List 3

1. Show that ${\mathrm{End}}_{\mathbb{R}[G]}V$ is $\mathbb{C}$ for $V$ of complex type, ${\mathrm{Mat}}_2(\mathbb{R})$ for $V$ of real type, and $\mathbb{H}$ for $V$ of quaternionic type.

\begin{proof} Assume that $V$ be of real type, then $V=V_{\mathbb{R}}\otimes \mathbb{C}\simeq W\oplus W$ where $W$ is an absolutely irreducible $\mathbb{R}G$-module. Assume that $\phi :G\to \mathrm{GL}(V)={\mathrm{GL}}_n(\mathbb{R})$ is the corresponding representation of $W$, then $\phi (\mathbb{R}G)={\mathrm{M}}_{n\times n}(\mathbb{R})$ by $W$ absolutely irreducible. Assume that $\psi :G\to W\oplus W$, then $\psi (\mathbb{R}G)=\left\{\left[\begin{array}{cc}A& \\ & A\end{array}\right]:A\in {\mathrm{M}}_{n\times n}(\mathbb{R})\right\}$. It deduces that

$${\mathrm{End}}_{\mathbb{R}G}(W\oplus W)\simeq \left\{x\in {\mathrm{M}}_{n\times n}(\mathbb{R}):x\left[\begin{array}{cc}A& \\ & A\end{array}\right]=\left[\begin{array}{cc}A& \\ & A\end{array}\right]x\right\}=\left\{x\in {\mathrm{M}}_{n\times n}(\mathbb{R}):x=\left[\begin{array}{cc}aI& bI\\ cI& dI\end{array}\right]\right\}\simeq {\mathrm{M}}_2(\mathbb{R}).$$

Assume that $V$ is not of real type, then $V$ is irreducible over $R$ and so ${\mathrm{End}}_{\mathbb{R}[G]}V$ is a division ring. Recall that

Let $V$ be a finite dimensional vector space over $\mathbb{R}$. If there exists a multiplication such that $(V,\cdot )$ is a division ring, then $V$ is $\mathbb{R}$, $\mathbb{H}$ or $\mathbb{C}$.

Hence ${\mathrm{End}}_{\mathbb{R}[G]}V\in \{\mathbb{C},\mathbb{R},\mathbb{H}\}$. Define $S:V\to V,v\mapsto iv$, then $S\in {\mathrm{End}}_{\mathbb{R}[G]}V$ and $S^2=-\mathrm{id}$. Hence, ${\mathrm{End}}_{\mathbb{R}[G]}V\in \{\mathbb{C},\mathbb{H}\}$.

Using the argument in exercise 2, if $V$ is of quaternionic type, there exists a linear transformation $\gamma :V\to V$ over $\mathbb{R}$ such that ${\gamma }^2=-\mathrm{id}$ and $\gamma \ne S$. Hence $S,\gamma \in {\mathrm{End}}_{\mathbb{R}[G]}V$ and so ${\mathrm{End}}_{\mathbb{R}[G]}=\mathbb{H}$.

Assume that $V$ is of complex type. If ${\mathrm{End}}_{\mathbb{R}G}(V)=\mathbb{H}$, since $S\in {\mathrm{End}}_{\mathbb{R}G}(V)$, there exists $T\in {\mathrm{End}}_{\mathbb{R}(G)}(V)$ such that $T^2=-\mathrm{id}$ and $ST=-TS$. Define $B(x,y)=H(x,Ty)$. Since $B(x,iy)=H(x,Tiy)=-iH(x,Ty)=H(-ix,Ty)=B(-ix,y)$, $B$ is a bilinear form of $V$, which is a contradiction. \end{proof}

2. Show that $V$ is of real type if and only if $V$ is the complexification of a representation $V_{\mathbb{R}}$ over the field of real numbers.

\begin{proof} Assume that $V$ is the complexification of a representation $V_{\mathbb{R}}$ over the field of real numbers, then there exists a basis $v_1,\cdots ,v_n$ of $V$ such that the matrix of $\rho (g)$ with respect to this basis has real entries. Let $\{v_1,\cdots ,v_n\}$ be a set of basis of $V_{\mathbb{R}}$, then there exists a positive definite symmetric $\mathbb{R}$-bilinear form $B_0$ as $B_0({\sum }_{i=1}^n{s}_i{v}_i,{\sum }_{i=1}^n{t}_i{v}_i)={\sum }_{i=1}^n{s}_i{t}_i$ with $s_i,t_i\in \mathbb{R}$. Then define $B:V\times V\to \mathbb{C}$ as

$$B_0(v,w)=\sum _{i,j}{k}_i{\ell }_j{B}_0(v_i,v_j)$$

where $v={\sum }_{i=1}^n{k}_i{v}_i$, $w={\sum }_{j=1}^n{\ell }_j{v}_j$, $k_i,{\ell }_j\in \mathbb{C}$ and $B(v_i,v_j)=\frac{1}{|G|}{\sum }_{g\in G}{B}_0(g{v}_i,g{v}_j)$. It is easy to verify $B$ is a symmetric bilinear form on $V$ and $\chi$ is real-valued. Thus, $V$ is of real type.

Conversely, assume that $V$ is of real type, then $\chi$ is real-valued and there exists a symmetric bilinear form $B$ on $V$ preserved by $G$.

Define $\phi :V\to V^{\ast },x\mapsto B(x,\cdot )$. Note that

$$\phi (gx)(y)=B(gx,y)=B(x,g^{-1}y)=\phi (x)(g^{-1}y)=(g\phi (x))(y),$$

then $\phi$ is an isomorphism of modules. Let $H_0:V\times V\to \mathbb{C}$ be a positive definite non-degenerated Hermitian form. Define $H(x,y):=\frac{1}{|G|}{\sum }_{g\in G}{H}_0(gx,gy)$ for all $x,y\in V$, then $H$ is $G$-invariant. Furthermore, define

$$\psi :V\to V^{\ast },x\mapsto H(x,\cdot )$$

and $\alpha ={\psi }^{-1}\circ \phi$, then $\alpha$ is $G$-invariant, and it satisfies that $\alpha (v+w)=\alpha (v)+\alpha (w)$ and $\alpha (\lambda v)=\overline{\lambda }\alpha (v)$ for all $\lambda \in \mathbb{C}$. Thus, $\alpha$ is an isomorphism of real vector spaces. Consider ${\alpha }^2:V\to V$. Since ${\alpha }^2$ is an isomorphism of vector space and ${\alpha }^2(gv)=g{\alpha }^2(v)$, by Schur’s lemma ${\alpha }^2=\lambda I$ for some $\lambda \in \mathbb{C}$.

Using the definition of $\alpha$, we have that

$$H(\alpha (v),\cdot )=B(v,\cdot )\text{mbox}andH(\lambda v,\cdot )=B(\alpha (v),\cdot ),$$

which deduces

$$H(\alpha (v),\alpha (v))=B(v,\alpha (v))=B(\alpha (v),v)=\lambda H(v,v)$$

and $\lambda \in {\mathbb{R}}_{+}$. Define $\gamma ={\lambda }^{-1/2}\alpha$, then $\gamma :V\to V$ is a isomorphism of real vector spaces and ${\gamma }^2=\mathrm{id}$. Hence, $V$ is a sum of real eigenspaces $V_{+}$ and $V_{-}$ corresponding to eigenvalues $1$ and $-1$. Finally, $\gamma (iv)=-iv$ yields that $i{V}_{+}=V_{-}$ and so $V=V_{+}\otimes \mathbb{C}$. \end{proof}

3. Show that any nontrivial finite group of odd order has an irreducible representation which is not defined over $\mathbb{R}$.

\begin{proof} Assume that $G$ is a nontrivial group of odd order and $\chi$ is a nontrivial irreducible character. Since $|G|$ is odd, the map $\phi :G\to G,g\mapsto g^2$ is injective. It follows that

$$FS(\chi )=\frac{1}{|G|}\sum _{g\in G}\chi (g^2)=\frac{1}{|G|}\sum _{g\in G}\chi (g)\psi (g^{-1})=⟨\chi ,\psi ⟩$$

where $\psi$ is the trivial representation of $G$. Note that $FS(\chi )=0$ as $\chi$ is irreducible, then $\chi$ is not real-valued.

Let $\rho$ be the corresponding representation of $\chi$. There exists $g\in G$ such that $\mathrm{tr}(\rho (g))$ is complex and so the sum of eigenvalues of $\rho (g)$ is complex. It deduces that $\rho (g)$ can not be defined over $\mathbb{R}$. \end{proof}

4. Let $D_5$ denote the dihedral group with ten elements which has presentation

$$D_5=⟨x,y\mid x^5=1,y^2=1,yx=x^{-1}y⟩.$$

Identify the subgroup generated by $x$ with $C_5$.

• a) Find the characters values of ${\mathrm{Ind}}_{C_5}^{D_5}{V}_a$ for the one-dimensional representations of $C_5$ defined by ${\rho }_a(x)={\omega }^a,a=0,1,2,3,4,\omega =\exp \left(\frac{2\pi i}{5}\right)$. Which of these induced characters are irreducible?
• b) Use part a) to determine the character table of $D_5$.
• c) Calculate the Frobenius-Schur indicator for each irreducible character. For the real-valued characters find $D_5$-invariant bilinear forms on the representation spaces.

\begin{proof} a) The conjugacy classes of $D_5$ are

$$\{1\},\{x,x^4\},\{x^2,x^3\},\{x^{i}y:0⩽i⩽4\}.$$

Assume that ${\rho }_a(x)={\omega }^a$ is a representation of $C_5$, and assume that $\varphi :=\psi {\uparrow }_{C_5}^{D_5}$ is the corresponding induced representation. Then $\varphi (1)=2$, $\varphi (a)=\frac{1}{5}(5\cdot \psi (a)+5\cdot \psi (a^{-1}))=2\mathrm{re}({\omega }^a)$, $\varphi (a^2)=\psi (a^2)+\psi (a^3)=2\mathrm{re}({\omega }^{2a})$ and $\varphi (b)=\frac{1}{5}({\sum }_{k=0}^4{\omega }^{ak})$. Thus the characters values of ${\mathrm{Ind}}_{C_5}^{D_5}{V}_a$ are

	$1$	$a$	$a^2$	$b$
${\varphi }_0$	$2$	$2$	$2$	$1$
${\varphi }_1$	$2$	$2\mathrm{re}(\omega )$	$2\mathrm{re}({\omega }^2)$	$0$
${\varphi }_2$	$2$	$2\mathrm{re}({\omega }^2)$	$2\mathrm{re}({\omega }^4)$	$0$
${\varphi }_3$	$2$	$2\mathrm{re}({\omega }^3)$	$2\mathrm{re}(\omega )$	$0$
${\varphi }_4$	$2$	$2\mathrm{re}({\omega }^4)$	$2\mathrm{re}({\omega }^3)$	$0$

Note that $(\frac{\sqrt{5}+1}{2}{)}^2+(\frac{\sqrt{5}-1}{2}{)}^2=3/4$, ${\varphi }_i$ are irreducible for $i\in \{1,2,3,4\}$.

ii) By i) there are two irreducible $2$-dimensional representation. Then $D_5$ has two irreducible $1$-dimensional representation by $10=2^2\cdot 2+1^2\cdot 1$, and one of them is the trivial one.

	$1$	$a$	$a^2$	$b$
${\chi }_1$	$1$	$1$	$1$	$1$
${\chi }_2$	$1$	$m$	$n$	$k$
${\chi }_3$	$2$	$\frac{\sqrt{5}-1}{2}$	$-\frac{\sqrt{5}+1}{2}$	$0$
${\chi }_4$	$2$	$-\frac{\sqrt{5}+1}{2}$	$\frac{\sqrt{5}-1}{2}$	$0$

By Schur’s orthogonality relations $m,n,k$ satisfy that $1+m-2=0$, $1+n-2=0$ and $1+k=0$. Therefore, $m=n=1$ and $k=-1$.

iii) The Frobenius-Schur indicators of ${\chi }_i$ are $FS({\chi }_i)=1$ for $i\in \{1,2,3,4\}$, as they are all real-valued.

When $\dim \chi =1$, it is easy to verify $B(e_1,e_1)=1$ is a $G$-invariant bilinear form. When $\dim \chi =2$, define $B_0(e_i,e_j)={\delta }_{ij}$ for $i,j\in \{1,2\}$ and define $B(u,v)=\frac{1}{|G|}{\sum }_{g\in G}{B}_0(u,v)$. It deduces that $B(e_1,e_1)=B(e_2,e_2)=1$ and $B(e_1,e_2)=0$. So $B=B_0$ is the $D_5$-invariant bilinear form. \end{proof}

5. By the theorem on Frobenius-Schur indicator, any self-dual simple $G$-module $V$ possesses a unique (up to scalar) non-zero $G$-invariant bilinear form. Prove that this form is non-degenerate, that is, $B\left(v,v^{\prime }\right)=0$ for all $v^{\prime }\in V$ implies that $v=0$.

\begin{proof} Assume that $B(v,v^{\prime })=0$ for all $v^{\prime }\in V$, then it follows that $B(v^g,v^{\prime g})=B(v,v^{\prime })=0$ for all $w^{\prime }=v^{\prime g}\in V$. Since $V$ is a simple module, $v^G=V$ and so for any $w\in V$, $B(w,w^{\prime })=0$ for all $w^{\prime }\in V$. It yields that $B$ is zero, contradiction. Therefore, $B$ is non-degenerate. \end{proof}

6. Consider $S_3$ as a natural subgroup of $S_4$. Find the irreducible decomposition of the restriction ${\mathrm{Res}}_{S_3}^{S_4}$ for each irreducible character ${\chi }_i$ of $S_4$.

\begin{proof} The irreducible character table of $S_4$ is

	$(1)$	$(12)(34)$	$(12)$	$(123)$	$(1234)$
${\rho }_1$	1	1	1	1	1
${\rho }_2$	1	1	-1	1	-1
${\rho }_3$	2	2	0	-1	0
${\rho }_4$	3	-1	-1	0	1
${\rho }_5$	3	-1	1	0	-1

Then the restriction ${\mathrm{Res}}_{S_3}^{S_4}$ for each irreducible character ${\chi }_i$ are

	$(1)$	$(12)$	$(123)$
${\rho }_1$	1	1	1
${\rho }_2$	1	-1	1
${\rho }_3$	2	0	-1
${\rho }_4$	3	-1	0
${\rho }_5$	3	1	0

Note that ${\rho }_i$ are irreducible for $i\in \{1,2,3\}$, and ${\rho }_4={\rho }_2+{\rho }_3$, ${\rho }_5={\rho }_1+{\rho }_3$. \end{proof}

7. Suppose that a finite group $G$ has a left action on a finite set $X$, and obtain a linear representation $\rho$ of $G$ of degree $|X|$ by identifying $X$ with the basis elements of a vector space.

• a) Show that if $\chi$ is the character of $\rho$ then the value $\chi$ takes on an arbitrary $g\in G$ equals the number of $x\in X$ such that $gx=x$.
• b) Show that the inner product of $\chi$ and the trivial character of $G$ is equal to ${\Sigma }_{x\in X}\frac{1}{\left[G:G_x\right]}$, where $\left[G:G_x\right]$ is the index of the subgroup $G_x=\{g\in G\mid gx=x\}$ in $G$.

\begin{proof} a) Assume that $|X|=n$. For any $g\in G$, if $g{x}_i=x_{\sigma (i)}$, then $\rho (g)={\sum }_{i=1}^n{E}_{i\sigma (i)}$ and so $\chi (g)=\mathrm{tr}(\rho (g))=\#\{i:i=\sigma (i)\}=\#\{x:gx=x\}$. Therefore, the value $\chi$ takes on an arbitrary $g\in G$ equals the number of $x\in X$ such that $gx=x$.

b) The inner product of $\chi$ and the trivial character of $G$ is

$$\begin{array}{rl}⟨\chi ,\mathrm{tr}⟩& =\frac{1}{|G|}\sum _{g\in G}\chi (g)\mathrm{tr}(g^{-1})=\frac{1}{|G|}\sum _{g\in G}\chi (g)\\ & =\frac{1}{|G|}\sum _{g\in G}\#\{x:gx=x\}=\frac{1}{|G|}\#\{(g,x):gx=x\}\\ & =\frac{1}{|G|}\sum _{x\in X}|G_x|=\sum _{x\in X}\frac{1}{[G:G_x]}.\end{array}$$

Now we finish the proof. \end{proof}

8. Let $V_1,\dots ,V_r$ be a complete list of irreducible representations of a group $G$ and let ${\chi }_1,\dots ,{\chi }_r$ be their respective characters.

• a) Show that for any element $g\in G$ such that $g\ne 1$ the following identity holds ${\sum }_{i=1}^r\dim V_i{\chi }_i(g)=0$.
• b) An element $g\in G$ is called an involution if $g^2=1$. Show that the number of involutions in the group $G$ coincides with $\frac{1}{|G|}{\sum }_{g\in G}{\sum }_{i=1}^r\dim V_i{\chi }_i\left(g^2\right)$.
• c) Hence show that the number of involutions in $G$ equals the difference ${\sum }_{V_i\in \text{ Sym }}\dim V_i-{\sum }_{V_i\in \text{ Skew }}\dim V_i$, where $\mathrm{Sym}$ is the set of $G$-modules of symmetric type and $\mathrm{Skew}$ is the set of $G$-modules of skew-symmetric type.

\begin{proof} a) By Schur’s orthogonality relations, ${\sum }_{i=1}^r\mathrm{tr}(h){\chi }_i(g)=0$ for any $g\ne h$. Take $h=1$ and $g\ne 1$, then we get ${\Sigma }_{i=1}^r\dim V_i{\chi }_i(g)=0$.

b) If $g^2\ne 1$, then ${\sum }_{i=1}^r\dim V_i{\chi }_i\left(g^2\right)=0$ by a). If $g^2=1$, then ${\sum }_{i=1}^r\dim V_i{\chi }_i\left(g^2\right)=|G|/|C_{\mathrm{id}}|$ where $C_{\mathrm{id}}$ is the conjugacy class and $|C_{\mathrm{id}}|=1$. Therefore, $\frac{1}{|G|}{\sum }_{i=1}^r\dim V_i{\chi }_i\left(g^2\right)=1$ for any $g^2=1$. Hence,

$$\frac{1}{|G|}\sum _{g\in G}\sum _{i=1}^r\dim V_i{\chi }_i\left(g^2\right)=\sum _{g\in G}\left(\frac{1}{|G|}\sum _{i=1}^r\dim V_i{\chi }_i\left(g^2\right)\right)=\#\{g\in G:g^2=1\}.$$

c) By Frobenius-Schur indicator, $\frac{1}{|G|}{\sum }_{g\in G}{\chi }_i(g^2)=1$ if $V_i\in \mathrm{Sym}$, and $\frac{1}{|G|}{\sum }_{g\in G}{\chi }_i(g^2)=-1$ if $V_i\in \mathrm{Skew}$. Thus there is

$$\frac{1}{|G|}\sum _{g\in G}\sum _{i=1}^r\dim V_i{\chi }_i\left(g^2\right)=\sum _{i=1}^r\dim V_i\left(\frac{1}{|G|}\sum _{g\in G}{\chi }_i(g^2)\right)=\sum _{V_i\in \text{ Sym }}\dim V_i-\sum _{V_i\in \text{ Skew }}\dim V_i$$

and so we finish the proof. \end{proof}

9. Check that if $K\subset H\subset G$ are groups and if $V$ is a representation of $K$, then

$${\mathrm{Ind}}_H^G{\mathrm{Ind}}_K^{H}V\simeq {\mathrm{Ind}}_K^{G}V.$$

\begin{proof} Assume that $[H:K]=m$, $[G:H]=n$, $H=t_{1}K\bigsqcup \cdots \bigsqcup t_{m}K$ and $G=s_{1}H\bigsqcup \cdots \bigsqcup s_{n}H$. Then we have

$$G=\underset{1⩽i⩽n,1⩽j⩽m}{⨆}{s}_i{t}_{j}K.$$

Recall that $\mathbb{C}G\otimes V$ is a $G$-module with respect to the action $g(u\otimes w)=gu\otimes w$. By ^hp5t8s,

$$\begin{array}{rl}{\mathrm{Ind}}_H^G{\mathrm{Ind}}_K^{H}V& \simeq {\mathrm{Ind}}_H^G((t_1\otimes V)\oplus \cdots \oplus (t_m\otimes V))\\ & \simeq (s_1(t_1\otimes V)\oplus \cdots \oplus (t_m\otimes V))\oplus \cdots \oplus (s_n(t_1\otimes V)\oplus \cdots \oplus (t_m\otimes V))\\ & \simeq (s_1{t}_1\otimes V)\oplus \cdots \oplus (s_n{t}_m\otimes V)\\ & \simeq {\mathrm{Ind}}_K^G.\end{array}$$

Now we finish the proof. \end{proof}

10. Compute the decomposition into irreducible modules of all the representations of $A_5$ induced from the irreducible representations of $C_5,A_4$ and $C_2\times C_2$.

I find a solution of 10: see here. The character table of $A_5$ is not easy to find, in fact, it is obtained from the induced representation of $A_4$.

\begin{proof}

We compute the irreducible representations of $C_5$, $A_4$ and $C_2\times C_2$.

i) Let $a=(12345)$ and $C_5=⟨a⟩$, then all irreducible representations of $C_5$ are determined by ${\rho }_k(a)={\omega }^k$ where $k\in \{0,1,2,3,4\}$ and denote the corresponding characters as ${\psi }_k$. The induced characters are ${\psi }_k{\uparrow }_{C_5}^{A_5}$ and the values are

	$(1)$	$(123)$	$(12)(34)$	$(12345)$	$(21345)$
${\psi }_0$	12	0	0	2	2
${\psi }_1$	12	0	0	$2\mathrm{re}({\omega }^1)$	$2\mathrm{re}({\omega }^2)$
${\psi }_2$	12	0	0	$2\mathrm{re}({\omega }^2)$	$2\mathrm{re}({\omega }^4)$
${\psi }_3$	12	0	0	$2\mathrm{re}({\omega }^3)$	$2\mathrm{re}({\omega }^1)$
${\psi }_4$	12	0	0	$2\mathrm{re}({\omega }^4)$	$2\mathrm{re}({\omega }^3)$

where $\omega =e^{2\pi i/5}$.

ii) Recall the character table of $A_4$ is

	$(1)$	$(123)$	$(132)$	$(12)(34)$
Class size	1	4	4	3
${\psi }_1$	1	1	1	1
${\psi }_2$	1	$\omega$	${\omega }^2$	1
${\psi }_3$	1	${\omega }^2$	$\omega$	1
${\psi }_4$	3	0	0	-1

where $\omega =e^{2\pi i/3}$. Then the induces characters are

	$(1)$	$(123)$	$(12)(34)$	$(12345)$	$(21345)$
${\psi }_1$	5	2	1	0	0
${\psi }_2$	5	-1	1	0	0
${\psi }_3$	5	-1	1	0	0
${\psi }_4$	15	0	-1	0	0

and ${\psi }_2,{\psi }_3$ are irreducible.

Remark. “We claim by pure thought that their inductions must be irreducible!”: See here.

• If ${\psi }_i$ is reducible with $\dim {\psi }_i=5$, then it must contain a trivial representation.
• By 9 Frobenius Reciprocity, the multiplicity of the trivial representation of $A_5$ in ${\mathrm{Ind}}_{A_4}^{A_5}({\psi }_i)$ is the same as that of ${\psi }_i$ in the trivial representation of $A_4$, and so this multiplicity is zero if $\psi \ne 1$.

iii) Now the character table of $A_5$ can be written as

	$(1)$	$(123)$	$(12)(34)$	$(12345)$	$(21345)$
Class size	1	20	15	12	12
${\chi }_1$	1	1	1	1	1
${\chi }_2$	3	0	X	Y	Z
${\chi }_3$	3	0	X’	Y’	Z’
${\chi }_4$	4	1	0	-1	-1
${\chi }_5$	5	-1	1	0	0

and we need to find $X,Y$ and $Z$. By Schur orthogonality relations, we can solve that $X=-1$ and $Y=Z^{\prime }$ and $Y^{\prime }=Z$ with $Y+Z=1$, $|Y{|}^2+|Z{|}^2=3$.

Note that $g$ and $g^{-1}$ are conjugate for all $g\in A_5$, then we know $Y,Z\in \mathbb{R}$. Then we can solve that $Y=\frac{1+\sqrt{5}}{2}$ and $Z=\frac{1-\sqrt{5}}{2}$ and the character table of $S_5$ is completed, that is

	$(1)$	$(123)$	$(12)(34)$	$(12345)$	$(21345)$
Class size	1	20	15	12	12
${\chi }_1$	1	1	1	1	1
${\chi }_2$	3	0	-1	$\frac{1+\sqrt{5}}{2}$	$\frac{1-\sqrt{5}}{2}$
${\chi }_3$	3	0	-1	$\frac{1-\sqrt{5}}{2}$	$\frac{1+\sqrt{5}}{2}$
${\chi }_4$	4	1	0	-1	-1
${\chi }_5$	5	-1	1	0	0

iv) Let $C_2\times C_2=⟨1,(12)(34),(13)(24),(14)(23)⟩$, then the character table of $C_2\times C_2$ is

	$(1)$	$(12)(34)$	$(13)(24)$	$(14)(23)$
${\psi }_1$	1	1	1	1
${\psi }_2$	1	1	-1	-1
${\psi }_3$	1	-1	1	-1
${\psi }_4$	1	-1	-1	1

and the induced characters are

	$(1)$	$(123)$	$(12)(34)$	$(12345)$	$(21345)$
${\psi }_1$	15	0	3	0	0
${\psi }_2$	15	0	-1	0	0
${\psi }_3$	15	0	-1	0	0
${\psi }_4$	15	0	-1	0	0

And the decomposition is easy to compute. \end{proof}

11. Let $G$ be a finite group and $H$ its subgroup. If $U$ is a representation of $G$ and $W$ a representation of $H$, show that (with all tensor products over $\mathbb{C}$ )

$$U\otimes \mathrm{IndW}\simeq \mathrm{Ind}(\mathrm{Res}(U)\otimes W)$$

In particular, $\mathrm{Ind}(\mathrm{Res}(U)\simeq U\otimes P$, where $P$ is the permutation representation of $G$ on $G/H$.

\begin{proof}

Error

Assume that $G=g_{1}H\bigsqcup \cdots \bigsqcup g_{m}H$, then

$$U\otimes \mathrm{Ind}W\simeq U\otimes ((g_1\otimes W)\oplus \cdots \oplus (g_m\otimes W))\simeq (U\otimes (g_1\otimes W))\oplus \cdots \oplus (U\otimes (g_m\otimes W)):=A$$

and with $\mathrm{Res}(U):=U_0$ we have

$$\mathrm{Ind}(\mathrm{Res}(U)\otimes W)\simeq (g_1\otimes (U_0\otimes W))\oplus \cdots \oplus (g_m\otimes (U_0\otimes W)):=B.$$

Hence, it suffices to show $A\simeq B$. Define $\phi :A\to B,u_i\otimes g_k\otimes w_j\mapsto g_k\otimes u_i\otimes w_j$ where $\{u_i{\}}_{i\in I}$ and $\{w_j{\}}_{j\in J}$ are bases of $U$ and $W$, respectively. For any $g\in G$, $g(u_i\otimes (g_k\otimes w_j))=g{u}_i\otimes (g{g}_k\otimes w_j)$ and $g(g_k\otimes (u_i\otimes w_j))=g{g}_k\otimes (u_i\otimes w_j)$. Unfortunately, they are not equal and this method fails.

Let $\chi$ be the character of $U$, and let $\psi$ be the character of $W$. To show $U\otimes \mathrm{IndW}\simeq \mathrm{Ind}(\mathrm{Res}(U)\otimes W)$, it suffices to show for any irreducible character $\alpha$ there is $⟨\chi \psi {\uparrow }^G,\alpha ⟩=⟨(\chi {\downarrow }_H\psi ){\uparrow }^G,\alpha ⟩$.

Note that by Frobenius reciprocity, there is

$$⟨\chi \psi {\uparrow }^G,\alpha ⟩=⟨\psi {\uparrow }^G,\alpha \overline{\chi }⟩=⟨\psi ,(\alpha \overline{\chi }){\downarrow }_H⟩=⟨\chi {\downarrow }_H\psi ,\alpha {\downarrow }_H⟩=⟨(\chi {\downarrow }_H\psi ){\uparrow }^G,\alpha ⟩$$

and so we finish the proof. \end{proof}

12. Let $G$ be a finite group and $H$ its subgroup. If $C$ is a conjugacy class of $G$, and $C\cap H$ decomposes into conjugacy classes $D_1,\dots ,D_r$ of $H$, then the character formula for induced representation can be rewritten as the value of the character of $\mathrm{IndW}$ on $C$ is

$${\chi }_{IndW}(C)=\frac{|G|}{|H|}{\Sigma }_{i=1}^r\frac{\left|D_i\right|}{|C|}{\chi }_W\left(D_i\right)$$

\begin{proof} Note that

$$\begin{array}{rl}{\chi }_{\mathrm{Ind}W}(C)& =\frac{1}{|C|}\sum _{c\in C}{\chi }_{\mathrm{Ind}W}(c)=\frac{1}{|C||H|}\sum _{c\in C}\sum _{g\in G}{\chi }_W(g^{-1}cg)\\ & =\frac{1}{|C||H|}\sum _{i=1}^r{\chi }_W(D_i)(\#\{(c,g,x):g^{-1}cg\in D_i\})=\frac{1}{|C||H|}\sum _{i=1}^r{\chi }_W(D_i)\left(\sum _{g\in G,d\in D_i}\#\{c:c=gd{g}^{-1}\}\right)\\ & =\frac{1}{|C||H|}\sum _{i=1}^r{\chi }_W(D_i)|G||D_i|\end{array}$$

and we finish the proof. \end{proof}