Soal dan Pembahasan – UAS Struktur Aljabar (Teori Grup) Tahun Ajaran 2018/2019

Jawaban a)
Banyaknya elemen di $S_4$ adalah
$|S_4| = 4! = 24$
Misalkan $S_4 = \{a_1, a_2, a_3, \cdots, a_{24}\}$
dengan:
$$\begin{array}{cc} a_1 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{pmatrix} & a_2 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 3 & 2 & 4 \end{pmatrix} \\ a_3 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 4 & 3 & 2 \end{pmatrix} & a_4 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 4 & 3 \end{pmatrix} \\ a_5 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 2 & 3 & 1 \end{pmatrix} & a_6 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 2 & 1 & 4 \end{pmatrix} \\ a_7 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 3 & 4 \end{pmatrix} & a_8 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 3 & 4 & 2 \end{pmatrix} \\ a_9 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 4 & 2 & 3 \end{pmatrix} & a_{10} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \end{pmatrix} \\ a_{11} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 1 & 4 \end{pmatrix} & a_{12} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{pmatrix} \\ a_{13} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 1 & 3 \end{pmatrix} & a_{14} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 3 & 1 \end{pmatrix} \\ a_{15} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 2 & 4 & 1 \end{pmatrix} & a_{16} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 1 & 2 & 4 \end{pmatrix} \\ a_{17} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 1 & 4 & 2 \end{pmatrix} & a_{18} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{pmatrix} \\ a_{19} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix} & a_{20} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 2 & 1 & 3 \end{pmatrix} \\ a_{21} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{pmatrix} & a_{22} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 3 & 2 \end{pmatrix} \\ a_{23} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 3 & 2 & 1 \end{pmatrix} & a_{24} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4& 3 & 1 & 2 \end{pmatrix} \end{array}$$
Anggota $S_4$ yang berorder $4$ artinya $a \in S_4$ yang memenuhi persamaan $a^4 = e = a_1 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{pmatrix}$, di mana $a_1$ merupakan elemen identitas dalam $S_4$.
Hasil pengoperasian komposisi fungsi pangkat $4$ pada ke-$24$ elemen $S_4$ menunjukkan bahwa ada $6$ elemen yang berorder $4$, yaitu $\{a_{12}, a_{13}, a_{17}, a_{19}, a_{21}, a_{24}\}$
Sebagai ilustrasi, misalnya kita memilih $a_{12}$.
$$\begin{aligned} (a_{12})^2 & = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{pmatrix} \\ & = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{pmatrix} \end{aligned}$$
$$\begin{aligned} (a_{12})^3 & = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{pmatrix} \\ & = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix} \end{aligned}$$
$$\begin{aligned} (a_{12})^4 & = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 2 & 3 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{pmatrix} \\ & = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4 \end{pmatrix} = a_{1} \end{aligned}$$
Perhatikan bahwa himpunan beranggotakan anggota $S_4$ yang berorder $4$, yakni $\{a_{12}, a_{13}, a_{17}, a_{19}, a_{21}, a_{24}\}$ memiliki subset tak kosong sebanyak $2^6 -1 = 63$.
Pilih satu di antaranya, misalnya $H = \{a_{12}\}$.
Jawaban b)
Semua koset kiri dari $H$ dalam $S_4$ adalah
$\begin{aligned} & a_1H = \{a_1a_{12}\} = \{a_{12}\} \\ & a_2H = \{a_2a_{12}\} = \{a_{15}\} \\ & a_3H = \{a_3a_{12}\} = \{a_{23}\} \\ & a_4H = \{a_4a_{12}\} = \{a_{14}\} \\ & a_5H = \{a_5a_{12}\} = \{a_{11}\} \\ & a_6H = \{a_6a_{12}\} = \{a_{10}\} \\ & a_7H = \{a_7a_{12}\} = \{a_{8}\} \\ & a_8H = \{a_8a_{12}\} = \{a_{19}\} \\  & a_9H = \{a_9a_{12}\} = \{a_{5}\} \\  & a_{10}H = \{a_{10}a_{12}\} = \{a_{3}\} \\ & a_{11}H = \{a_{11}a_{12}\} = \{a_{17}\} \\ & a_{12}H = \{a_{12}a_{12}\} = \{a_{18}\} \\ & a_{13}H = \{a_{13}a_{12}\} = \{a_{22}\} \\ & a_{14}H = \{a_{14}a_{12}\} = \{a_{24}\} \\ & a_{15}H = \{a_{15}a_{12}\} = \{a_{13}\} \\ & a_{16}H = \{a_{16}a_{12}\} = \{a_{4}\} \\ & a_{17}H = \{a_{17}a_{12}\} = \{a_{9}\} \\ & a_{18}H = \{a_{18}a_{12}\} = \{a_{21}\} \\ & a_{19}H = \{a_{19}a_{12}\} = \{a_{20}\} \\ & a_{20}H = \{a_{20}a_{12}\} = \{a_{7}\} \\ & a_{21}H = \{a_{21}a_{12}\} = \{a_{1}\} \\ & a_{22}H = \{a_{22}a_{12}\} = \{a_{2}\} \\ & a_{23}H = \{a_{23}a_{12}\} = \{a_{6}\} \\ & a_{24}H = \{a_{24}a_{12}\} = \{a_{16}\} \end{aligned}$