# Computing the centraliser for elements in $S_4$.

Given the group $$S_4$$ we know that it has 5 cycle types of type. I want to pick one element from each of the conjugacy classes and find its centralizer.

$$\begin{matrix} type & no.elements & permutation \\ 1,1,1,1 & 1 & id \\ 2,1,1 & 6 & (12),(13),(14),(23),(24),(34)\\ 3,1&8&(123),(124),(134),(234),(132),(142),(143),(243) \\4&6&(1234),(1342),(1423),(4321),(2431),(3241)\\2,2&3&(12)(34),(13)(24),(14)(23) \\ \end{matrix}$$

Relevant knowledge:

I know that conjugacy classes are determined by cycle type, which means that $$S_4$$ has 5 conjugacy classes of sizes 1,3,6,6 and 8.

where the centraliser is given by

$$C_{S_4}(\tau):=\{x \in S_4|x\tau=\tau x\}$$

i.e. the set of elements which commute with $$\tau$$. i.e. the elements x s.t.$$x\tau x^{-1}=\tau$$

and the conjugacy class is a non-empty subset X of $$S_4$$ s.t. for x,y$$\in X ,\exists g\in S_4 s.t. gxg^{-1}=y$$ and if x $$\in X, g \in S_4$$ then $$gxg^{-1}=y \in X$$.

I also know that disjoint cycles commute so we can use this fact for the 2,1,1 cyle types.

Eg. for (1,2) every two cyle that doesn't contain 1 or 2 is an element of the centraliser. also for the identity cycle $$C(id)=S_4$$.

My question :

How do we go about finding the centraliser for one element of each conjugacy class ?

Is it a computation or is there any more relevant theorems ( i missed some class due to illness)

what is the method of computation?

• Here are a few things which helps: 1) $|C(\tau)| = |S_4|/|O(\tau)|$, where $O(\tau)$ is the congruence class containing $\tau$; 2) $\sigma(a_1 a_2 \cdots a_m)\sigma^{-1} = (\sigma(a_1) \sigma(a_2) \cdots \sigma(a_m))$. Oct 23 '18 at 19:20
• Do you know how to work out the size of the centraliser? Have you figured out that the cyclic group generated by the element itself is part of its centraliser? Oct 23 '18 at 19:21

I think about conjugation in general as a change in basis.

Conjugation doesn't change the structure of the permutation that is being conjugated.

Lets take the permutation (12)(34) and conjugate it with (123)

(123)(12)(34)(132)

What this says to me, is to take the 1 and replace it with a 2, the 2 and replace it with a 3, and the 3 and replace it with 1.

(23)(14)

Once you understand what conjugation does, it is easier to see which conjugations will leave an element unchanged.