How to use class equation for determining the center of $S_4$

Question

How to use class equation for determining the center of $S_4$

$$|G|=|Z(G)|+\sum_x [G:C_G(x)]$$ So I guess I need to find $$|G|-\sum_x [G:C_G(x)]=|Z(G)|$$

Well $|S_4|=4!=24$

and $C_G(x)$ is the set of all group elements that commute with $x$. This seems like I would need to use brute force, and that would take a very long time, since each elements has numerous elements in it's centralizer. How do I work this out?

Answer 1

G= disjoint union of classes [x] under the conjugate relation. Any x aside from the identity element has only one of the forms: (12) with 6 elements , (123) with 8 elements , (1234) with 6 elements, (12)(34) with 3 elements elements. Thus the cardinalities of the nontrivial classes are 6,8,6,and 3. We are left with only one trivial class of one element and it is [e]={e}. Thus Z(G)={e}.

Answer 2

As stated in the comments, the elements of $S_n$ of same cycle type are conjugate. And conversely.

So, there's only one odd man out: $e$.

There's more than one of each (non-trivial) cycle type.

We get immediately that the center is trivial. Actually for all $n\ge3$. ($S_2$ is abelian of course. )

That's the class equation for $S_n,\,n\ge3$ is $$\underbrace {1}_{\text{center}}+\underbrace{\text{junk}}_{ \text{classes of size $\gt1$}}$$.

When you look at the class equation, adding up the $1$'s gives the order of the center.