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:
$$\begin{array}{cc}{a}_1=\left(\begin{array}{cccc}1& 2& 3& 4\\ 1& 2& 3& 4\end{array}\right)& a_2=\left(\begin{array}{cccc}1& 2& 3& 4\\ 1& 3& 2& 4\end{array}\right)\\ a_3=\left(\begin{array}{cccc}1& 2& 3& 4\\ 1& 4& 3& 2\end{array}\right)& a_4=\left(\begin{array}{cccc}1& 2& 3& 4\\ 1& 2& 4& 3\end{array}\right)\\ a_5=\left(\begin{array}{cccc}1& 2& 3& 4\\ 4& 2& 3& 1\end{array}\right)& a_6=\left(\begin{array}{cccc}1& 2& 3& 4\\ 3& 2& 1& 4\end{array}\right)\\ a_7=\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 1& 3& 4\end{array}\right)& a_8=\left(\begin{array}{cccc}1& 2& 3& 4\\ 1& 3& 4& 2\end{array}\right)\\ a_9=\left(\begin{array}{cccc}1& 2& 3& 4\\ 1& 4& 2& 3\end{array}\right)& a_{10}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 1& 4& 3\end{array}\right)\\ a_{11}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 3& 1& 4\end{array}\right)& a_{12}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 3& 4& 1\end{array}\right)\\ a_{13}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 4& 1& 3\end{array}\right)& a_{14}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 4& 3& 1\end{array}\right)\\ a_{15}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 3& 2& 4& 1\end{array}\right)& a_{16}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 3& 1& 2& 4\end{array}\right)\\ a_{17}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 3& 1& 4& 2\end{array}\right)& a_{18}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 3& 4& 1& 2\end{array}\right)\\ a_{19}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 3& 4& 2& 1\end{array}\right)& a_{20}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 4& 2& 1& 3\end{array}\right)\\ a_{21}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 1& 2& 3& 4\end{array}\right)& a_{22}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 4& 1& 3& 2\end{array}\right)\\ a_{23}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 4& 3& 2& 1\end{array}\right)& a_{24}=\left(\begin{array}{cccc}1& 2& 3& 4\\ 4& 3& 1& 2\end{array}\right)\end{array}$$
Anggota $S_4$ yang berorder $4$ artinya $a\in S_4$ yang memenuhi persamaan $a^4=e=a_1=\left(\begin{array}{cccc}1& 2& 3& 4\\ 1& 2& 3& 4\end{array}\right)$, 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 $a_{12}$.
$$\begin{array}{rl}(a_{12}{)}^2& =\left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 3& 4& 1\end{array}\right)\circ \left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 3& 4& 1\end{array}\right)\\ & =\left(\begin{array}{cccc}1& 2& 3& 4\\ 3& 4& 1& 2\end{array}\right)\end{array}$$
$$\begin{array}{rl}(a_{12}{)}^3& =\left(\begin{array}{cccc}1& 2& 3& 4\\ 3& 4& 1& 2\end{array}\right)\circ \left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 3& 4& 1\end{array}\right)\\ & =\left(\begin{array}{cccc}1& 2& 3& 4\\ 4& 1& 2& 3\end{array}\right)\end{array}$$
$$\begin{array}{rl}(a_{12}{)}^4& =\left(\begin{array}{cccc}1& 2& 3& 4\\ 4& 1& 2& 3\end{array}\right)\circ \left(\begin{array}{cccc}1& 2& 3& 4\\ 2& 3& 4& 1\end{array}\right)\\ & =\left(\begin{array}{cccc}1& 2& 3& 4\\ 1& 2& 3& 4\end{array}\right)=a_1\end{array}$$
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
$\begin{array}{rl}& a_{1}H=\{a_1{a}_{12}\}=\{a_{12}\}\\ & a_{2}H=\{a_2{a}_{12}\}=\{a_{15}\}\\ & a_{3}H=\{a_3{a}_{12}\}=\{a_{23}\}\\ & a_{4}H=\{a_4{a}_{12}\}=\{a_{14}\}\\ & a_{5}H=\{a_5{a}_{12}\}=\{a_{11}\}\\ & a_{6}H=\{a_6{a}_{12}\}=\{a_{10}\}\\ & a_{7}H=\{a_7{a}_{12}\}=\{a_8\}\\ & a_{8}H=\{a_8{a}_{12}\}=\{a_{19}\}\\ & a_{9}H=\{a_9{a}_{12}\}=\{a_5\}\\ & a_{10}H=\{a_{10}{a}_{12}\}=\{a_3\}\\ & a_{11}H=\{a_{11}{a}_{12}\}=\{a_{17}\}\\ & a_{12}H=\{a_{12}{a}_{12}\}=\{a_{18}\}\\ & a_{13}H=\{a_{13}{a}_{12}\}=\{a_{22}\}\\ & a_{14}H=\{a_{14}{a}_{12}\}=\{a_{24}\}\\ & a_{15}H=\{a_{15}{a}_{12}\}=\{a_{13}\}\\ & a_{16}H=\{a_{16}{a}_{12}\}=\{a_4\}\\ & a_{17}H=\{a_{17}{a}_{12}\}=\{a_9\}\\ & a_{18}H=\{a_{18}{a}_{12}\}=\{a_{21}\}\\ & a_{19}H=\{a_{19}{a}_{12}\}=\{a_{20}\}\\ & a_{20}H=\{a_{20}{a}_{12}\}=\{a_7\}\\ & a_{21}H=\{a_{21}{a}_{12}\}=\{a_1\}\\ & a_{22}H=\{a_{22}{a}_{12}\}=\{a_2\}\\ & a_{23}H=\{a_{23}{a}_{12}\}=\{a_6\}\\ & a_{24}H=\{a_{24}{a}_{12}\}=\{a_{16}\}\end{array}$