Exercise 21.1
Posted: Thu May 16, 2013 8:26 pm
$\newcommand{\inv}[1]{{#1}^{-1}}\newcommand{\CH}{\mathbb CH}\newcommand{\CG}{\mathbb CG}\newcommand{\spn}{\operatorname{span}}$Exercise 21.1. Let $G=D_8=\langle a,b; a^4=b^2=1, \inv bab=\inv a\rangle$ and let $H$ be the subgroup $\langle a^2,b \rangle$. Define $U$ to be the 1-dimensional subspace of $\CH$ spanned by
$$1-a^2+b-a^2b.$$
(a) Check that $U$ is a $\CH$-submodule of $\CH$.
(b) Find a basis of the induced $\CG$-module $U\uparrow G$.
(c) Write down the character of the $\CH$-module $U$ and the character of the $\CG$-module $U\uparrow G$. Is $U\uparrow G$ irreducible?
Pripomeňme, že triedy konjugácie sú $\{1\}$, $\{a^2\}$, $\{a,a^3\}$, $\{b,ba^2\}$, $\{ba,ba^3\}$.
(a) Stačí si všimnúť, že:
$ua^2=-u$
$ub=u$
(b) $U\uparrow G=U(\CG)=\spn \{ug; g\in G\}$
$u=1-a^2+b-a^2b$
$ua=a-a^3+a^3b-ab$
Zjavne $ua^2,ub,ua^2b\in U$ a $ua^3=(ua^2)a,uab=u(ba^2)a, ua^3b=(ub)a \in \spn \langle ua \rangle$
Takže vektory $u$ a $ua$ generujú $U\uparrow G$.
(c) Modulu $U$ zodpovedá charakter
$$\begin{array}{c|cccc}
& 1 & a^2 & b & a^2b \\\hline
\psi & 1 & -1 & 1 & -1
\end{array}$$
$U\uparrow G=\spn \langle u,v \rangle$, kde $v=ua$.
Pri tejto báze dostaneme reprezentáciu:
$$
\begin{align*}
ua&=v &va&=-u &a\mapsto\begin{pmatrix}0&1\\-1&0\end{pmatrix} \\
ua^2&=-u &va^2&=-v &a^2\mapsto\begin{pmatrix}-1&0\\0&-1\end{pmatrix} \\
ub&=u &vb&=v &b\mapsto\begin{pmatrix}1&0\\0&-1\end{pmatrix} \\
uab&=v &vab&=-u &ab\mapsto\begin{pmatrix}0&1\\-1&0\end{pmatrix} \
\end{align*}
$$
Dostávame teda charakter
$$\begin{array}{c|ccccc}
& 1 & a & a^2 & b & ab \\\hline
\psi\uparrow G & 2 & 0 & -2 & 0 & 0
\end{array}$$
Pre reprezentantov tried konjugácie máme
$$\begin{array}{c|cccccc}
g & 1 & a & a^2 & b & a^2b & ab \\\hline
|C_G(g)| & 8 & 4 & 8 & 4 & 4 & 4 \\\hline
|C_G(g)| & 4 & - & 8 & 4 & 4 & - \\\hline
\end{array}$$
Čiže keď počítame $\psi\uparrow G$ podľa vzorca z tejto kapitoly, tak dostaneme
$(\psi\uparrow G)(b) = \frac44 (\psi(b)+\psi(a^2b))$
$(\psi\uparrow G)(a^2) = \frac84 \psi(a^2)$
$(\psi\uparrow G)(1) = \frac84 \psi(1)$
čo súhlasí s hodnotami, ktoré nám vyšli.
$$1-a^2+b-a^2b.$$
(a) Check that $U$ is a $\CH$-submodule of $\CH$.
(b) Find a basis of the induced $\CG$-module $U\uparrow G$.
(c) Write down the character of the $\CH$-module $U$ and the character of the $\CG$-module $U\uparrow G$. Is $U\uparrow G$ irreducible?
Pripomeňme, že triedy konjugácie sú $\{1\}$, $\{a^2\}$, $\{a,a^3\}$, $\{b,ba^2\}$, $\{ba,ba^3\}$.
(a) Stačí si všimnúť, že:
$ua^2=-u$
$ub=u$
(b) $U\uparrow G=U(\CG)=\spn \{ug; g\in G\}$
$u=1-a^2+b-a^2b$
$ua=a-a^3+a^3b-ab$
Zjavne $ua^2,ub,ua^2b\in U$ a $ua^3=(ua^2)a,uab=u(ba^2)a, ua^3b=(ub)a \in \spn \langle ua \rangle$
Takže vektory $u$ a $ua$ generujú $U\uparrow G$.
(c) Modulu $U$ zodpovedá charakter
$$\begin{array}{c|cccc}
& 1 & a^2 & b & a^2b \\\hline
\psi & 1 & -1 & 1 & -1
\end{array}$$
$U\uparrow G=\spn \langle u,v \rangle$, kde $v=ua$.
Pri tejto báze dostaneme reprezentáciu:
$$
\begin{align*}
ua&=v &va&=-u &a\mapsto\begin{pmatrix}0&1\\-1&0\end{pmatrix} \\
ua^2&=-u &va^2&=-v &a^2\mapsto\begin{pmatrix}-1&0\\0&-1\end{pmatrix} \\
ub&=u &vb&=v &b\mapsto\begin{pmatrix}1&0\\0&-1\end{pmatrix} \\
uab&=v &vab&=-u &ab\mapsto\begin{pmatrix}0&1\\-1&0\end{pmatrix} \
\end{align*}
$$
Dostávame teda charakter
$$\begin{array}{c|ccccc}
& 1 & a & a^2 & b & ab \\\hline
\psi\uparrow G & 2 & 0 & -2 & 0 & 0
\end{array}$$
Pre reprezentantov tried konjugácie máme
$$\begin{array}{c|cccccc}
g & 1 & a & a^2 & b & a^2b & ab \\\hline
|C_G(g)| & 8 & 4 & 8 & 4 & 4 & 4 \\\hline
|C_G(g)| & 4 & - & 8 & 4 & 4 & - \\\hline
\end{array}$$
Čiže keď počítame $\psi\uparrow G$ podľa vzorca z tejto kapitoly, tak dostaneme
$(\psi\uparrow G)(b) = \frac44 (\psi(b)+\psi(a^2b))$
$(\psi\uparrow G)(a^2) = \frac84 \psi(a^2)$
$(\psi\uparrow G)(1) = \frac84 \psi(1)$
čo súhlasí s hodnotami, ktoré nám vyšli.