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

$${\rho }_j(g)=\{\begin{array}{ll}1,& \text{mbox}ifg\in K_4\\ {\omega }^j,& \text{mbox}ifg\in (123)K_4\\ {\omega }^{2j},& \text{mbox}ifg\in (132)K_4\end{array},$$

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 $S_4\to$ $S_3$ 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 $\ker \pi =⟨(12)(34),(13)(24),(14)(23)⟩\cong K_4$, then it yields that $S_4/K_4\simeq S_3$. Since $S_4/K_4=\{K_{4}g:g\in S_3\}$, we define $\rho (h)=\phi (g)$ for any $h\in K_{4}g$ and irreducible representation $\phi$ 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 $G^{\prime }\subseteq \ker \rho$, where $G^{\prime }=⟨[h,g]:\forall h,g\in G⟩$. Hence, $G^{\prime }\subseteq \mathrm{rad}(\mathbb{C}G)$. Since $\mathbb{C}G$ is semi-simple, we have that $G^{\prime }$ is trivial. Recall that $G^{\prime }⊲G$ and $G/G^{\prime }$ is abelian, then $G\simeq G/G^{\prime }$ 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 $\chi \psi :(g,h)\to \chi (g)\psi (h)$ 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 $⟨\chi \psi ,\chi \psi ⟩=1$, that is

$$⟨\chi \psi ,\chi \psi ⟩=\frac{1}{|G||H|}\sum _{(g,h)\in G\times H}\chi (g)\psi (h)\overline{\chi (g)\psi (h)}=\left(\frac{1}{|G|}\sum _{g\in G}\chi (g)\overline{\chi (g)}\right)\left(\frac{1}{|H|}\sum _{h\in H}\chi (h)\overline{\chi (h)}\right)=1.$$

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 $\{{\chi }_i{\psi }_j:1⩽i⩽n,1⩽j⩽m\}$ 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 ${\rho }_c:G\to G{L}_{\mathbb{C}}\left(\mathbb{C}\left[G_c\right]\right)$ denote the permutation representation associated to the conjugation action of $G$ on its own underlying set $G_c$, i.e., $g\cdot x=gx{g}^{-1}$. 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 $\dim V_x$ ?
• (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 ${\chi }_1,\dots ,{\chi }_r$ are the distinct irreducible characters of $G$ and ${\rho }_1,\dots ,{\rho }_r$ 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

$$⟨{\chi }_c,{\chi }_j⟩=\frac{1}{|G|}\sum _{g\in G}{\chi }_c(g)\overline{{\chi }_j(g)}=\frac{1}{|G|}\sum _{C\in \mathcal{C}}|C||C_G(C)|\overline{{\chi }_j(C)}=\sum _{C\in \mathcal{C}}\overline{{\chi }_j(C)}=\sum _{C\in C}{\chi }_j(C)$$

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 $H\le G$ a subgroup. Consider the set of cosets $G/H$ as a $G$-set with action given by $g\cdot xH=gxH$ and let $\rho$ be the associated permutation representation on $\mathbb{C}[G/H]$.

• (i) Show that for $g\in G,{\chi }_{\rho }(g)=\left|\left\{xH\in G/H:g\in xH{x}^{-1}\right\}\right|$.
• (ii) If $H◃G$, show that

$${\chi }_{\rho }(g)=\{\begin{array}{ll}0& g\notin H\\ |G/H|& g\in H\end{array}$$

• (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 $g\in xH{x}^{-1}$. Therefore, ${\chi }_{\rho }(g)=\left|\left\{xH\in G/H:g\in xH{x}^{-1}\right\}\right|$.

ii) If $H⊲G$, then $xH{x}^{-1}=H$ for any $x\in G$. It follows that ${\chi }_{\rho }(g)=|\{xH\in G/H:g\in H\}|$ and so

$${\chi }_{\rho }(g)=\{\begin{array}{ll}0& g\notin H\\ |G/H|& g\in H\end{array}.$$

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 $h\ne 1$. \end{proof}

12. Using the inclusion $A_5\to S_5$ 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,

$$⟨\chi ,\chi ⟩=\frac{1}{60}\sum _{g\in A_5}\chi (g)\overline{\chi (g)}=\frac{1}{60}(1\cdot 4^2+20\cdot 1^2+15\cdot 0+24\cdot (-1{)}^2)=1.$$

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