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Question Number 192342 by Mastermind last updated on 15/May/23
1) Compute in S_a , a^(−1) ba where a=(1 2)(1 3 5), b=(1 5 7 1) 2) Given permutation α = (1 2)(3 4), β = (1 3)(5 6). Find a permutation x∈S_6 ∃αx = β. help!
$$\left.\mathrm{1}\right)\:\mathrm{Compute}\:\mathrm{in}\:\mathrm{S}_{\mathrm{a}} \:,\:\mathrm{a}^{−\mathrm{1}} \mathrm{ba}\:\:\mathrm{where}\: \\ $$$$\mathrm{a}=\left(\mathrm{1}\:\mathrm{2}\right)\left(\mathrm{1}\:\mathrm{3}\:\mathrm{5}\right),\:\mathrm{b}=\left(\mathrm{1}\:\mathrm{5}\:\mathrm{7}\:\mathrm{1}\right) \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Given}\:\mathrm{permutation}\:\alpha\:=\:\left(\mathrm{1}\:\mathrm{2}\right)\left(\mathrm{3}\:\mathrm{4}\right), \\ $$$$\beta\:=\:\left(\mathrm{1}\:\mathrm{3}\right)\left(\mathrm{5}\:\mathrm{6}\right).\:\mathrm{Find}\:\mathrm{a}\:\mathrm{permutation} \\ $$$$\mathrm{x}\in\mathrm{S}_{\mathrm{6}} \:\exists\alpha\mathrm{x}\:=\:\beta. \\ $$$$ \\ $$$$\mathrm{help}! \\ $$
Answered by aleks041103 last updated on 15/May/23
1) b=(1 5 7 1) = (1 5 7) a=(1 2)(1 3 5)=(2 1)(1 3 5)=(2 1 3 5) ⇒a^(−1) =(5 3 1 2) ⇒a^(−1) ba= =(5 3 1 2)(1 5 7)(2 1 3 5)= = (((1 2 3 4 5 6 7)),((1 3 7 4 5 6 2)) ) = (2 3 7)=a^(−1) ba 2) αx=β, α,β,x∈S_6 , x=? Since S_6 is a group, then ∃α^(−1) ⇒α^(−1) αx=x=α^(−1) β α=(1 2)(3 4) ⇒α^(−1) =(4 3)(2 1)=(3 4)(1 2)=(1 2)(3 4)=α ⇒x=αβ=(1 2)(3 4)(1 3)(5 6)= =(1 2)(4 3)(3 1)(5 6)= =(1 2)(4 3 1)(5 6)= =(2 1)(1 4 3)(5 6)= =(2 1 4 3)(5 6)= =(1 4 3 2)(5 6)=x
$$\left.\mathrm{1}\right) \\ $$$${b}=\left(\mathrm{1}\:\mathrm{5}\:\mathrm{7}\:\mathrm{1}\right)\:=\:\left(\mathrm{1}\:\mathrm{5}\:\mathrm{7}\right) \\ $$$${a}=\left(\mathrm{1}\:\mathrm{2}\right)\left(\mathrm{1}\:\mathrm{3}\:\mathrm{5}\right)=\left(\mathrm{2}\:\mathrm{1}\right)\left(\mathrm{1}\:\mathrm{3}\:\mathrm{5}\right)=\left(\mathrm{2}\:\mathrm{1}\:\mathrm{3}\:\mathrm{5}\right) \\ $$$$\Rightarrow{a}^{−\mathrm{1}} =\left(\mathrm{5}\:\mathrm{3}\:\mathrm{1}\:\mathrm{2}\right) \\ $$$$\Rightarrow{a}^{−\mathrm{1}} {ba}= \\ $$$$=\left(\mathrm{5}\:\mathrm{3}\:\mathrm{1}\:\mathrm{2}\right)\left(\mathrm{1}\:\mathrm{5}\:\mathrm{7}\right)\left(\mathrm{2}\:\mathrm{1}\:\mathrm{3}\:\mathrm{5}\right)= \\ $$$$=\begin{pmatrix}{\mathrm{1}\:\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\:\mathrm{6}\:\mathrm{7}}\\{\mathrm{1}\:\mathrm{3}\:\mathrm{7}\:\mathrm{4}\:\mathrm{5}\:\mathrm{6}\:\mathrm{2}}\end{pmatrix}\:=\:\left(\mathrm{2}\:\mathrm{3}\:\mathrm{7}\right)={a}^{−\mathrm{1}} {ba} \\ $$$$ \\ $$$$\left.\mathrm{2}\right) \\ $$$$\alpha{x}=\beta,\:\alpha,\beta,{x}\in{S}_{\mathrm{6}} ,\:{x}=? \\ $$$${Since}\:{S}_{\mathrm{6}} \:{is}\:{a}\:{group},\:{then}\:\exists\alpha^{−\mathrm{1}} \\ $$$$\Rightarrow\alpha^{−\mathrm{1}} \alpha{x}={x}=\alpha^{−\mathrm{1}} \beta \\ $$$$\alpha=\left(\mathrm{1}\:\mathrm{2}\right)\left(\mathrm{3}\:\mathrm{4}\right) \\ $$$$\Rightarrow\alpha^{−\mathrm{1}} =\left(\mathrm{4}\:\mathrm{3}\right)\left(\mathrm{2}\:\mathrm{1}\right)=\left(\mathrm{3}\:\mathrm{4}\right)\left(\mathrm{1}\:\mathrm{2}\right)=\left(\mathrm{1}\:\mathrm{2}\right)\left(\mathrm{3}\:\mathrm{4}\right)=\alpha \\ $$$$\Rightarrow{x}=\alpha\beta=\left(\mathrm{1}\:\mathrm{2}\right)\left(\mathrm{3}\:\mathrm{4}\right)\left(\mathrm{1}\:\mathrm{3}\right)\left(\mathrm{5}\:\mathrm{6}\right)= \\ $$$$=\left(\mathrm{1}\:\mathrm{2}\right)\left(\mathrm{4}\:\mathrm{3}\right)\left(\mathrm{3}\:\mathrm{1}\right)\left(\mathrm{5}\:\mathrm{6}\right)= \\ $$$$=\left(\mathrm{1}\:\mathrm{2}\right)\left(\mathrm{4}\:\mathrm{3}\:\mathrm{1}\right)\left(\mathrm{5}\:\mathrm{6}\right)= \\ $$$$=\left(\mathrm{2}\:\mathrm{1}\right)\left(\mathrm{1}\:\mathrm{4}\:\mathrm{3}\right)\left(\mathrm{5}\:\mathrm{6}\right)= \\ $$$$=\left(\mathrm{2}\:\mathrm{1}\:\mathrm{4}\:\mathrm{3}\right)\left(\mathrm{5}\:\mathrm{6}\right)= \\ $$$$=\left(\mathrm{1}\:\mathrm{4}\:\mathrm{3}\:\mathrm{2}\right)\left(\mathrm{5}\:\mathrm{6}\right)={x} \\ $$
Commented by Mastermind last updated on 18/May/23
Thank you boss, you really tried
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{boss},\:\mathrm{you}\:\mathrm{really}\:\mathrm{tried} \\ $$

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