Algebraic structures

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)

Oh I understand it! 

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