We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
113
2
avatar+324 

Hi I need your help:

Find the solution:  \(τ=(1,2,5,7)(3,4,6,7)(1,4,8,9,3)(4,7,2)\)

The answer is \((1,7)(2,8,9,4,3,6)(5) => (2,8,9,4,3,6)(1,7)\)

 

An other ex. is this:\(σ=(1,3,5)(2,7)(11,1,4)(5,6,2,8,9)(7,3)(4,5,2,1)\)

The answer is \((1,11,3,2,4,6,7,5,8,9)(10) =>(1,11,3,2,4,6,7,5,8,9)\)

I don't know/understand the steps to find this answer.

Thank you!

 Jun 2, 2019
 #1
avatar+22884 
+2

2.)

An other ex. is this: \(\sigma=(1,3,5)(2,7)(11,1,4)(5,6,2,8,9)(7,3)(4,5,2,1)\)
The answer is \((1,11,3,2,4,6,7,5,8,9)(10) \Rightarrow (1,11,3,2,4,6,7,5,8,9)\)

 

Start with \(x = 1 \text{ in } k(x)\)

 

\(\small{ \begin{array}{|ccccccccccc|c|cccccc|} \hline f &\circ& g &\circ& h &\circ& i &\circ& j &\circ& k & x & k(x) & j(k(x)) & i(j(k(x))) & \tiny{h(i(j(k(x))))} & \tiny{g(h(i(j(k(x)))))} & \tiny{f(g(h(i(j(k(x))))))} \\ \hline (1,3,5)&\circ&(2,7)&\circ&(11,1,4)&\circ&(5,6,2,8,9)&\circ&(7,3)&\circ&(4,5,2,1) &\color{red} 1 & 4 &4 & 4 & 11 &11 & 11 \\ & & & & & & & & & & & 11 & 11 &11 & 11 & 1 &1 &3 \\ & & & & & & & & & & & 3 & 3 &7 & 7 & 7 &2 &2 \\ & & & & & & & & & & & 2 & 1 &1 & 1 & 4 &4 &4 \\ & & & & & & & & & & & 4 & 5 &5 & 6 & 6 &6 &6 \\ & & & & & & & & & & & 6 & 6 &6 & 2 & 2 &7 &7 \\ & & & & & & & & & & & 7 & 7 &3 & 3 & 3 &3 &5 \\ & & & & & & & & & & & 5 & 2 &2 & 8 & 8 &8 &8 \\ & & & & & & & & & & & 8 & 8 &8 & 9 & 9 &9 &9 \\ & & & & & & & & & & & 9 & 9 &9 & 5 & 5 &5 &\color{red}1 \\ \hline & & & & & & & & & & & \color{red}10 & 10 &10 & 10 & 10 &10 &\color{red}10 \\ \hline \end{array} } \)

 

Permutation cycles:
(1, 11, 3, 2, 4, 6, 7, 5, 8, 9) (10)

 

laugh

 Jun 3, 2019
 #2
avatar+324 
+1

Oh I understand it! 

Thank you very much.The first \(τ\) it's wrong.

 Jun 5, 2019

10 Online Users

avatar