Lecture 13

Solution: Let $v_{1},\ldots,v_{d}$ be an eigenbasis of $V$ for $\pi(gN)$, i.e. $\pi(gN)v_{i}=\lambda_{i}v_{i}$, and $\{v_{i}\}$ is a basis of $V$. Then $\chi_{\pi}(gN)=\sum_{i}\lambda_{i}$, and

and hence $\{v_{i}\}$ is an eigenbasis for $V$ w.r.t. $\widetilde{\pi}(g)$, giving $\chi_{\widetilde{\pi}}(g)=\sum_{i}\lambda_{i}=\chi_{\pi}(gN)$.

Prop. 13.7. Every normal subgroup $N$ of a finite group $G$ may be expressed as

for some collection $(\widetilde{\rho}_{i},V_{i})$ of irreducible representations of $G$.

Proceed as follows:

1. (a)
   ​

   Let $\{(\rho_{i},V_{i})\}$ denote the irreducible representations of $\widetilde{G}$, and let $(\widetilde{\rho}_{i},V_{i})$ be the lift of $(\rho_{i},V_{i})$ to $G$. Define

   $$K=\bigcap_{i=1}^{n}\ker\widetilde{\rho}_{i}.$$

   Show that $K$ is normal in $G$.

2. (b)
   ​

   Use Proposition 13.2 to show that $N\subset K$, and hence $|G/K|\leq|G/N|$.

3. (c)
   ​

   Use Theorem 6.6 to show that $|G/N|=\sum_{i}(\operatorname{dim}V_{i})^{2}$.

4. (d)
   ​

   Use Propositions 13.8 to show that each $(\widetilde{\rho}_{i},V_{i})$ is the lift of an irreducible representation representation $(\sigma_{i},V_{i})$ of $G/K$.

5. (e)
   ​

   Use Theorem 6.6 to show that $\sum_{i}(\operatorname{dim}V_{i})^{2}\leq|G/K|$.

6. (f)
   ​

   Conclude that $N=K$.

Solution:

1. (a)
   ​

   This is a direct consequence of the two following facts from group theory (i) If $\psi:G\rightarrow H$ is a group homomorphism, then $\ker\psi$ is a normal subgroup of $G$ (ii) The intersection of two normal subgroups is a normal subgroup.

2. (b)
   ​

   By Proposition 13.2, $N\subset\ker\widetilde{\rho}_{i}$ for each $i$, hence $N\subset\bigcap_{i}\ker\widetilde{\rho}_{i}=K$, hence $|G/K|\leq|G/N|$.

3. (c)
   ​

   By Theorem 6.6, $|G/N|=|\widetilde{G}|=\sum_{i}(\operatorname{dim}V_{i})^{2}$.

4. (d)
   ​

   Since $(\rho_{i},V_{i})$ is an irreducible representation of $\widetilde{G}$, its lift $(\widetilde{\rho}_{i},V_{i})$ is an irreducible representation of $G$. By construction $K\subset\ker\widetilde{\rho}_{i}$, hence by Proposition 13.2, we may express $(\widetilde{\rho}_{i},V_{i})$ as the lift of a representation $(\sigma_{i},V_{i})$ of $G/K$. Since $(\widetilde{\rho}_{i},V_{i})$ is an irreducible representation of $G$, Proposition 13.3 gives that $(\sigma_{i},V_{i})$ is an irreducible representation of $G/K$.

5. (e)
   ​

   We have shown that each $(\sigma_{i},V_{i})$ is an irreducible representation of $G/K$. Theorem 6.6 then gives $\sum_{i}(\operatorname{dim}V_{i})^{2}\leq|G/K|$, provided that the representations $(\sigma_{i},V_{i})$ are all pairwise non-isomorphic to each other. The representations $(\widetilde{\rho}_{i},V_{i})$ are non-isomorphic representations of $G/N$, hence for each pair $i,j$, we can find $g\in G/N$ such that

   $$\chi_{\widetilde{\rho}_{i}}(gN)\neq\chi_{\widetilde{\rho}_{j}}(gN).$$

   Then by Problem 77,

   $$\chi_{\widetilde{\rho}_{i}}(gN)=\chi_{\rho_{i}}(g)=\chi_{\sigma_{i}}(gK),$$

   and similarly for $\sigma_{j}$, giving

   $$\chi_{\sigma_{i}}(gK)\neq\chi_{\sigma_{j}}(gK),$$

   showing that the representations are non-isomorphic.

6. (f)
   ​

   By (b), (c), and (e), we have

   $$|G/K|\leq|G/N|=\sum_{i}(\operatorname{dim}V_{i})^{2}\leq|G/K|,$$

   hence $|G/N|=|G/K$, i.e. $|N|=|K|$. Since $N\subset K$, we then have that $N=K=\bigcap_{i=1}^{n}\ker\widetilde{\rho}_{i}$, as desired.

Solution: The character table of $S_{5}$ is

We recall that the conjugacy classes of $A_{5}$ are given as follows (cf. Problem 51 Lecture 10)

We know the trivial representation, and compute the norm of the restriction of $(\pi|_{A_{5}},W_{0})$:

so $(\pi|_{A_{5}},W_{0})$ is irreducible. We also consider $(\rho|_{A_{5}},V)$:

so this is also irreducible. The table so far reads

From this, we see that there must be two 3-dimensional irreducible representations. We now consider $(\pi\wedge\pi)|_{A_{5}}$. Observe that

so $\pi\wedge\pi$ is the direct sum of two distinct irreducible representations of $A_{5}$. We compute the following inner products:

so the two remaining irreducible representations we are looking for are both subrepresentations of $(\pi\wedge\pi)|_{A_{5}}$. The character table is then of the form

Since each element of $A_{5}$ is conjugate to its inverse $\chi_{\sigma_{i}}(g)=\chi_{\sigma_{i}}(g^{-1})=\overline{\chi_{\sigma_{i}}(g)}$, hence $a,b,c,d\in{\mathbb{R}}$. Using Theorem 19 for the norms of the columns:

Solving these gives $a=-1$, $b=0$, $c=\frac{1\pm\sqrt{5}}{2}$, $d=\frac{1\pm\sqrt{5}}{2}$. Column orthogonality gives that $d$ is the quadratic conjugate of $c$, giving the table

We see that each irreducible representation is faithful, i.e. has trivial kernel. By Proposition 13.7, we then have that $\{e\}$ is the only proper normal subgroup of $A_{5}$.

Solution: We outline two different solutions:

1. (i)
   ​

   Using analysis: We start by showing the following lemma:

   Lemma: If $|\chi_{\pi}(h)|=\operatorname{dim}V$, then $\pi(h)=\lambda{\,\mathrm{Id}}$ for some $\lambda\in{\mathbb{C}}^{*}$.

   Proof: $\chi_{\pi}(h)=\lambda_{1}+\lambda_{2}+\ldots+\lambda_{\operatorname{dim}V}$, where each $\lambda_{i}$ is an eigenvalue of $\pi(h)$, and hence a root of unity. Since

   $$\operatorname{dim}V=|\chi_{\pi}(h)|=|\lambda_{1}+\ldots+\lambda_{\operatorname{dim}V}|\leq|\lambda_{1}|+\ldots+|\lambda_{\operatorname{dim}V}|=\operatorname{dim}V,$$

   equality holds in the triangle equality if the numbers are colinear, we must have $\lambda_{1}=\lambda_{2}=\ldots+\lambda_{\operatorname{dim}V}$, hence $\pi(h)=\lambda_{1}{\,\mathrm{Id}}$.

   We now define $H\subset G$ by

   $$H=\{h\in G\,|\,|\chi_{\pi}(h)|=\operatorname{dim}V\}.$$

   Observe that by the Lemma above, $H$ is a subgroup of $G$: if $h_{1},h_{2}\in H$, then $\pi(h_{1})=\lambda_{1}{\,\mathrm{Id}}$, $\pi(h_{2})=\lambda_{2}{\,\mathrm{Id}}$, hence $\pi(h_{2}^{-1})=\lambda_{2}^{-1}{\,\mathrm{Id}}$, and so $\pi(h_{1}h_{2}^{-1})=\lambda_{1}\lambda_{2}^{-1}{\,\mathrm{Id}}$, giving $\chi_{\pi}(h_{1}h_{2}^{-1})=\lambda_{1}\lambda_{2}^{-1}\cdot\operatorname{dim}V$, and so $|\chi_{\pi}(h_{1}h_{2}^{-1})|=\operatorname{dim}V$, hence $h_{1}h_{2}^{-1}\in H$.

   Let $c=\max_{g\in G\setminus H}|\chi_{\pi}(g)|<\operatorname{dim}V$. Then for any irreducible representation $(\rho,W)$ of $G$, we have

   	$\displaystyle\langle\chi_{\rho},\chi_{\pi^{\otimes n}}\rangle_{G}$	$\displaystyle=\langle\chi_{\rho},\chi_{\pi}^{n}\rangle_{G}=\frac{|H|}{|G|}\langle\chi_{\rho|_{H}},\chi_{\pi|_{H}}^{n}\rangle_{H}+\frac{1}{|G|}\sum_{g\in G\setminus H}\chi_{\rho}(g)\chi_{\pi}(g)^{n}$
   		$\displaystyle=\frac{|H|}{|G|}\langle\chi_{\rho|_{H}},\chi_{\pi|_{H}}^{n}\rangle_{H}+O\left(\frac{|G|-|H|}{|G|}\cdot\operatorname{dim}W\cdot c^{n}\right)$

   For $h\in H$, we write $\pi(h)=\lambda_{h}{\,\mathrm{Id}}$, hence $\chi_{\pi|_{H}}^{n}(h)=(\operatorname{dim}V)^{n}\lambda_{h}^{n}$, and so

   $$\langle\chi_{\rho},\chi_{\pi^{\otimes n}}\rangle_{G}=(\operatorname{dim}V)^{n}\left(\sum_{h\in H}\chi_{\rho|_{H}}(h)\overline{\lambda_{h}}^{n}+O\left(\frac{|G|-|H|}{|G|}\cdot\operatorname{dim}W\cdot\left(\frac{c}{\operatorname{dim}V}\right)^{n}\right)\right)$$

   Since $\pi$ is faithful, we must have $\lambda_{h_{1}}\neq\lambda_{h_{2}}$ for $h_{1}\neq h_{2}\in H$ (otherwise $\lambda_{h_{1}h_{2}^{-1}}=1$, hence $h_{1}h_{2}^{-1}\in\ker\pi$). Thus $\psi:H\rightarrow{\mathbb{C}}^{\times}$, $\psi(h)=\lambda_{h}$ is an isomorphism from $H$ to a finite subgroup of the unit circle; $H$ is therefore cyclic, and the values taken by $\psi$ must be $1,\omega,\omega^{2},\ldots,\omega^{|H|-1}$, where $\omega=e^{2\pi i/|H|}$. Note that every other irreducible character of $H$ is of the form $\psi^{k}$ for some $k\in{\mathbb{Z}}/|H|{\mathbb{Z}}$. There then exist $k_{1},k_{2},\ldots k_{N}$ such that

   $$\chi_{\rho|H}=\sum_{i=1}^{N}\psi^{k_{i}},$$

   giving

   $$\sum_{h\in H}\chi_{\rho|_{H}}(h)\overline{\lambda_{h}}^{n}=\sum_{i=1}^{N}\sum_{h\in H}\psi^{k_{i}-n}(h).$$

   Letting $n=k_{1}+n^{\prime}|H|$ gives

   $$\sum_{h\in H}\psi^{k_{i}-(k_{1}+n^{\prime}|H|)}(h)=\begin{cases}|H|\quad&\mathrm{if}\;k_{i}\equiv k_{1}\quad(\mathrm{mod}\;|H|)\\ 0\quad&\mathrm{otherwise.}\end{cases}$$

   Summing over the $k_{i}$, we get

   $$\sum_{h\in H}\chi_{\rho|_{H}}(h)\overline{\lambda_{h}}^{k_{1}+n^{\prime}|H|}=\#\{1\leq i\leq N\,|\,k_{i}\equiv_{H}k_{1}\}$$

   We enter this into the expression for $\langle\chi_{\rho},\chi_{\pi^{\otimes(k_{1}+n^{\prime}|H|)}}\rangle_{G}$ obtained above, giving

   $$\langle\chi_{\rho},\chi_{\pi^{\otimes(k_{1}+n^{\prime}|H|)}}\rangle_{G}=(\operatorname{dim}V)^{k_{1}+n^{\prime}|H|}\left(\#\{1\leq i\leq N\,|\,k_{i}\equiv_{H}k_{1}\}+O\left(\left(\frac{c}{\operatorname{dim}V}\right)^{k_{1}+n^{\prime}|H|}\right)\right).$$

   Since $c<\operatorname{dim}V$, taking $n^{\prime}$ large enough gives

   $$\langle\chi_{\rho},\chi_{\pi^{\otimes(k_{1}+n^{\prime}|H|)}}\rangle_{G}=(\operatorname{dim}V)^{k_{1}+n^{\prime}|H|}\left(\#\{1\leq i\leq N\,|\,k_{i}\equiv_{H}k_{1}\}+\eta_{n^{\prime}}\right),$$

   with $|\eta_{n^{\prime}}|\leq\frac{1}{2}$. The right-hand side is then non-zero, giving $\langle\chi_{\rho},\chi_{\pi^{\otimes(k_{1}+n^{\prime}|H|)}}\rangle_{G}>0$, i.e. $\rho$ occurs as a subrepresentation of $\pi^{\otimes(k_{1}+n^{\prime}|H|)}$.

2. (ii)
   ​

   Using linear algebra: Denote the distinct non-zero values $\chi_{\pi}$ takes by $r_{1},r_{2},\ldots r_{m}$. We may assume that $r_{1}=\operatorname{dim}V$. For $i=1,\ldots,m$, define

   $$a_{i}=\sum_{g\in G\,|\,\chi_{\pi}(g)=r_{i}}\chi_{\sigma}(g)$$

   Using that $\pi$ is faithful, we have $a_{1}=\chi_{\pi}(e)=\operatorname{dim}W$. For each $n\in{\mathbb{N}}$, we have

   $$\langle\chi_{\rho},\chi_{\pi^{\otimes n}}\rangle_{G}=\frac{1}{|G|}\big{(}a_{1},a_{2},\ldots,a_{m}\big{)}\cdot\big{(}r_{1}^{n},r_{2}^{n},\ldots r_{m}^{n}\big{)}$$

   (here “$\cdot$” denotes the standard inner product on ${\mathbb{C}}^{m}$). We compute the determinant of the following Vandermonde matrix:

   $$\det\left(\begin{matrix}r_{1}&r_{2}&\ldots&r_{m}\\ r_{1}^{2}&r_{2}^{2}&\ldots&r_{m}^{2}\\ \vdots&\vdots&&\vdots\\ r_{1}^{m}&r_{2}^{m}&\ldots&r_{m}^{m}\end{matrix}\right)=r_{1}r_{2}\cdots r_{m}\prod_{1\leq i<j\leq m}(r_{j}-r_{i})\neq 0.$$

   The vectors $(r_{1},\ldots r_{m})$, $(r_{1}^{2},\ldots r_{m}^{2})$, $\ldots$ $(r_{1}^{m},\ldots r_{m}^{m})$ then form a basis of ${\mathbb{C}}^{m}$. Since $a_{1}=\operatorname{dim}W$, the vector $(a_{1},a_{2},\ldots,a_{m})$ is non-zero, and thus cannot be orthogonal to every vector in a basis of ${\mathbb{C}}^{m}$. There therefore exists $1\leq n\leq m$ such that

   $$\langle\chi_{\rho},\chi_{\pi^{\otimes n}}\rangle_{G}=\frac{1}{|G|}\big{(}a_{1},a_{2},\ldots,a_{m}\big{)}\cdot\big{(}r_{1}^{n},r_{2}^{n},\ldots r_{m}^{n}\big{)}\neq 0.$$