List 2

1. Determine the character table for the group $D_{2n}$.

Easy.

2. Compute a complete character table for quaternion group $Q$.

\begin{proof} Let $Q=\{\pm 1,\pm i,\pm j,\pm k\}$ with $j^2=j^2=k^2=-1$ and $ij=k,jk=i,ki=j$. Note that the conjugacy classes of $Q$ are

$$\{1\},\{-1\},\{i,-i\},\{j,-j\},\{k,-k\},$$

so $Q$ has $5$ irreducible representations on $\mathbb{C}$.

Assume that $\rho$ is a $1$-dimensional representation of $Q$, then

$$\rho (-1)=\rho (ij{i}^{-1}{j}^{-1})=\rho (i)\rho (i^{-1})\rho (j)\rho (j^{-1})=1.$$

It follows that $\rho (i),\rho (j),\rho (k)\in \{1,-1\}$ as $\rho (i{)}^2=\rho (j{)}^2=\rho (k{)}^2=\rho (-1)=1$. By $\rho (k)=\rho (i)\rho (k)$, we can determine the following $4$ characters.

	$1$	$-1$	$i$	$j$	$k$
${\chi }_1$	$1$	$1$	$1$	$1$	$1$
${\chi }_2$	$1$	$1$	$1$	$-1$	$-1$
${\chi }_3$	$1$	$1$	$-1$	$1$	$-1$
${\chi }_4$	$1$	$1$	$-1$	$-1$	$1$

Since $8=1^2\cdot 4+2^2\cdot 1$, $Q$ has an irreducible representation of dimension $2$. By Schur’s orthogonality relations, ${\chi }_5$ is determined as follows.

	$1$	$-1$	$i$	$j$	$k$
${\chi }_1$	$1$	$1$	$1$	$1$	$1$
${\chi }_2$	$1$	$1$	$1$	$-1$	$-1$
${\chi }_3$	$1$	$1$	$-1$	$1$	$-1$
${\chi }_4$	$1$	$1$	$-1$	$-1$	$1$
${\chi }_5$	$2$	$-2$	$0$	$0$	$0$

Now we get a complete character table for $Q$. \end{proof}

3. Identify the rotations of the tetrahedron in ${\mathbb{R}}^3$ with the alternation group $A_4$. Construct an irreducible 3-dimensional representation of $A_4$. Let $K_4$ be Klein subgroup of $A_4(K_4$ consists of $e,(12)(34),(13)(24),(14)(23)$, then show that $A_4/K_4\simeq C_3$. Use the projection $A_4\to C_3$ to determine three 1-dimensional representations of $A_4$.

\begin{proof} i) Assume that $\{1,2,3,4\}$ are vertices of the tetrahedron, then any rotation fixing a vertex $m$ is a permutation of $i,j,k$ and can be written as $(ijk)$. Hence the automorphism group is

$$⟨(ijk):i,j,k\in \{1,2,3,4\}⟩\cong A_4.$$

ii) A $3$-dimension irreducible representation of $A_4$ is the standard representation, where

$$V=\{a_1{v}_1+\cdots +a_4{v}_4:\sum a_i=0\}.$$

iii) It is easy to show that $K_4⊲A_4$. Since $|K_4|=4$ and $|A_4|=12$, $A_4/K_4$ is a group of order $3$ and so $A_4/K_4\cong C_3$.

iv) Define $\pi :A_4\to C_3$. Note $A_4/K_4=\{K_4,(123)K_4,(132)K_4\}$, then define ${\rho }_j:A_4\to \mathbb{C}$ as

1,&\mbox{if }g\in K_4\\ \omega^j,&\mbox{if }g\in (123)K_4 \\ \omega^{2j},&\mbox{if } g\in (132)K_4 \end{cases},$$ where $\omega=e^{2\pi i/3}$. Hence, $\rho_j\,(j=1,2,3)$ are three $1$-dimensional representations. `\end{proof}` **4. Show that $S_4$ can be realized as the automorphisms of the cube, construct an irreducible 3-dimensional representation of $S_4$. Determine its character.** `\begin{proof}` Consider $4$ body diagonals of the cube. Each permutation of them is an automorphism, which yields that $S_4$ is the automorphism group. The standard representation is $3$-dimensional and representation, and it is easy to determine its character. `\end{proof}` **Remark of 3 and 4.** There is another way to construct the $3$-dimensional representation. Each automorphism of tetrahedron (rep. cube) induces a permutation of "pairs of edges" (rep. "pairs of faces") and it is a permutation representation. **5. Let $K_4$ be the Klein group from Ex.3. Show that $S_4 / K_4 \simeq S_3$. Using projection $S_4 \rightarrow$ $S_3$ construct irreducible representations of $S_4$ corresponding to the three irreducible representations of $S_3$. Determine their characters.** `\begin{proof}` Since each automorphism induces a permutation of pairs of face, there is a natural map $\pi:S_4\to S_3$. Note that $\ker\pi=\left\langle(12)(34),(13)(24),(14)(23)\right\rangle\cong K_4$, then it yields that $S_4/K_4\simeq S_3$. Since $S_4/K_4=\{K_4g:g\in S_3\}$, we define $\rho(h)=\varphi(g)$ for any $h\in K_4g$ and irreducible representation $\varphi$ of $S_3$. Since $S_3$ has $3$ irreducible representations of dimension $1,1,2$, $S_4$ also has $3$ irreducible representations of dimension $1,1,2$. `\end{proof}` **6. Using Ex. 4 and Ex. 5 determine character table of $S_4$ containing four irreducible modules from Ex. 4 and Ex. 5 and $V$. Can you construct $V$ explicitly using its character?** `\begin{proof}` Note that $|S_4|=24=3^2+3^2+1^2+1^2+2^2$, so we have determined all irreducible modules from Ex. 4 and Ex. 5, and it is easy to determine its character table. `\end{proof}` **7. Prove that if all irreducible representations of finite group $G$ are 1-dimensional then $G$ is abelian.** `\begin{proof}` For any character $\chi$ of $G$ and its corresponding representation $\rho$, by $\dim\rho=1$, we have that $\rho(g)=[\mathrm{tr}(g)]$ for all $g\in G$ and so

\rho(gh)=\rho(g)\rho(h)=\rho(h)\rho(g)=\rho(hg)

for all $g,h\in G$. It follows that $[h,g]=h^{-1}g^{-1}hg\in\ker\rho$ for all $g,h\in G$ and $G'\subseteq\ker \rho$, where $G'=\left\langle[h,g]:\forall h,g\in G\right\rangle$. Hence, $G'\subseteq\mathrm{rad}(\mathbb{C}G)$. Since $\mathbb{C}G$ is semi-simple, we have that $G'$ is trivial. Recall that $G'\lhd G$ and $G/G'$ is abelian, then $G\simeq G/G'$ is abelian. `\end{proof}` **8. Let $G \times H=\{(g, h) \mid g \in G, h \in H\}$, the direct product of the groups $G$ and $H$.** - **(i) Prove that if $\chi$ is an irreducible character of $G$ and $\psi$ an irreducible character of $H$ then $\chi \psi:(g, h) \rightarrow \chi(g) \psi(h)$ is an irreducible character of $G \times H$.** - **(ii) Prove that this construction gives all the irreducible characters of $G \times H$.** `\begin{proof}` i) To show $\chi\psi$ is irreducible, it is enough to show that $\left\langle \chi\psi,\chi\psi\right\rangle=1$, that is

\left\langle\chi\psi,\chi\psi\right\rangle =\frac{1}{|G||H|}\sum_{(g,h)\in G\times H}\chi(g)\psi(h)\overline{\chi(g)\psi(h)}=\left(\frac{1}{|G|}\sum_{g\in G}\chi(g)\overline{\chi(g)}\right)\left(\frac{1}{|H|}\sum_{h\in H}\chi(h)\overline{\chi(h)}\right)=1.

Therefore, by Schur's orthogonality relations, $\chi\psi$ is an irreducible character of $G\times H$. ii) Suppose that $\chi_1,\cdots,\chi_n$ are all the irreducible characters of $G$, and suppose that $\psi_1,\cdots,\psi_m$ are all the irreducible characters of $H$. Then $\{\chi_i\psi_j:1\leqslant i\leqslant n,1\leqslant j\leqslant m\}$ is a set of irreducible characters of $G\times H$. Note that

\sum_{i,j}(\chi_i\psi_j(1))^2=\sum_{i,j}\chi_i^2(1)\psi_j^2(1)=(\sum_i\chi_i^2(1))(\sum_j\psi_j^2(1))=|G||H|=|G\times H|.

Therefore, this construction gives all the irreducible characters of $G\times H$. `\end{proof}` **9. Let $\rho_c: G \rightarrow G L_{\mathbb{C}}\left(\mathbb{C}\left[G_c\right]\right)$ denote the permutation representation associated to the conjugation action of $G$ on its own underlying set $G_c$, i.e., $g \cdot x=g x g^{-1}$. Let $\chi_c=\chi_{\rho_c}$ be the character of $\rho_c$.** - **(i) For $x \in G$ show that the vector subspace $V_x$ spanned by all the conjugates of $x$ is a $G$-subspace. What is $\operatorname{dim} V_x$ ?** - **(ii) For $g \in G$ show that $\chi_c(g)=\left|C_G(g)\right|$ where $C_G(g)$ is the centralizer of $g$ in $G$.** - **(iii) If $\chi_1, \ldots, \chi_r$ are the distinct irreducible characters of $G$ and $\rho_1, \ldots, \rho_r$ are the corresponding irreducible representations, determine the multiplicity of $\rho_j$ in $\mathbb{C}\left[G_c\right]$.** `\begin{proof}` i) Note that $V_x=\mathrm{span}\{x^g:g\in G\}$ and $|\{x^g:g\in G\}|=|G|/|C_G(x)|$, then $\dim V_x=|G|/|C_G(x)|$. ii) For each $g\in G$, $\chi_c(g)=\mathrm{tr}(\rho_c(g))=|\{h\in G:h^g=h\}|=|C_G(g)|$. iii) Note that

\left\langle\chi_c,\chi_j\right\rangle =\frac{1}{|G|}\sum_{g\in G}\chi_c(g)\overline{\chi_j(g)}=\frac{1}{|G|}\sum_{C\in\mathcal C}|C||C_G(C)|\overline{\chi_j(C)}=\sum_{C\in\mathcal C}\overline{\chi_j(C)}=\sum_{C\in C}{\chi_j(C)}

where $C\in\mathcal C$ runs over conjugacy classes $\mathcal C$ of $G$. It follows that the multiplicity of $\rho_j$ equals $\sum_{C\in C}{\chi_j(C)}$. `\end{proof}` **10. Let $G$ be a finite group and $H \leq G$ a subgroup. Consider the set of cosets $G / H$ as a $G$-set with action given by $g \cdot x H=g x H$ and let $\rho$ be the associated permutation representation on $\mathbb{C}[G / H]$.** - **(i) Show that for $g \in G, \chi_\rho(g)=\left|\left\{x H \in G / H: g \in x H x^{-1}\right\}\right|$.** - **(ii) If $H \triangleleft G$, show that**

\chi_\rho(g)= \begin{cases}0 & g \notin H \ |G / H| & g \in H\end{cases}

- **(iii) Determine the character $\chi_\rho$ when $G=S_4$ and $H=S_3$ (viewed as the subgroup of all permutations fixing 4 ).** `\begin{proof}` i) Note that $\chi_\rho(g)$ equals the number of fixed point of $\rho(g)$ on $G/H$. If $xH\in G/H$ satisfies $gxH=xH$, then $g\in xHx^{-1}$. Therefore, $\chi_\rho(g)=\left|\left\{x H \in G / H: g \in x H x^{-1}\right\}\right|$. ii) If $H\lhd G$, then $xHx^{-1}=H$ for any $x\in G$. It follows that $\chi_\rho(g)=|\{xH\in G/H:g\in H\}|$ and so

\chi_\rho(g)= \begin{cases}0 & g \notin H \ |G / H| & g \in H\end{cases}.

iii) Now $H$ is not a normal subgroup of $G$. The cosets $G/H=\{S_3,(1234)S_3,(13)(24)S_3,(1432)S_3\}$. We only need to consider representative elements of a conjugacy class. By computation - $\chi_\rho(1)=4$ - $\chi_\rho((12))=2$ - $\chi_\rho((123))=1$ - $\chi_\rho((12)(34))=0$ - $\chi_\rho((1234))=0$ Now we determine the character $\chi_\rho$. `\end{proof}` **11. Show that the orthogonality of the rows of the character table is equivalent to an orthogonality for the columns. Namely,** - **(i) for $g \in G$**

\sum_\chi \overline{\chi(g)} \chi(g)=\frac{|G|}{c(g)},

**where the sum is over all irreducible characters, and $c(g)$ is the number of elements in the conjugacy class of $g$.** - **(ii) If $g$ and $h$ are elements of $G$ that are not conjugate, then $\sum_\chi \overline{\chi(g)} \chi(h)=0$.** - **(iii) Write both formulas for $g=e$.** `\begin{proof}` i), ii) Easy. iii) $\sum_\chi (\chi(1))^2=|G|$ for i) and $\sum_\chi \chi(1)\chi(h)=0$ for any $h\neq 1$. `\end{proof}` **12. Using the inclusion $A_5 \rightarrow S_5$ construct a 4-dimensional irreducible representation of $A_5$ which we again call standard.** `\begin{proof}` Note that each element of $A_5$ is also an element of $S_5$. Suppose $\rho$ is the standard representation of $S_5$, then $\rho|_{A_5}$ is the representation of $A_5$ with dimension $4$. Now we aim to show $\rho|_{A_5}$ is irreducible. Note that elements of $A_5$ are in the form of $(1)$, $(ijk)$, $(ij)(kl)$ or $(ijklm)$, and we can compute that

|{(ijk):(ijk)\in A_5}|=20,,|{(ij)(kl):(ij)(kl)\in A_5}|=15,,|{(ijklm):(ijklm)\in A_5}|=24.

Assume that the corresponding character of $\rho|_{A_5}$ is $\chi$, then $\chi((ijk))=1$, $\chi((ij)(kl))=0$, $\chi((ijklm))=-1$. Therefore,

\left\langle\chi,\chi\right\rangle = \frac{1}{60}\sum_{g\in A_5}\chi(g)\overline{\chi(g)}=\frac{1}{60}(1\cdot 4^2+20\cdot1^2+15\cdot 0+24\cdot (-1)^2)=1.

By Schur's orthogonality relations, $\rho|_{A_5}$ is a $4$ dimensional irreducible representation. `\end{proof}`