Compute Centralizer of $G=S_{3}$ and $a=(1,2,3)$

Question

I remember any permutation group with has 3 or more element,It can't be compute. Am I wrong ? the centralizer in this problem is not exist? since it need it must be commute. Anyone can confirm me It has centralizer or not

Answer 1

Considering the defenition of $C_{S_3}(a)$ which is $$\{x\in S_3|~xa=ax\}$$ we can guess which possible elements of the group would commute with $a$. Obviously, $aa=aa$ so $a\in C_{S_3}(a)$. This shows that -as you read in Zelos' comment- the centralzer is not empty and so it exists. If $x=(1,2)$ so $$xa=(1,3)\neq(2,3)=ax$$ You can check the above inequality when $x=(3,2)$. If $x=(1,3,2)$ it is not that hard seeeing that $$x=a^{-1}$$and so it commutes with $a$. We have nothing to do with $e_{S_3}$ -since it is a good boy- so it is contained in $C_{S_3}(a)$. Sorry if this computational way is emply of any abstract face.