Permutation in $S_n$ By book states the following:

"Given the permutation $( 1 , 2)$ in $S_n$, what elements commute with
  it ? Certainly any permutation leaving both $1$ and $2$ fixed does.
  There are $(n - 2) !$ such. Also $( 1 , 2)$ commutes with itself. This
  way we get $2 (n - 2) !$ elements in the group generated by $( 1 , 2)$
  and the $(n- 2) !$ permutations leaving $1$ and $2$ fixed. Are there
  others? There are $n(n - 1 ) /2$ transpositions and these are
  precisely all the conjugates of $( 1 , 2).$ Thus the conjugate class
  of $( 1 , 2)$ has in it $n(n - 1 ) /2$ elements."

How did they get those results? 
 A: One way to do this is the following :
If $\sigma = (1 2)$, then for any $\tau \in S_n$,
$$
\eta := \tau\sigma\tau^{-1} = (\tau(1) \tau(2))
$$
In particular, $\eta$ is also a a transposition. Furthermore, if $\eta$ is any transposition, then there is a $\tau \in S_n$ such that $\eta = \tau\sigma\tau^{-1}$. Hence, the conjugacy class of $\sigma$ consists of all the transpositions in $S_n$.
Count the number of transpositions in $S_n$. This is the cardinality of $C(\sigma)$, the conjugacy class of $\sigma$.
By the orbit-stabilizer theorem, the centralizer $Z(\sigma)$ and $C(\sigma)$ are related by the equation
$$
|Z(\sigma)|C(\sigma)| = |S_n| = n!
$$
Hence,
$$
|Z(\sigma)| = \frac{n!}{|C(\sigma)|}
$$
A: This is an excerpt from Dummit and Foote... The only logical step is that the size of the conjugacy class is the index of the centralizer, i.e. $|S_{n} :\; C_{S_{n}}((12))|$ = size of conjugacy class. Hence, $|C_{S_{n}}((12))| = \frac{n!}{\frac{n\cdot (n-1)}{2}} = 2\cdot (n-2)!$. Hence, we conclude that there are no other elements of $S_{n}$ that commute with $(12)$.
