What is the order of $C_{S_n}((12)(34)) \; \forall n\geq 4$. Determine the elements of this centralizer explicitly
Since $(12)(34)$ is a conjugacy class in $S_n \; \forall n\geq 4$, denote this class as $K$
Then we have $$|K| = \frac{|S_n|}{|C_{S_n}((12)(34))|}=\frac{(C^{n}_{2}\cdot\frac{2!}{2})(C^{n-2}_{2}\cdot\frac{2!}{2})}{2!} = \frac{n(n-1)(n-2)(n-3)}{8}$$
$$\Rightarrow |C_{S_n}((12)(34))|=8\cdot(n-4)!$$
But how do I determine these elements explicitly?
So we want to have $\sigma(12)(34)\sigma^{-1}=(12)(34)$. We have the following set $S$ which are in the stabilizer of $(12)(34)$: $S:=\{(12)(34),e,(13)(24),(14)(23),(1324),(1423),(12),(34)\}$. So let $S'_{n-4} \subset S_n$ denote the permutations in $S_n$ that fixes $\{1,2,3,4\}$. The centralizer should be $S\times S'_{n-4}:=\{s\sigma|s \in S, \sigma\in S'_{n-4}\}$.