I am trying to calculate the following cycle permutation:
$$(x_d,x_a,x_c)^{-1}(x_c,x_e,x_b)^{-1}(x_d,x_a,x_c)(x_c,x_e,x_b).$$ But I yield the wrong answer; according to the proof (the complete proof is given at the bottom of the post), it should be: $$(x_a,x_b,x_c)=(x_d,x_a,x_c)^{-1}(x_c,x_e,x_b)^{-1}(x_d,x_a,x_c)(x_c,x_e,x_b).$$ I started by writing $(x_d,x_a,x_c)^{-1}(x_c,x_e,x_b)^{-1}=(x_c,x_a,x_d)(x_b,x_e,x_c)$, then we have - $$(x_c,x_a,x_d)(x_b,x_e,x_c)(x_d,x_a,x_c)(x_c,x_e,x_b)$$
I start from the right most cycle, and the rightmost element $x_b$ which maps to $x_c$, in the next cycle from right $x_c$ maps to $x_d$, in the next cycle there is no $x_d$, in the next cycle $x_d$ maps to $x_c$.
Similarly $x_c$ maps to $x_a$, but the middle element of the rightmost cycle $x_e$ maps to $x_e$ not to $x_b$, so my result is $(x_d,x_a,x_c)^{-1}(x_c,x_e,x_b)^{-1}(x_d,x_a,x_c)(x_c,x_e,x_b)=(x_a,x_e,x_c)$.
The source of the problem is the article "Galois for beginners" by John Stillwell, see the complete proof below:
I do not understand exactly which part of the following routine verification is the problem, but here it is in full: The product of the four $3$-cycles acts on the $x_i$ as follows: \begin{eqnarray*} x_a\ \mapsto\ x_a\ \mapsto\ x_c\mapsto\ x_b\ \mapsto\ x_b,\\ x_b\ \mapsto\ x_c\ \mapsto\ x_d\mapsto\ x_d\ \mapsto\ x_c,\\ x_c\ \mapsto\ x_e\ \mapsto\ x_e\mapsto\ x_c\ \mapsto\ x_a,\\ x_d\ \mapsto\ x_d\ \mapsto\ x_a\mapsto\ x_a\ \mapsto\ x_d,\\ x_e\ \mapsto\ x_b\ \mapsto\ x_b\mapsto\ x_e\ \mapsto\ x_e. \end{eqnarray*} In cycle notation this is $(x_a\ x_b\ x_c)$.