HW12

1. Assume that $G$ is the cyclic (additive) group $\mathbb{Z}/p\mathbb{Z}$ where $p$ is a prime number, and let $k$ be a field of characteristic $p$. Consider the morphism

$$\phi :G\to {\mathrm{GL}}_2(k),\lambda \mapsto \left(\begin{array}{cc}1& \lambda \\ 0& 1\end{array}\right)$$

Prove that the line generated by the first standard basis vector of $k^2$ is $G$-stable but has no $G$-stable complement.

\begin{proof} Define $e_1,e_2$ as the set of standard basis. For any $\lambda \in G$, there is $\phi (\lambda )\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right)$ and so the line generated by the first standard basis vector is $G$-stable.

Assume that $k^2$ has a $G$-stable complement W, then $\dim W=1$ and there exists $(a,b{)}^T$ such that $W=⟨(a,b{)}^T⟩$. Since $W$ is $G$-stable, one can check

$$\left(\begin{array}{cc}1& \lambda \\ 0& 1\end{array}\right)\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}a+\lambda b\\ b\end{array}\right)\in W$$

and $\lambda b=0$ for all $\lambda \in G$, which yields that $b=0$ and $W=⟨(1,0{)}^T⟩$, contradiction. Therefore, the line generated by the first standard basis does not have $G$-stable complement. \end{proof}

2. Let $G$ be the dihedral group of order 8, generated by elements $\sigma$ and $\tau$ subject to relations

$${\sigma }^2={\tau }^2=(\sigma \tau {)}^4=1$$

• (1) Check that the map

  $$\rho :\{\begin{array}{rl}G& \to {\mathrm{GL}}_2(\mathbb{C})\\ \sigma & \mapsto \left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right),\tau \mapsto \left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\end{array}$$

  defines a complex representation of $G$.

• (2) Prove that this representation is rational over $\mathbb{Q}$.

\begin{proof} Let $S=\rho (\sigma )=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)$ and $T=\rho (\tau )=\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)$.

i) To verify that $\rho$ is a representation, we must check if the images $S$ and $T$ satisfy the defining relations of $G$. One can easily check that

$$\begin{array}{rl}& S^2=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)=\left(\begin{array}{cc}(-i)(i)& (0)(-i)+(-i)(0)\\ (i)(0)+(0)(i)& (i)(-i)\end{array}\right)=\left(\begin{array}{cc}-i^2& 0\\ 0& -i^2\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I\\ & \\ & T^2=\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right)=\left(\begin{array}{cc}(-1)(-1)& (0)(-1)+(-1)(0)\\ (-1)(0)+(0)(-1)& (-1)(-1)\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I\end{array}$$

and $(ST{)}^4=I$. Since the matrices $S=\rho (\sigma )$ and $T=\rho (\tau )$ satisfy all the defining relations for $G$, the map $\rho :G\to {\mathrm{GL}}_2(\mathbb{C})$ defines a complex representation of $G$.

ii) It suffices to find an invertible matrix $P\in {\mathrm{GL}}_2(\mathbb{C})$ such that for all $g\in G$, the matrix $P\rho (g)P^{-1}$ has entries only in $\mathbb{Q}$. Let $P=\left(\begin{array}{cc}1& -i\\ -i& 1\end{array}\right)$, which is a invertible matrix. Note that

$$\begin{array}{rl}& S^{\prime }=PS{P}^{-1}=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),\\ & T^{\prime }=PT{P}^{-1}=\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right).\end{array}$$

Thus, with $P=\left(\begin{array}{cc}1& -i\\ -i& 1\end{array}\right)$, the representation $\rho$ is equivalent to ${\rho }^{\prime }$ where

$${\rho }^{\prime }(\sigma )=S^{\prime }=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),{\rho }^{\prime }(\tau )=T^{\prime }=\left(\begin{array}{cc}0& -1\\ -1& 0\end{array}\right).$$

Hence ${\rho }^{\prime }$ is rational, and so $\rho$ is rational over $\mathbb{Q}$. \end{proof}

3. Let $(M,\rho )$ and $(N,\sigma )$ be two $k$-representations of $G$. Fix basis ${\left(e_i\right)}_{1\le i\le m}$ and ${\left(f_j\right)}_{1\le j\le n}$ of $M$ and $N$ respectively. Then ${\left(e_i\otimes f_j\right)}_{1\le i\le m,1\le j\le n}$ is a basis of $M{\otimes }_{k}N$. Denote by $(m,\rho ),(n,\sigma ),(mn,\rho \otimes \sigma )$ be the matrix representations associated with $(M,\rho ),(N,\sigma ),\left(M{\otimes }_{k}N,\rho \otimes \sigma \right)$ respectively. Determine the relation between $(\rho \otimes \sigma )(g)$ and $\rho (g),\sigma (g)$.

\begin{proof} Note that the tensor product $M{\otimes }_{k}N$ has a basis ${\left(e_i\otimes f_j\right)}_{\begin{array}{c}1⩽i⩽m\\ 1⩽j⩽n\end{array}}$. Let $\rho (g)$ also denote the $m\times m$ matrix representation of the linear transformation $\rho (g)$ : $M\to M$ with respect to the basis $\left(e_i\right)$. The entries of this matrix $[\rho (g){]}_{si}$, are defined by the action on the basis vectors

$$\rho (g)\left(e_i\right)=\sum _{s=1}^m[\rho (g){]}_{si}{e}_s.$$

Similarly, let $\sigma (g)$ also denote the $n\times n$ matrix representation of the linear transformation $\sigma (g):N\to N$ with respect to the basis $\left(f_j\right)$. The entries of this matrix $[\sigma (g){]}_{tj}$, are defined by

$$\sigma (g)\left(f_j\right)=\sum _{t=1}^n[\sigma (g){]}_{tj}{f}_t.$$

The tensor product representation $(\rho \otimes \sigma )$ acts on $M{\otimes }_{k}N$. For any $g\in G$, the linear transformation $(\rho \otimes \sigma )(g):M{\otimes }_{k}N\to M{\otimes }_{k}N$ is defined by its action on simple tensors $u\otimes v\in M{\otimes }_{k}N:$

$$(\rho \otimes \sigma )(g)(u\otimes v)=\rho (g)(u)\otimes \sigma (g)(v)$$

We want to determine the $mn\times mn$ matrix representation of $(\rho \otimes \sigma )(g)$ with respect to the basis $\left(e_i\otimes f_j\right)$. Let this matrix also be denoted by $(\rho \otimes \sigma )(g)$. Note that the image of a basis vector $e_i\otimes f_j$ under $(\rho \otimes \sigma )(g)$ is

$$(\rho \otimes \sigma )(g)\left(e_i\otimes f_j\right)=\rho (g)\left(e_i\right)\otimes \sigma (g)\left(f_j\right)=\left(\sum _{s=1}^m[\rho (g){]}_{si}{e}_s\right)\otimes \left(\sum _{t=1}^n[\sigma (g){]}_{tj}{f}_t\right)=\sum _{s=1}^m\sum _{t=1}^n\left([\rho (g){]}_{si}[\sigma (g){]}_{tj}\right)\left(e_s\otimes f_t\right).$$

If we denote the matrix for the tensor product representation as $M_P(g)$, its entries are

$${\left[M_P(g)\right]}_{(s,t),(i,j)}=[\rho (g){]}_{si}[\sigma (g){]}_{tj}$$

which is precisely the definition of the Kronecker product. Therefore, the relation between the matrix $(\rho \otimes \sigma )(g)$ and the matrices $\rho (g)$ and $\sigma (g)$ is $(\rho \otimes \sigma )(g)=\rho (g)\otimes \sigma (g)$. \end{proof}

4. Let $(M,\rho )$ be a $k$-representations of $G$. Fix basis ${\left(e_i\right)}_{1\le i\le m}$ of $M$. Let $\left(M^{\ast },{\rho }^{\ast }\right)$ be the contragredient representation of $(M,\rho )$. Denote by $(m,\rho )$, $\left(m,{\rho }^{\ast }\right)$ be the matrix representations associated with $(M,\rho )$, $(M^{\ast },{\rho }^{\ast })$. Determine the relation between $\rho (g)$ and ${\rho }^{\ast }(g)$.

\begin{proof} Recall that $M^{\ast }={\mathrm{Hom}}_k(M,k)$ and $g\cdot \alpha =\alpha \cdot \rho (g{)}^{-1}$ for any $\alpha \in {\mathrm{Hom}}_k(M,k)$. Let $f_1,\cdots ,f_m$ be a set of basis of $M^{\ast }$, where $f_i(e_j)={\delta }_{ij}$. Then for any $g\in G$, one can check that

$${\rho }^{\ast }(g)(f_i)(e_j)=f_i(\rho (g{)}^{-1}{e}_j)=a_{ij},\text{ where }\rho (g{)}^{-1}=(a_{ij}{)}_{m\times m}$$

and so ${\rho }^{\ast }(g)(f_i)={\sum }_{j=1}^m{a}_{ij}{f}_j$. Hence, the matrix representation is

$${\rho }^{\ast }(g)=\left(\begin{array}{cccc}{a}_{11}& a_{21}& \cdots & a_{n1}\\ ⋮& ⋮& & ⋮\\ a_{1n}& a_{2n}& \cdots & a_{nn}\end{array}\right).$$

Therefore, the relation between $\rho (g)$ and ${\rho }^{\ast }(g)$ is ${\rho }^{\ast }(g)=(\rho (g{)}^{-1}{)}^T$. \end{proof}

5.

• (1) Let $G$ be a finite group acting on a finite set $\Omega$. Let ${\mathrm{Fix}}^G(\Omega )$ denote the set of fixed points of $\Omega$ under $G$, and let ${\Omega }^{\#}$ denote the set of orbits of $G$ on $\Omega \backslash {\mathrm{Fix}}^G(\Omega )$. Prove that

  $$|\Omega |=\left|{\mathrm{Fix}}^G(\Omega )\right|+\sum _{\omega \in {\Omega }^{\#}}|\omega |.\text{\hspace{1em}(*)}$$

  From now on, we assume that $p$ is a prime number and that $G$ is a nontrivial $p$-group.

• (2) Assume now that $|\Omega |$ is a power of $p$ different from 1. Prove that ${\mathrm{Fix}}^G(\Omega )$ is divisible by $p$.

• (3) Let $k$ be a (commutative) field of characteristic $p$. Let $M$ be a $kG$ module. Prove that ${\mathrm{Fix}}^G(M)\ne 0$. (HINT. (a) Let ${\left(e_i\right)}_{i=1,\dots ,r}$ be a $k$-basis of $M$. Prove that the abelian group generated by ${\left(g{e}_i\right)}_{(i=1,\dots ,r)(g\in G)}$ is an ${\mathbb{F}}_{p}G$-module. (b) Apply question (2) above to conclude.)

• (4) Prove that, up to isomorphism, the trivial representation of $G$ is the unique irreducible $kG$-module.

\begin{proof} i) The set can be partitioned into two disjoint subsets $\Omega ={\mathrm{Fix}}^G(\Omega )+{\Omega }^{\prime }$ where ${\Omega }^{\prime }$ is the set of points that are not fixed by $G$. Note that $|{\Omega }^{\prime }|={\sum }_{\omega \in {\Omega }^{♯}}|\omega |$, then $(\ast )$ holds.

ii) Since each $|\omega |\in {\Omega }^{♯}$ equals $|G|/|G_{{\omega }_0}|$, where ${\omega }_0\in \omega$. Note that $|G_{{\omega }_0}|\ne |G|$ by ${\omega }_0\in {\Omega }^{\prime }$, then $p\mid |\omega |$ for all $\omega \in {\Omega }^{♯}$. As $|\Omega |$ is a power of $p$ different from 1, by $(\ast )$ we know $|{\mathrm{Fix}}^G(\Omega )|$ is divisible by $p$.

iii) Let $(e_i{)}_{i=1}^r$ be a $k$-basis of $M$, and let $A=⟨g{e}_i:g\in G,e_i\in (e_i)⟩$. It is easy to check $A$ is an abelian group and is an ${\mathbb{F}}_{p}G$-module. Note that $A$ is a $p$-group and $0\in {\mathrm{Fix}}^G(A)$, then by ii) $p\mid |{\mathrm{Fix}}^G(A)|$ and so ${\mathrm{Fix}}^G(A)\ne 0$. For nonzero $v={\sum }_{i=1}^r{a}_i{e}_i\in {\mathrm{Fix}}^G(A)$, it is easy to check $v\in {\mathrm{Fix}}^G(M)$ and so ${\mathrm{Fix}}^G(M)\ne 0$.

iv) For any irreducible $kG$-module $M$, by iii) we have ${\mathrm{Fix}}^G(M)\ne 0$, which is a stable $kG$-submodule and so ${\mathrm{Fix}}^G(M)=M$. It yields that $M$ is trivial. Therefore, the trivial representation of $G$ is the unique irreducible $kG$-module. \end{proof}