Exercise 21.2. Let $G=S_4$ and let $H$ be the subgroup $\langle(1,2,3)\rangle\cong C_3$.
(a) If $\chi_1,\dots,\chi_5$ are the irreducible characters of $G$, as given in Section 18.1, work out the restrictions $\chi_i \downarrow H$ as sums of the irreducible characters $\psi_1,\psi_2,\psi_3$ of $C_3$.
(b) Calculate the induced characters $\psi_j\uparrow G$ ($1\le j \le 3$) as sums of the irreducible characters $\chi_i$ of $G$.
Tabuľka charakterov grupy $G=S_4$ je (podľa toho, čo sme videli v Section 18.1):
$$
\begin{array}{c|ccccc}
& 1 & (12) & (123) & (12)(34) & (1234) \\\hline
\chi_1 & 1 & 1 & 1 & 1 & 1 \\
\chi_2 & 1 & -1 & 1 & 1 & -1 \\
\chi_3 & 2 & 0 & -1 & 2 & 0 \\
\chi_4 & 3 & 1 & 0 & -1 & -1 \\
\chi_5 & 3 & -1 & 0 & -1 & 1
\end{array}
$$
Pre podgrupu $H$ máme charaktery
$$\begin{array}{c|ccc}
& 1 & (123) & (132) \\\hline
\psi_1 & 1 & 1 & 1 \\
\psi_2 & 1 & \omega & \omega^2 \\
\psi_3 & 1 & \omega^2 & \omega
\end{array}
$$
(a) Najprv máme spraviť zúženia $\chi_1,\dots,\chi_5$ na $H$:
$$\begin{array}{c|ccc|c}
& 1 & (123) & (132) \\\hline
\chi_1\downarrow H & 1 & 1 & 1 & \psi_1\\
\chi_2\downarrow H & 1 & 1 & 1 & \psi_1 \\
\chi_3\downarrow H & 2 &-1 &-1 & \psi_2+\psi_3 \\
\chi_4\downarrow H & 3 & 0 & 0 & \psi_1+\psi_2+\psi_3 \\
\chi_5\downarrow H & 3 & 0 & 0 & \psi_1+\psi_2+\psi_3 \\\hline
\end{array}$$
(b) Teraz by sme mali vypočítať charaktery indukované z $\psi_j$ na $G$. Vieme, že
$$(\psi_j\uparrow G)(1)=\frac{|G|}{|H|}\psi_j(1)$$
teda pre všetky tri charaktery dostaneme $(\psi_j\uparrow G)(1)=8$.
Ďalej pre $g=(123)$ sa $(123)^G \cap H$ rozdelí na dve jednoprvkové triedy konjugácie v $H$ s reprezentantmi $x_1=(123)$ a $x_2=(132)$. Ďalej pre tieto prvky platí $|C_G(g)|=3$ a $|C_H(x_i)|=3$. Dostaneme teda
$$(\psi_j\uparrow G)(123)=\psi_j(123)+\psi_j(132).$$
$$
\begin{array}{c|ccccc|c}
& 1 & (12) & (123) & (12)(34) & (1234) & \\\hline
\psi_1\uparrow G & 8 & 0 & 2 & 0 & 0 & \chi_1+\chi_2+\chi_4+\chi_5\\
\psi_2\uparrow G & 8 & 0 &-1 & 0 & 0 & \chi_3+\chi_4+\chi_5\\
\psi_3\uparrow G & 8 & 0 &-1 & 0 & 0 & \chi_3+\chi_4+\chi_5\\\hline
\end{array}
$$
Keď porovnáme koeficienty v časti (a) a (b), vidíme, že vyšli presne také, ako nám hovorí Frobeniova veta.
Exercise 21.2 - $C_3$ v $S_4$
Moderator: Martin Sleziak