1}=\overline{\lambda_i}$ because $\lambda_i$ are roots of unity. Thus

$\chi(g^{-1})=\sum_{i=1}^n \lambda_i^{-1}=\sum_{i=1}^n \overline{\lambda_i}=\overline{\sum_{i=1}^n \lambda_i}=\overline{\chi(g)}.$  $\Box$