How to show that the cycle $(2 5) = (2 3) (3 4) (4 5) (4 3) (3 2)$ I generally do not have any problem multiplying cycles, but I've seen on Wikipedia that
$$(2 5) = (2 3) (3 4) (4 5) (4 3) (3 2). $$
I started following the path of $2$ on the right:
$$2\to3\to4\to5\to \ ?$$
Where does $5$ go? I should stop here, right? Then $2\to 5$, that is, $(25)$. But what about $(23)(34)$?
 A: This is another conjugation problem in disguise:
$(2\ 3)(3\ 4)(4\ 5)(4\ 3)(3\ 2) = (2\ 3)[(3\ 4)(4\ 5)(3\ 4)^{-1}](3\ 2)$
$= (2\ 3)(3\ 5)(2\ 3)^{-1}$ (since $(3\ 4)$ takes $4 \to 3$ and fixes $5$)
$= (2\ 5)$ (since $(2\ 3)$ takes $3 \to 2$ and fixes $5$).
A: Let's go through the product(s) 
$$
(2 3) (3 4) (4 5) (4 3) (3 2)
$$
step by step.
Start with $2$, by writing $(2$, and see how the elements map out:


*

*$2\mapsto 3\mapsto 4\mapsto 5\mapsto 5\mapsto 5\qquad:\qquad(2\; 5$

*$5\mapsto 5\mapsto 5\mapsto 4\mapsto 3\mapsto 2\qquad:\qquad(2\; 5)(3$

*$3\mapsto 2\mapsto 2\mapsto 2\mapsto 2\mapsto 3\qquad:\qquad(2\; 5)(3)(4$

*$4\mapsto 4\mapsto 3\mapsto 3\mapsto 4\mapsto 4\qquad:\qquad(2\;5)(3)(4)$


The process has terminated and we know $(2\;5)(3)(4)=(2\;5)$. That is, we can see that
$$
(2\;5) = (2 3) (3 4) (4 5) (4 3) (3 2).
$$
Did all of those steps make sense?
A: Assuming there are cycles in $S_5$, $$(2\ 3) = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 1 & 3 & 2 & 4 & 5 \end{pmatrix}=(2\ 3)(1)(4)(5) \\
(3\ 4) = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 4 & 3 & 5 \end{pmatrix}=(3\ 4)(1)(2)(5)$$
Hence, $5\mapsto 5$ in both of those cycles.
However, to conclude that $5 \mapsto 2$ you need to follow it through the cycles as you followed 2, starting from the right to the left.
See my answer here for a detailed explanation of composing cycles: Shortcut for composing cycles
A: Starting from the right, we see that $2\to3\to4\to5$ and $5\to4\to3\to2$. On the other hand, $3\to2\to3$. Every other number just maps to itself, which you can check for yourself.
Note that if a number is fixed, then we don't write it in the shorthand notation for the permutation. Thus we just have $(2\ 5)$
