So I have $g = (1, 2)(3, 4) = \left(\begin{array}{cccc}1 & 2 & 3 & 4 \cr 2 & 1 & 4 & 3 \end{array}\right).$ I want to find $C_{S_4}(g)$. I know that I can just ``bruteforce'' the group, but I thought there might be a simplier way.
So far I think that $A = (1,2)^k(3,4)^n \in C_{S_4}(g)$ where $k, n \in \mathbb{Z}$, but I am not sure if $A = C_{S_4}(g)$
