## 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

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

## Induced representations; an example

Posted: February 23, 2011 in Representations of Finite Groups
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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 $\langle g_2 \rangle =H \subset \mathbb{C}[H].$

## Degree of induced representations

Posted: February 23, 2011 in Representations of Finite Groups
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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

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

## Induced representations

Posted: February 21, 2011 in Representations of Finite Groups
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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 $\deg \rho.$

## The dual representation

Posted: February 11, 2011 in Representations of Finite Groups
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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! $\Box$