Size of the conjugacy class of the element σ = (1, 2, 3)(4, 5) in the group Sym5 . Order of the centraliser CSym5 (σ)

Question

I am a little confused with this question. Am i right in looking for all the elements in Sym5, that are conjugate to $\sigma$ , in other words $a = b\sigma b^{-1}$ for $a,b \in $ S5

ther will be 5 single cycles correct? I am not sure what else EDIT: Do I have to work out EVERY single permutation.. from (1) ,(12) ,(12)(34) ,(12)(34), (123) , (1234), (12345)
and multiply a cycle above by $ \sigma $ and its inverse??

Also, the definition of a centraliser is that of a subset of a group, so is the second part saying $\sigma$ is a subset?

Answer 1

Conjugate permutations have the same cycle type. And permutations with the same cycle type are conjugate. Thus you want to count the number of $3$-cycles. There are $20$, I believe.

Next you can use the orbit-stabilizer theorem. Consider the conjugation action. We get that the stabilizer, which is the centralizer, has order the index of the order of the orbit. So $5!/20=6$.