HW14

1. Let $k$ be a field and $\Omega$ be a $G$-set. Recall that $k\Omega ={⨁}_{\omega \in \Omega }k\omega$ denotes the permutation $kG$-module. Prove:

• (i) Suppose $G$ acts on $\Omega$ transitively. If we pick any element $\omega \in \Omega$ and let $H={\mathrm{Stab}}_G(\omega )$ then $H$ is also the stabilizer of the space $k\omega$, and $k\Omega \cong {\mathrm{Ind}}_H^G\left(k_H\right)$, where $k_H$ denotes the trivial $kH$-module.

  This shows that permutation modules on transitive $G$-sets are exactly the modules that are induced from the trivial module on some subgroup.

• (ii) Let $k=\mathbb{C}$. For the permutation module $\mathbb{C}\Omega$, the value of its character on $g$ equals the number of fixed points of $g$ on $\Omega$:

  $${\chi }_{\mathbb{C}\Omega }(g)=\left|{\Omega }^{⟨g⟩}\right|.$$

\begin{proof} i) Assume that $[G:H]=\{g_{1}H,\cdots ,g_{n}H\}$, where $g_i\in G$. Note that $g_i\omega \ne g_j\omega$ if $i\ne j$, otherwise $g_j^{-1}{g}_i\in H$ leads to a contradiction. Since $G$ is transitive on $\Omega$, one can check $\{g_i\omega :1⩽i⩽n\}=\Omega$. Now we can set $\Omega =\{g_1\omega ,\cdots ,g_n\omega \}$. By ^hp5t8s, there is

$${\mathrm{Ind}}_H^G(k_H)\simeq (g_1\otimes k_H)\oplus \cdots \oplus (g_n\otimes k_H),$$

where $k_H$ is the trivial $kH$-module. Define

$$\phi :(g_1\otimes k_H)\oplus \cdots \oplus (g_n\otimes k_H)\to k\Omega ,g_i\otimes 1_H\mapsto g_i\omega .$$

Then one can check $\phi$ is an isomorphism between $kG$-modules and so $k\Omega \simeq {\mathrm{Ind}}_H^G(k_H)$.

ii) Define $\Omega =\{{\omega }_1,\cdots ,{\omega }_n\}$. Let $M_g$ be the matrix of $g$ of the permutation representation. Then $M_g=(a_{ij})$ is a $0$-$1$ matrix, and $a_{ii}=1$ iff $g{\omega }_i={\omega }_i$. Thus ${\chi }_{\mathbb{C}\Omega }(g)=\left|{\Omega }^{⟨g⟩}\right|$. \end{proof}

2. Let $H$ and $K$ be subgroups of $G$ with $HK=G$ and $H\cap K=1$. Show that for any $kH$-module $U$, the module ${\mathrm{Res}}_K^G{\mathrm{Ind}}_H^G(U)$ is a direct sum of copies of the regular module $kK$.

\begin{proof} Define $K=\{k_1,\cdots ,k_n\}$. Note that $|G|=|HK|=|H||K|/|H\cap K|=|H||K|$, then $|[G:H]|=|K|$ and so $[G:H]=\{k_{i}H:1⩽i⩽n\}$. It deduces that

$${\mathrm{Ind}}_H^G(U)=(k_1\otimes U)\oplus \cdots \oplus (k_n\otimes U).$$

For any $k\in K$ and $k_i\otimes u$, there is $k(k_i\otimes u)=k{k}_i\otimes u$. Suppose $U=\mathrm{span}\{u_1,\cdots ,u_t\}$, then as $k$-vector space

$${\mathrm{Ind}}_H^G(U)={\oplus }_{i=1}^n(k_i\otimes U)={\oplus }_{i=1}^n(k_i\otimes ({\oplus }_{j=1}^{t}k{u}_j))={\oplus }_{j=1}^t({\oplus }_{i=1}^{n}k(k_i\otimes u_j)).$$

Define $M_j={\oplus }_{i=1}^{n}k(k_i\otimes u_j)$, which is a $k$-vector space and for any $k_0\in K$ there is $k_0\cdot (k_i\otimes u_j)=(k_0{k}_i)\otimes u_j$. Therefore, $M_j\simeq kK$ as $kK$-module and so

$${\mathrm{Res}}_K^G{\mathrm{Ind}}_H^G(U)={\oplus }_{j=1}^t{M}_j\simeq (kK{)}^t$$

is a direct sum of copies of the regular module $kK$. \end{proof}

3. We say that a $G$-set $\Omega$ is $n$-transitive (or, more properly, the action of $G$ on $\Omega$ is $n$-transitive) if $\Omega$ has at least $n$ elements and for every pair of $n$-tuples $\left(a_1,\dots ,a_n\right)$ and $\left(b_1,\dots ,b_n\right)$ in which the $a_i$ are distinct elements of $\Omega$ and the $b_i$ are distinct elements of $\Omega$, there exists $g\in G$ with $g{a}_i=b_i$ for every $i$. Prove the following statements.

• (i) $S_n$ acts $n$-transitively on $\{1,\dots ,n\}$.
• (ii) Let $\Omega$ be a $G$-set with at least $n$ elements (where $n\ge 1$ ) and let $\omega \in \Omega$. Then $G$ acts $n$-transitively on $\Omega$ if and only if $G$ acts transitively on $\Omega$ and ${\mathrm{Stab}}_G(\omega )$ acts $(n-1)$-transitively on $\Omega -\{\omega \}$.

\begin{proof} i) For any distinct $n$-tuples $\left(a_1,\dots ,a_n\right)$ and $\left(b_1,\dots ,b_n\right)$ with $a_i,b_i\in \{1,\cdots ,n\}$, define $\sigma :\{1,\cdots ,n\}\to \{1,\cdots ,n\},a_i\mapsto b_i$. Then $\sigma$ is a permutation and so $\sigma \in S_n$. By the arbitrary of $a_i$ and $b_i$, $S_n$ acts $n$-transitively on $\{1,\dots ,n\}$.

ii) We first assume that $G$ acts $n$-transitively on $\Omega$, then $G$ is transitive on $\Omega$. Then for $(n-1)$-tuples $(a_2,\cdots ,a_n)$ and $(b_2,\cdots ,b_n)$ with $a_1=b_1=\omega$, there exists $g\in G$ such that $g((a_1,\cdots ,a_n))=(b_1,\cdots ,b_n)$ and so $g\in {\mathrm{Stab}}_G(\omega )$. By the arbitrary of $(n-1)$-tuples, ${\mathrm{Stab}}_G(\omega )$ acts $(n-1)$-transitively on $\Omega -\{\omega \}$.

Conversely, assume that $G$ is transitive on $\Omega$ and ${\mathrm{Stab}}_G(\omega )$ acts $(n-1)$-transitively on $\Omega -\{\omega \}$. For any given $n$-tuples $\left(a_1,\dots ,a_n\right)$ and $\left(b_1,\dots ,b_n\right)$, we aim to find $g\in G$ such that $g((a_1,\cdots ,a_n))=(b_1,\cdots ,b_n)$. Since $G$ is transitive on $\Omega$, there exists $g_0$ such that $g_0(a_1)=b_1$. Then there exists $h_0\in {\mathrm{Stab}}_G(b_1)$ such that $h_0((g_0{a}_2,\cdots ,g_0{a}_n))=(b_2,\cdots ,b_n)$. Thus

$$h_0{g}_0((a_1,\cdots ,a_n))=h_0(b_1,g_0{a}_2,\cdots ,g_0{a}_n)=(b_1,\cdots ,b_n)$$

and so $g:=h_0{g}_0$ is what we desire. Now we finish the proof. \end{proof}

4.

• (i) Let $\Omega$ be a $G$-set. Prove: the permutation module $\mathbb{C}\Omega$ may be written as a direct sum of $\mathbb{C}G$-modules

  $$\mathbb{C}\Omega =\mathbb{C}\oplus V$$

  for some module $V$.

• (ii) Under the hypothesis of (i), suppose further that $G$ acts transitively on $\Omega$ and $|\Omega |\ge 2$. Let $H={\mathrm{Stab}}_G(\omega )$ for some $\omega \in \Omega$. Prove the following statements.

  • (a) ${⟨{\mathrm{Ind}}_H^G(1),{\mathrm{Ind}}_H^G(1)⟩}_G=|H\backslash G/H|$.
  • (b) $V$ is simple if and only if ${⟨{\mathrm{Ind}}_H^G(1),{\mathrm{Ind}}_H^G(1)⟩}_G=2$.
  • (c) $V$ is simple if and only if $G$ acts 2-transitively on $\Omega$. In that case, $V$ is not the trivial representation.

\begin{proof} i) Define $\Omega =\{{\omega }_1,\cdots ,{\omega }_n\}$, then $W:=⟨{\sum }_{i=1}^n{\omega }_i⟩\simeq \mathbb{C}$ is a $\mathbb{C}G$-submodule of $\mathbb{C}\Omega$. Let $V:=\mathrm{span}\{{\omega }_i-{\omega }_j:1⩽i,j⩽n\}$, then $V$ is also a $\mathbb{C}G$-submodule of $\mathbb{C}\Omega$. Since $\dim W+\dim V=n$ and $W\cap V=\{0\}$, we have $\mathbb{C}\Omega =W\oplus V\simeq \mathbb{C}\oplus V$.

ii) a) By Frobenius reciprocity and Mackey decomposition formula, we have

$$\begin{array}{rl}{⟨{\mathrm{Ind}}_H^G(1),{\mathrm{Ind}}_H^G(1)⟩}_G& ={⟨1,{\mathrm{Res}}_H^G{\mathrm{Ind}}_H^G(1)⟩}_H={⟨1,{\oplus }_{g\in [H\backslash G/H]}{\mathrm{Ind}}_{H\cap H^g}^H({}^g{\mathrm{Res}}_{H^g\cap H}^{H}1)⟩}_H\\ & =\sum _{g\in [H\backslash G/H]}{⟨1,{\mathrm{Ind}}_{H\cap H^g}^H(1_{H\cap H^g})⟩}_H=\sum _{g\in [H\backslash G/H]}{⟨1,1⟩}_{H\cap H^g}\\ & =|H\backslash G/H|.\end{array}$$

b) By exercise 1, we have ${\mathrm{Ind}}_H^G(1)\simeq \mathbb{C}\Omega =\mathbb{C}\oplus V$. It deduces that

$${⟨{\mathrm{Ind}}_H^G(1),{\mathrm{Ind}}_H^G(1)⟩}_G=⟨\mathbb{C}\oplus V,\mathbb{C}\oplus V⟩=1+2⟨\mathbb{C},V⟩+⟨V,V⟩.$$

Since $G$ is transitive on $\Omega$, one can check $⟨\mathbb{C},V⟩=0$. Then $V$ is simple if and only if ${⟨{\mathrm{Ind}}_H^G(1),{\mathrm{Ind}}_H^G(1)⟩}_G=2$.

c) Assume that $G$ is $2$-transitive, then $H$ is transitive on $\Omega \setminus \{\omega \}$. For any distinct $g_1,g_2\in G\setminus H$, there exists $h_1\in H$ such that $({\omega }^{g_1}{)}^{h_1}={\omega }^{g_2}$ and so $g_1{h}_1{g}_2^{-1}\in H$. It deduces that $g_1\in H{g}_{2}H$. By the arbitrary of $g_1,g_2$, we know $G=H\bigsqcup H{g}_{2}H$ and so $|H\backslash G/H|=2$. By a) and b), $V$ is simple.

Conversely, assume that $V$ is simple, then by a) and b), there is $|H\backslash G/H|=2$. Since $|H\backslash G/H|$ is the number of orbits of $H$ on $\Omega$ by ^vdlh0x and $H={\mathrm{Stab}}_G(\omega )$, we know $H$ is transitive on $\Omega \setminus \{\omega \}$ and so $G$ is $2$-transitive on $\Omega$ by Exercise 3. Furthermore, since $V$ is simple and $⟨\mathbb{C},V⟩=0$, $V$ is not trivial. \end{proof}