a question on symmetric group $S_5$[NBHM_2006_PhD Screening Test_algebra] 
Given that $x=(1 2)(3 4 5)\in S_5$ so its order $6$, and its a product
  of two cycle,I want to know whether $x$ commutes with all elements of
  $S_5$ and is it conjugate to $(4 5)(2 3 1)$?

Thank you for the help. 
 A: There is a rule for conjugating in $S_n$.  If you have two permutations $\sigma$ and $\omega$, then to compute $\sigma\omega\sigma^{-1}$, you just permute the list of numbers in $\omega$ by $\sigma$.  For instance
$$
(234)(12)(34)(234)^{-1}=(13)(42)
$$
because $(234)$ sends $2$ to $3$, $3$ to $4$ and so forth.  You should be able to work out that those two elements you have are conjugates.  Once you know that you can answer the second question.  If $x$ commutes with all elements of $S_5$, how big is its set of conjugates?
A: If you want to prove $\,x=(12)(345)\,$ conjugate to $\,(45)(231)\,$ choose an element $\,\sigma\in S_5\,$ s.t.
$$\sigma(1)=4\,,\,\sigma(2)=5\,,\,\sigma(3)=2\,,\,\sigma(4)=3\,,\,\sigma(5)=1\Longrightarrow \sigma=(14325)$$and then
$$\sigma(12)(345)\sigma^{-1}=(14325)(12)(345)(15234)=(123)(45)$$
From here that your second question's answer is yes. About the first one: choose some easy permutation, say $\,(13)\,$ , and check whether $\,x(13)x^{-1}=(13)\,$ ...
A: There's also a rule for recognizing conjugates in $S_n$: two permutations are conjugate if and only if they have the same cycle structure.  So for example $(12)(345)$ is conjugate to $(45)(231)$ and to $(14)(235)$, but not to $(12)(34)$ or $(1234)$.
A: Another hopefully-intuitive way of understanding how you might find a counterexample for whether $x$ commutes with all other elements of $S_5$ is by thinking of the action of $S_5$ and what your permutation $x$ represents.  You know that if $x$ commutes with all elements, then $x$ conjugates each element to itself: $xcx^{-1} = c$ for all elements $c$, by right-multiplying the definition of commutation $xc=cx$ by $x^{-1}$ on both sides.  (Similarly, left-multiplying both sides of the equation by $c^{-1}$ gives you $c^{-1}xc = x$, which shows why the answer to your second question is an answer to the first).  The mental image of a 15-puzzle or a Rubik's cube might help here: imagine two 'rings', one with the numbers $1$ and $2$ on it and one with the numbers $3, 4, 5$.  Then applying $x$ simultaneously 'rotates' both of these rings clockwise (taking $1$ to $2$ and vice versa, and rolling $3\rightarrow 4\rightarrow 5\rightarrow 3$).  Applying $x^{-1}$ is equivalent to rotating these rings counterclockwise, so in the absence of any interim terms the two rings get back to their original positions: $xx^{-1} = e$; your challenge is to find some operation $c$ so that inserting $c$ in between the two applications of $x$ will shuffle things up a bit.  Trying a $c$ that explicitly jumbles the two 'rings' seems like the best way of throwing a gear in the works, so take for instance $c=(13)$ and run the calculations...
