How can I prove that $C_{S_4}((12)(34))$ is a subgroup of order 8? I would like to know if there is a smart method (i.e. avoid to make all by hand) to understand the order of $C_{S_4}((12)(34))$. The only thing I thought is that $|S_4:C_{S_4}((12)(34))|=|\mathcal{O}((12)(34))|$ and then one can  try to reason about the order of the orbit. But this didn't helped me!
Thank you

The conjugacy class of $(12)(34)$ consists of all products of disjoint transpositions, since two permutations are conjugate if and only if they have the same cycle type. So, only $(13)(24)$ and $(14)(32)$ are conjugate to $(12)(34)$. As you mention, the size of the conjugacy class equals the index of the stabilizer of $(12)(34)$. Hence, the stabilizer has index $3$, and therefore order $8$.
