Archive for the ‘Representations of Finite Groups’ Category

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.

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.

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

We’re now going to use induced representations to find m degree two representations for D_{2m}, the dihedral group of order 2m. We already gave these representations in Example 1 (also see the remark after that!) but we didn’t explain how we got them!

Example. Let g_1,g_2 be the generators of D_{2m}, where o(g_1)=2, \ o(g_2)=m and g_1g_2=g_2^{-1}g_1. Let H = \langle g_2 \rangle. Show that every induced representation from H to G has degree two and find all of them.

Solution. Since H is a cyclic group, it has m representations and all of them have degree one (see Example 1). Thus, by the theorem in this post, if \rho is a representation of H, then \deg \overline{\rho}=[D_{2m}:H] \deg \rho=2 \times 1=2. Now let \rho : H \longrightarrow \text{GL}(\mathbb{C}) be a representation of H. Then \rho(g_2)=\zeta, where \zeta is an m-th root of unity. Let \overline{\rho} = \text{Ind}_H^G \rho : G \longrightarrow \text{GL}(V) be the induced representation of \rho. So V = \mathbb{C}[G] \otimes_{\mathbb{C}[H]} \mathbb{C} \cong \mathbb{C}^2, where \mathbb{C} in the tensor product is considered as a left \mathbb{C}[H]-module and the action is defined by \rho, i.e. rw=\rho(r)(w), for all r \in \mathbb{C}[H] and w \in \mathbb{C}. Now, \{H,g_1H\} is a set of all left cosets of H in G. Thus if we let v_1=1 \otimes_{\mathbb{C}[H]} 1 and v_2=g_1 \otimes_{\mathbb{C}[H]} 1, then \mathcal{B}=\{v_1,v_2\} is a \mathbb{C}-basis for V. Note that \mathcal{B} corresponds to the standard basis of \mathbb{C}^2. Now, to find \overline{\rho}, I just need to find, for all g \in G, the matrix of \overline{\rho}(g) with respect to \mathcal{B}. We know that an element g \in G is either in the form g_1g_2^j, \ 0 \leq j < m, or in the form g_2^j, \ 0 \leq j < m. So we will consider two cases:

Case 1 . g = g_1g_2^j, \ 0 \leq j < m. In order to find the matrix of \overline{\rho}(g) with respect to \mathcal{B}, we only need to find \overline{\rho}(g)(v_i), \ i=1,2. We have

\overline{\rho}(g)(v_1) = g \otimes_{\mathbb{C}[H]}1=g_1 \otimes_{\mathbb{C}[H]}(g_2^j \cdot 1) = g_1 \otimes_{\mathbb{C}[H]} \rho(g_2^j)=g_1 \otimes_{\mathbb{C}[H]} \zeta^j=\zeta^jv_2.

Similarly, since gg_1=g_2^{-j}, we have

\overline{\rho}(g)(v_2)=g_2^{-j} \otimes_{\mathbb{C}[H]} 1 = 1 \otimes_{\mathbb{C}[H]} (g_2^{-j} \cdot 1)=1 \otimes_{\mathbb{C}[H]} \rho(g_2^{-j})=\zeta^{-j}v_1.

So in this case the matrix of \overline{\rho}(g) with respect to \mathcal{B} is  \begin{pmatrix} 0 & \zeta^{-j} \\ \zeta^j & 0 \end{pmatrix}.

Case 2 . g=g_2^j, \ 0 \leq j < m. A similar argument as case 1 shows that the matrix in this case is \begin{pmatrix} \zeta^j & 0 \\ 0 & \zeta^{-j} \end{pmatrix}. \ \Box

In case 1 (and case 2) in the above, the reason that we were allowed to move powers of g_2 to the right side of tensor product is that the tensor product is over \mathbb{C}[H] and any power of g_2 is in \mathbb{C}[H] because \langle g_2 \rangle =H \subset \mathbb{C}[H].

Note. We will keep all the notation and hypothesis given in this Definition.

Lemma 1. If W is a left \mathbb{C}[H]-module and g \in G, then g \mathbb{C}[H] \otimes_{\mathbb{C}[H]} W \cong W as \mathbb{C}-vector spaces.

Proof.  Clearly both g\mathbb{C}[H] \otimes_{\mathbb{C}[H]} W and W are \mathbb{C}-vector spaces. Define \varphi_1 : g \mathbb{C}[H] \times W \longrightarrow W and \psi_1 : W \longrightarrow g \mathbb{C}[H] \otimes_{\mathbb{C}[H]} W by \varphi_1(gr,w)=rw and \psi(w)=g \otimes_{\mathbb{C}[H]}w, for all r \in \mathbb{C}[H] and w \in W. Since \varphi_1 is \mathbb{C}[H]-balanced, it induces an abelian group homomorphism \varphi : g \mathbb{C}[H] \otimes_{\mathbb{C}[H]} W \longrightarrow W defined by \varphi(gr \otimes_{\mathbb{C}[H]}w)=rw, for all r \in \mathbb{C}[H] and w \in W. Obviously \varphi is also a \mathbb{C}-linear and \varphi \psi and \psi \varphi are identity maps. So \varphi is a \mathbb{C}-linear isomorphism. \Box

Lemma 2. Let [G:H]=m and suppose that \{g_1H, g_2H, \cdots , g_mH \} are the set of left cosets of H in G. Then \mathbb{C}[G] = g_1 \mathbb{C}[H] \oplus g_2 \mathbb{C}[H] \oplus \cdots \oplus g_m \mathbb{C}[H], as right \mathbb{C}[H]-modules.

Proof. By definition, every element of \mathbb{C}[G] is uniquely written as a finite \mathbb{C}-linear combination of elements of G and every element of G is in g_i H \subset g_i \mathbb{C}[H], for some 1 \leq i \leq m. So \mathbb{C}[G]=\sum_{i=1}^m g_i \mathbb{C}[H]. To show that this sum is direct, suppose that

\sum_{i=1}^m g_i u_i = 0, \ \ \ \ \ \ \ \ \ \ \ \ \ (1)

for some u_i \in \mathbb{C}[H]. We need to prove that u_i=0, for all i. To do that, let H=\{h_1, \cdots , h_k \} and u_i=\sum_{j=1}^k c_{ij} h_j, where 1 \leq i \leq m and c_{ij} \in \mathbb{C}. Then

0 = \sum_{i=1}^m g_iu_i = \sum_{i,j}c_{ij}g_ih_j. \ \ \ \ \ \ \ \ \ \ \ (2)

Now if in (2), g_ih_j = g_rh_s, for some i,j,r,s, then g_i \in g_rH and hence i=r, because the cosets are disjoint. So h_j=h_s, i.e. j=s. Thus, in (2), the elements g_ih_j are pairwise distinct and hence, since every element of \mathbb{C}[G] is uniquely written as a \mathbb{C}-linear combination of elements of G, we get c_{ij}=0, for all i,j. Therefore u_i=0 for all i. \ \Box

Theorem. If \rho: H \longrightarrow \text{GL}(W) is a representation of H, then \deg \text{Ind}_H^G \rho = [G:H] \deg \rho.

Proof.  Let [G:H]=m and suppose that \{g_1H, g_2H, \cdots , g_mH \} are the set of left cosets of H in G. Recall that tensor product distributes over direct sum. Thus by Lemma 2

\mathbb{C}[G] \otimes_{\mathbb{C}[H]} W \cong \bigoplus_{i=1}^m (g_i \mathbb{C}[H] \otimes_{\mathbb{C}[H]} W)

and hence by Lemma 1

\deg \text{Ind}_H ^G \rho = \dim_{\mathbb{C}} (\mathbb{C}[G] \otimes_{\mathbb{C}[H]} W) = \sum_{i=1}^m \dim_{\mathbb{C}} W = \sum_{i=1}^m \deg \rho = m \deg \rho. \ \Box

We are now going to explain how we can find a representation of G from a representation of a subgroup of G.

Definition. Let G be a finite group and let H be a subgroup of G. Let \rho : H \longrightarrow \text{GL}(W) be a representation of H. Let V = \mathbb{C}[G] \otimes_{\mathbb{C}[H]} W and define \overline{\rho} : G \longrightarrow \text{GL}(V) by \overline{\rho}(g)(x \otimes_{\mathbb{C}[H]} w)=(gx) \otimes_{\mathbb{C}[H]} w, for all g \in G, \ x \in \mathbb{C}[G] and w \in W. We call \overline{\rho} the representation induced from H to G and we write \overline{\rho} = \text{Ind}_H^G \rho.

The first thing to do is to see why the definition basically makes sense. I will explain this in the following remarks.

Remark 1. Recall that if R and S are rings, M is an (S,R)-bimodule and N is a left R-module, then M \otimes_R N is a left S-module. The left action of S on M \otimes_R N is defined by s(x \otimes_R y)=(sx) \otimes_R y, for all s \in S, \ x \in M and y \in N. Of course, as usual, the action extends to all elements of M \otimes_R N by linearity. Now, in the above definition, we know that W is a left \mathbb{C}[H]-module (see Remark 1 in here if you have already forgotten!) and obviously \mathbb{C}[G] is an (\mathbb{C}[G],\mathbb{C}[H]-bimodule because \mathbb{C}[H] is a subalgebra of \mathbb{C}[G]. Thus V = \mathbb{C}[G] \otimes_{\mathbb{C}[H]} W is indeed a \mathbb{C}[G]-module and the left action of \mathbb{C}[G] on V is defined by u(x \otimes_{\mathbb{C}[H]} w) = (ux) \otimes_{\mathbb{C}[H]} w, for u,x \in \mathbb{C}[G] and w \in W. This is why we defined \overline{\rho} as you see.

Remark 2. It is Clear that \overline{\rho}(1) = \text{id}_V. Also, we have \overline{\rho}(g_1g_2)(x \otimes_{\mathbb{C}[H]} w)=(g_1g_2x) \otimes_{\mathbb{C}[H]} w and \overline{\rho}(g_1) \overline{\rho}(g_2)(x \otimes_{\mathbb{C}[H]} w) = \overline{\rho}(g_1)(g_2x \otimes_{\mathbb{C}[H]} w)=(g_1g_2x) \otimes_{\mathbb{C}[H]} w, for all g_1,g_2 \in G, \ x \in \mathbb{C}[G] and w \in W. Thus \overline{\rho}(g_1g_2)=\overline{\rho}(g_1) \overline{\rho}(g_2) and so \overline{\rho} is a group homomorphism, i.e. \overline{\rho} is a representation of G.

In the second part of this note, I will find a formula for \deg \overline{\rho} in terms of \deg \rho.

We introduced the dual representation \rho^* of a representation \rho of a finite group G in the third construction in this post. We will keep notations we used in that post. I’m now going to show that \rho^* is indeed a representation and it is irreducible if and only if \rho is irreducible.

Remark 1\rho^* is a representation of G.

Proof. First we need to show that \rho^* is well-defined, i.e. \rho^*(g) \in \text{GL}(V^*) for all g \in G. The fact that \rho^*(g) is \mathbb{C}-linear follows trivially from the fact that the elements of V^* are linear. Also, if \rho^*(g)(f)=0, for some g \in G and f \in V^*, then f \rho(g^{-1})(v) = 0, for all v \in V. But, since \rho(g^{-1}) is an isomorphism, \rho(g^{-1}) is onto and hence f = 0. Finally, \rho^* is a group homomorphism because if g_1,g_2 \in G and f \in V^*, then

\rho^*(g_1g_2)(f) = f \rho(g_2^{-1}g_1^{-1})=f \rho(g_2^{-1}) \rho(g_1^{-1})=\rho^*(g_1)(f \rho(g_2^{-1}))= \rho^*(g_1) \rho^*(g_2)(f). \Box

Remark 2. Suppose that U is a \mathbb{C}[G]-submodule of V^* and let W = \{v \in V: \ f(v)=0, \ \forall f \in U \}.

1) W is a \mathbb{C}[G]-submodule of V.

2) If \rho is irreducible, then \rho^* is irreducible.

Proof. 1) f \rho(g)=\rho^*(g^{-1})(f) =0 over W for all f \in U, because \rho^*(g^{-1})(U) \subseteq U. So \rho(g)(W) \subseteq W.

2) Since \rho is irreducible, either W=(0) or W=V. Clearly if W=V, then U=(0). Now, supppose that W=(0). Let \dim_{\mathbb{C}}V = \dim_{\mathbb{C}}V^*=n and choose a basis \{f_1, \cdots , f_m \} for U. Consider the natural linear map \varphi : V \longrightarrow \bigoplus_{i=1}^m (V/\ker f_i), which is injective because W=(0). But \dim_{\mathbb{C}} (V/\ker f_i) \leq 1, because V/\ker f_i are all \mathbb{C}-subspaces of \mathbb{C}. Hence

n=\dim_{\mathbb{C}}V \leq \sum_{i=1}^m \dim_{\mathbb{C}} (V/\ker f_i) \leq m,

which implies that m=n and so U=V^*. \Box

Remark 3.  Suppose that W is a \mathbb{C}[G]-submodule of V and let U=\{f \in V^* : \ f(W)=(0) \}.

1) U is a \mathbb{C}[G]-submodule of V^*.

2) If \rho^* is irreducible, then \rho is irreducible.

Proof.  1) Since \rho(g^{-1})(W) \subseteq W, we have \rho^*(g)(f)(W)=f \rho(g^{-1})(W)=(0), for all f \in U.

2) Since \rho^* is irreducible, either U=(0) or U=V^*. Clearly if U=V^*, then W=(0). So suppose that U=(0). If W \neq V, then there exists a \mathbb{C}-subspace W_1 \neq (0) of V such that V=W \oplus W_1. Choose 0 \neq v_1 \in W_1 and extend it to a basis \{v_1, \cdots , v_n \} for V. Now define f \in V^* by f(\sum_{i=1}^n c_i w_i) = c_1w_1. Clearly 0 \neq f \in U. Contradiction! \Box