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# Algebraic structures

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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
+23137
+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)

Jun 3, 2019
#2
+336
+1

Oh I understand it!

Thank you very much.The first $$τ$$ it's wrong.

Jun 5, 2019