HW13

1. Write out a proof of Maschke’s Theorem in the case of representations over $\mathbb{C}$ along the following lines.

• Given a representation $\rho :G\to GL(V)$ where $V$ is a vector space over $\mathbb{C}$, let $(\cdot ,\cdot )$ be any positive definite Hermitian form on $V$. Define a new form $(\cdot ,\cdot {)}_1$ on $V$ by

  $$(v,w{)}_1=\frac{1}{|G|}\sum _{g\in G}(gv,gw)$$

• Show that $(\cdot ,\cdot {)}_1$ is a positive definite Hermitian form, preserved under the action of $G$; that is, $(v,w{)}_1=(gv,gw{)}_1$ always.

• If $W$ is a subrepresentation of $V$, show that $V=W\oplus W^{\perp }$ as representations.

\begin{proof} For any $g\in G$, we have

$$(gv,gw{)}_1=\frac{1}{|G|}\sum _{h\in G}(hgv,hgw)=\frac{1}{|G|}\sum _{g\in G}(gv,gw)=(v,w{)}_1$$

and so $(\cdot ,\cdot {)}_1$ is preserved under the action of $G$. Now we check $(\cdot ,\cdot {)}_1$ is a positive definite Hermitian form. Firstly, it is easy to verify

$$\begin{array}{rl}& (u,\alpha v+\beta w{)}_1=\alpha (u,v{)}_1+\beta (u,w{)}_1\\ & (\alpha u+\beta v,w{)}_1=\bar{\alpha }(u,w{)}_1+\bar{\beta }(v,w{)}_1\end{array}$$

because $(\cdot ,\cdot )$ is a Hermitian form on $V$. Also $(v,v{)}_1=\frac{1}{|G|}{\sum }_{g\in G}(gv,gv)⩾0$ as $(gv,gv)⩾0$, and $(v,v{)}_1=0$ iff $(gv,gv)=0$ for all $g\in G$ iff $v=0$. Furthermore, one can easily check $(u,v{)}_1=\overline{(v,u{)}_1}$ by $(\cdot ,\cdot )$ Hermitian. Therefore, $(\cdot ,\cdot {)}_1$ is a positive definite Hermitian form.

Define $W^{\perp }=\{v\in V:(v,w{)}_1=0,\forall w\in W\}$. Since $(\cdot ,\cdot {)}_1$ is positive definite, for any nonzero $w\in W$, $(w,w)\ne 0$ and so $W\cap W^{\perp }=\{0\}$. It is easy to check $W^{\perp }$ is a vector subspace of $V$. Since $(\cdot ,\cdot {)}_1$ is a positive definite Hermitian form, there is $\dim W+\dim W^{\perp }=\dim V$ and so $V=W\oplus W^{\perp }$ as vector spaces. For any $g\in G$ and $v\in W^{\perp }$, we have $(gv,w{)}_1=(v,g^{-1}w)=0$ for all $w\in W$ and so $gv\in W^{\perp }$. Hence $W^{\perp }$ is a subrepresentation of $V$, and then $V=W\oplus W^{\perp }$ as representations. \end{proof}

2. Prove the following theorem of Burnside: let $G$ be a finite group, $k$ an algebraically closed field in which $|G|$ is invertible, and $\rho :G\to GL(V)$ be a representation over $k$. By taking a basis of $V$ write each endomorphism $\rho (g)$ as a matrix. Let $\dim V=n$. Show that the representation is simple if and only if there exist $n^2$ elements $g_1,\dots ,g_{n^2}$ of $G$ so that the matrices $\rho \left(g_1\right),\dots ,\rho \left(g_{n^2}\right)$ are linearly independent, and that this happens if and only if the algebra homomorphism $kG\to {\mathrm{End}}_k(V)$ is surjective. (Note that $\rho$ itself is generally not surjective.)

\begin{proof} We prove a lemma first.

Lemma

Let $V$ be an irreducible finite dimensional representation of algebra $A$, and let $v_1,\dots ,v_n\in V$ be any linearly independent vectors. Then for any $w_1,\dots ,w_n\in V$ there exists an element $a\in A$ such that $a{v}_i=w_i$.

Proof of lemma. Assume that $\dim V=n$. Define $\phi :\mathrm{End}(V)\to nV,x\mapsto (x{v}_1,\cdots ,x{v}_n)$ where $v_1,\cdots ,v_n$ is a set of basis of $V$. Since $\phi (ax)=a\phi (x)$ for all $a\in A$ and $\phi$ is a bijection, then $\mathrm{End}(V)$ and $nV$ are isomorphic $A$-modules. Let $\rho :A\to \mathrm{End}(V)$ be the corresponding representation. Then $\rho (A)⩽\mathrm{End}(V)$ is a sub-module of $\mathrm{End}(V)\simeq nV$. Since $V$ is irreducible, we have that $\rho (A)\simeq rV$ for some $r<n$. Suppose $\psi :\rho (A)\to rV$ is an isomorphisms between $A$-modules.

We claim that there exists a matrix $M_{r\times n}$ such that $\psi (\rho (a))M=\phi (\rho (a))$ for all $a\in A$. Now we prove the claim by induction. Consider the map $J=\phi \circ {\psi }^{-1},rV\to nV$, and it suffices to show $J(y)=yM$ for any $y\in rV$. When $r=1$, set $J_i={\pi }_i\circ J$ with ${\pi }_i:nV\to V_{(i)}$ with $nV=V_{(1)}\oplus \cdots \oplus V_{(n)}$. By Schur lemma, $J_i$ is a scalar multiple and $J_i$ are not all zero. Suppose $J_i(v)=v{M}_{1i}$ with $M_{1i}\in k$, then

$$J(y)=(y{M}_{11},\cdots ,y{M}_{1n})=y{M}_{(1)},\text{where }{M}_{(1)}=(M_{11},\cdots ,M_{1n})\in {\mathrm{M}}_{1\times n}(k).$$

Assume the statement hold for $r-1$. Write $rV$ as $rV=V_{(1)}\oplus W^{\prime }$, where $W^{\prime }\simeq (r-1)V$. Consider $J{|}_{V_{(1)}}$, by the argument above, there exists $Q\in {\mathrm{M}}_{1\times n}(k)$ such that $J((v,0,\cdots ,0))=vQ$. Note that there exists $g\in \mathrm{GL}(n,k)$ such that $Qg=(1,0,\cdots ,0)$, so WLOG we can assume $J((v,0,\cdots ,0))=(v,0,\cdots ,0)$. It is easy to check $\pi \circ J{|}_{W^{\prime }}:(r-1)V\to (n-1)V$ is injective, where $\pi :nV\to (n-1)V$ is a projective map with $\ker \pi =V_{(1)}$. By induction hypothesis, there exists $M_{W^{\prime }}$ such that $\pi \circ J{|}_{W^{\prime }}(y^{\prime })=y^{\prime }{M}_{W^{\prime }}$, where $y^{\prime }\in (r-1)V$. Then for any $y=(v,y^{\prime })\in rV$,

$$M=\left(\begin{array}{cccc}1& 0& \dots & 0\\ M_{21}& & & \\ ⋮& & M_{sub}& \\ M_{r1}& & & \end{array}\right)$$

is what we desired, where

$$W^{\prime }\to V_{(1)}^{\prime },y^{\prime }\to {\pi }_1(J(y^{\prime }))$$

determines $L:=(M_{21},\cdots ,M_{r1}{)}^T$ (the existence of $L$ is guaranteed by induction hypothesis). Now we prove the proof of the claim.

Note that there exists a non-zero vector $(w_1,\cdots ,w_n{)}^T$ such that $M(w_1,\cdots ,w_n{)}^T=\vec{0}$. It follows that for any $a\in A$, $\phi (\rho (a))(w_1,\cdots ,w_n{)}^T=0$. Take $a=1$, then $(v_1,\cdots ,v_n)(w_1,\cdots ,w_n{)}^T=0$, which contradicts with $v_1,\cdots ,v_n$ linearly independent. Thus $\rho (A)\simeq nV$ and so $\rho$ is surjective. For any $w_1,\cdots ,w_n\in V$, there exists $x\in \mathrm{End}(V)$ such that $x{v}_i=w_i$. Then there is an $a\in A$ such that $\rho (a)=x$ and $a{v}_i=w_i$. Now we finish the proof of the lemma. \end{proof}

It is clear that $\rho (g_1),\cdots ,\rho (g_{n^2})$ are linearly independent iff $\rho (kG)={\mathrm{End}}_k(V)$, i.e. $kG\to {\mathrm{End}}_k(V)$ is surjective.

Now we prove that $\rho$ simple iff $kG\to {\mathrm{End}}_k(V)$ surjective. Define $A=kG$, which is an algebra. Take any $c\in \mathrm{End}(V)$ and let $v_1,\cdots ,v_n$ be a basis of $V$. Let $w_i=c{v}_i$. Then by ^wha39z, there exists $a\in A$ such that $w_i=a{v}_i$. Thus $\rho (a)=c$ and so $kG\to {\mathrm{End}}_k(V)$ is surjective. Conversely, if $\rho (kG)={\mathrm{End}}_k(V)$, then all $kG$-submodules of $V$ are $\{0,V\}$ and so $\rho$ is simple. \end{proof}

3. Let $G=⟨x\mid x^n=1⟩$ be a cyclic group of order $n$, and let ${\zeta }_n\in \mathbb{C}$ be a primitive $n$-th root of unity. Then the simple complex characters of $G$ are the $n$ functions

$${\chi }_r\left(x^s\right)={\zeta }_n^{rs}$$

where $0\le r\le n-1$.

\begin{proof} Since $G$ is an abelian group, each simple representation over $\mathbb{C}$ is dimension $1$ and so we can assume that all simple complex characters are ${\chi }_0,\cdots ,{\chi }_{n-1}$. Assume that ${\chi }_i(x)=t\in \mathbb{C}$, then ${\chi }_i(x^n)=t^n=1$ yields $t={\zeta }_n^k$ for some integer $k$. Therefore, set ${\chi }_i(x)={\zeta }^i$, then we get all simple complex characters, where ${\chi }_r\left(x^s\right)={\zeta }_n^{rs}$ for $0⩽r⩽n-1$. \end{proof}

4.

• (i) By using characters, show that if $V$ and $W$ are $\mathbb{C}G$-modules then we have isomorphisms ${\left(V{\otimes }_{\mathbb{C}}W\right)}^{\ast }\cong V^{\ast }{\otimes }_{\mathbb{C}}{W}^{\ast }$, and $(\mathbb{C}G{)}^{\ast }\cong \mathbb{C}G$ as $\mathbb{C}G$-modules.
• (ii) If $k$ is any field and $V,W$ are $kG$-modules, show that there are isomorphisms ${\left(V{\otimes }_{k}W\right)}^{\ast }\cong V^{\ast }{\otimes }_k{W}^{\ast }$, and $(kG{)}^{\ast }\cong kG$ as $kG$ modules.

\begin{proof} i) Recall that for a representation $(M,\rho )$, there is ${\rho }^{\ast }(g)=(\rho (g{)}^{-1}{)}^T$. If $V$ and $W$ are $\mathbb{C}G$-modules with corresponding representations $\rho$ and $\sigma$, respectively, then

$$(\rho \otimes \sigma {)}^{\ast }(g)=((\rho (g)\otimes \sigma (g){)}^{-1}{)}^T\text{\hspace{1em}(*)}$$

and

$${\rho }^{\ast }(g)\otimes {\sigma }^{\ast }(g)=((\rho (g){)}^{-1}{)}^T\otimes ((\sigma (g){)}^{-1}{)}^T.\text{\hspace{1em}(**)}$$

Since $\mathrm{tr}(A\otimes B)=\mathrm{tr}(A)\mathrm{tr}(B)$ for matrices $A,B$. It deduces that values of $(\ast )$ and $(\ast \ast )$ are equal, as $\mathrm{tr}(((\rho (g){)}^{-1}{)}^T)=\overline{\mathrm{tr}(\rho (g))}$. Hence ${\left(V{\otimes }_{\mathbb{C}}W\right)}^{\ast }\cong V^{\ast }{\otimes }_{\mathbb{C}}{W}^{\ast }$.

Assume $\chi$ is the character of $\mathbb{C}G$, then $\chi (1)=|G|$ and $\chi (g)=0$ for all $g\ne 1$. Also we can check ${\chi }^{\ast }(1)=|G|$ and ${\chi }^{\ast }(g)=0$, thus $(\mathbb{C}G{)}^{\ast }\simeq \mathbb{C}G$ as $\mathbb{C}G$-modules.

ii) Assume $\{v_1,\cdots ,v_n\}$ and $\{w_1,\cdots ,w_m\}$ are bases of $V$ and $W$, respectively. Then $V^{\ast }$ and $W^{\ast }$ have bases $\{f_1,\cdots ,f_n\}$ and $\{g_1,\cdots ,g_m\}$, respectively, where $f_i(v_j)=g_i(w_j)={\delta }_{ij}$. Since $\{v_i\otimes w_j:1⩽i⩽n,1⩽j⩽m\}$ is a basis of $V\otimes W$, we define $(v_i\otimes w_j{)}^{\ast }\in (V\otimes W{)}^{\ast }$ such that $(v_i\otimes w_j{)}^{\ast }(v_{i^{\prime }},w_{j^{\prime }})={\delta }_{i{i}^{\prime }}{\delta }_{j{j}^{\prime }}$. Define

$$\phi :V^{\ast }\times W^{\ast }\to (V\otimes W{)}^{\ast },(f_i,g_j)\mapsto (v_i\otimes w_j{)}^{\ast }$$

and we can check $\phi$ is an $V^{\ast },W^{\ast }$-bilinear map. Then $\phi$ induces a map $\varphi :V^{\ast }\otimes W^{\ast }\to (V\otimes W{)}^{\ast }$, which is an isomorphism between vector spaces. Furthermore, it is easy to verify $\varphi$ is a $kG$-homomorphism, which yields ${\left(V{\otimes }_{k}W\right)}^{\ast }\cong V^{\ast }{\otimes }_k{W}^{\ast }$ as $kG$-modules.

For $kG$-module $kG$, it has a set of basis $\{g_1,\cdots ,g_n\}$, which are all elements of $G$ and $n=|G|$. Define $h_i\in (kG{)}^{\ast }$ such that $h_i(g_j)={\delta }_{ij}$. Then define

$$\psi :kG\to (kG{)}^{\ast },g_i\mapsto h_i$$

and it is cleat that $\psi$ is an isomorphism as $kG$-modules. Therefore, we have $(kG{)}^{\ast }\cong kG$ as $kG$ modules. \end{proof}

5. Let $G$ be the non-abelian group of order 21:

$$G=⟨x,y\mid x^7=y^3=1,yx{y}^{-1}=x^2⟩$$

Show that $G$ has 5 conjugacy classes, and find its character table.

\begin{proof} We can compute that the $5$ conjugacy classes of $G$ are

• $\{1\}$
• $\{x,x^2,x^4\}$
• $\{x^3,x^5,x^6\}$
• $\{x^{i}y:0⩽i⩽6\}$
• $\{x^i{y}^2:0⩽i⩽6\}$

Thus $G$ has $5$ irreducible characters. By Schur’s orthogonality relations, the character table can be written as

	$1$	$x$	$x^3$	$y$	$y^2$
${\chi }_1$	$1$	$1$	$1$	$1$	$1$
${\chi }_2$	$1$	$1$	$1$	$\omega$	${\omega }^2$
${\chi }_3$	$1$	$1$	$1$	${\omega }^2$	$\omega$
${\chi }_4$	$3$	$m$	$n$	$0$	$0$
${\chi }_5$	$3$	$n$	$m$	$0$	$0$

Then we determine values of $m,n$.

Since $C_7\simeq ⟨x⟩⊲G$ is a normal subgroup of $G$ and $C_7$ has character table as exercise 3 shows. Recall that

Let $W$ be a $H$-module with character $\psi$, then $\psi {\uparrow }_H^G(g)=\frac{1}{|H|}{\sum }_{x\in G}\psi (x^{-1}gx)$ with the convention that $\psi (g)=0$ if $g\notin H$.

Link to original

then we have

$$\begin{array}{r}m=\frac{1}{7}\sum _{h\in G}\psi (h^{-1}xh)={\zeta }^1+{\zeta }^2+{\zeta }^4\\ n=\frac{1}{7}\sum _{h\in G}\psi (h^{-1}{x}^{3}h)={\zeta }^3+{\zeta }^5+{\zeta }^6\end{array}$$

where $\zeta$ is the $7$-th root of unity. So the character table is

	$1$	$x$	$x^3$	$y$	$y^2$
${\chi }_1$	$1$	$1$	$1$	$1$	$1$
${\chi }_2$	$1$	$1$	$1$	$\omega$	${\omega }^2$
${\chi }_3$	$1$	$1$	$1$	${\omega }^2$	$\omega$
${\chi }_4$	$3$	${\zeta }^1+{\zeta }^2+{\zeta }^4$	${\zeta }^3+{\zeta }^5+{\zeta }^6$	$0$	$0$
${\chi }_5$	$3$	${\zeta }^3+{\zeta }^5+{\zeta }^6$	${\zeta }^1+{\zeta }^2+{\zeta }^4$	$0$	$0$

Now we finish the solution. \end{proof}

6. Find the character table of the following group of order 36 :

$$G=⟨a,b,c\mid a^3=b^3=c^4=1,ab=ba,ca{c}^{-1}=b,cb{c}^{-1}=a^2⟩.$$

(It follows from these relations that $⟨a,b⟩$ is a normal subgroup of $G$ of order 9.)

\begin{proof} Firstly, we can compute that all conjugacy classes of $G$ are

• $\{1\}$
• $\{a^i:1⩽i⩽2\}\cup \{b^j:1⩽j⩽2\}$
• $\{a^i{b}^j:1⩽i,j⩽2,(i,j)\ne (0,0)\}$
• $\{a^i{b}^{j}c:0⩽i,j⩽2\}$
• $\{a^i{b}^j{c}^2:0⩽i,j⩽2\}$
• $\{a^i{b}^j{c}^3:0⩽i,j⩽2\}$

Thus $G$ has $6$ irreducible representations.

Since $H=⟨a,b⟩⊲G$ and it has $9$ irreducible characters $\{{\chi }_{ij}:1⩽i,j⩽3\}$, where ${\chi }_{ij}(a)={\omega }^i$, ${\chi }_{ij}^b={\omega }^j$ and $\omega$ is the $3$-rd root of unity. Thus by ^aizvs9 (which is used in exercise 5), $G$ has two characters with $\chi (1)=4$, that is,

	$1$	$a$	$ab$	$abc$	$ab{c}^2$	$ab{c}^3$
${\chi }^{\prime }$	$4$	$-2$	$1$	$0$	$0$	$0$
${\chi }^{\prime \prime }$	$4$	$4$	$4$	$0$	$0$	$0$

where ${\chi }^{\prime }$ is irreducible but ${\chi }^{\prime \prime }$ is reducible. Since $36=1^2+4^2+t_1^2+t_2^2+t_3^2+t_4^2$, one can check it only has one solution $t_1=4$, $t_2=t_3=t_4=1$. Thus the character table is

	$1$	$a$	$ab$	$abc$	$ab{c}^2$	$ab{c}^3$
${\chi }_1$	$1$	$1$	$1$	$1$	$1$	$1$
${\chi }_2$	$1$
${\chi }_3$	$1$
${\chi }_4$	$1$
${\chi }_5$	$4$	$-2$	$1$	$0$	$0$	$0$
${\chi }_6$	$4$

Since $o(a)=3$, ${\chi }_2(a)\in \{1,\omega ,{\omega }^2\}$. However, if ${\chi }_2(a)=\omega$ or ${\chi }_2(a)={\omega }^2$, there are three distinct irreducible characters of dimension $4$: ${\chi }_5$, ${\chi }_2{\chi }_5$ and ${\chi }_2^2{\chi }_5$, which is impossible. Hence ${\chi }_2(a)={\chi }_3(a)={\chi }_4(a)=1$ and so for $ab$. Since ${\chi }_i(abc)\in \{1,i,-1,-i\}$, now we determine ${\chi }_2,{\chi }_3,{\chi }_4$. Then by Schur’s orthogonality relations we get ${\chi }_6$. The character table is as follows.

	$1$	$a$	$ab$	$abc$	$ab{c}^2$	$ab{c}^3$
${\chi }_1$	$1$	$1$	$1$	$1$	$1$	$1$
${\chi }_2$	$1$	$1$	$1$	$i$	$-1$	$-i$
${\chi }_3$	$1$	$1$	$1$	$-1$	$1$	$-1$
${\chi }_4$	$1$	$1$	$1$	$-i$	$-1$	$i$
${\chi }_5$	$4$	$-2$	$1$	$0$	$0$	$0$
${\chi }_6$	$4$	$1$	$-2$	$0$	$0$	$0$

Now we finish the solution. \end{proof}

Remark. By magma we can get the character table directly.

G:=Group<a,b,c|a^3,b^3,c^4,(a,b),c*a*c^-1*b^-1,c*b*c^-1*a>;
GroupName(G);
Order(G);
G:=PermutationGroup(G);
 
NumberOfClasses := #ConjugacyClasses(G);
print "Number of conjugacy classes:", NumberOfClasses;
 
ConjugacyClasses(G);
 
CharacterTable(G);