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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?

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  • $\begingroup$ Do you know that two elements in $S_n$ are conjugate if and only if they have the same cycle type? For instance $(12)(34)$ is conjugate to $(13)(24)$ but not $(123)$. $\endgroup$ – Prahlad Vaidyanathan Jun 20 '15 at 6:05
  • $\begingroup$ @PrahladVaidyanathan Yep, I have proven this in the past. Am I meant to use Orbits? $\endgroup$ – Permute Jun 20 '15 at 6:50
  • $\begingroup$ No, but you can count the number of elements in each conjugacy class. For instance, the number of elements in the conjugacy class of $(12)(34)$ is precisely the number of ways you can partition $\{1,2,3,4\}$ into 2 pairs. $\endgroup$ – Prahlad Vaidyanathan Jun 20 '15 at 9:16
  • $\begingroup$ @PrahladVaidyanathan I have done this now and I have: 1-1-1-1 type has 1 1-1-2 type has 6 1-3 type has 8 4 type has 6 2-2 type has 3 For the total of 24. How do I solve the problem with this though? $\endgroup$ – Permute Jun 20 '15 at 9:23
  • $\begingroup$ Now note that something is in the center if and only if its conjugacy class is a singleton. $\endgroup$ – Prahlad Vaidyanathan Jun 20 '15 at 10:51
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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}.

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  • $\begingroup$ So I take it that $\sum_{x}[G:C_G(x)]$ is the sum of conjugacy classes? I wasn't entirely sure how to read that one. It seems like the quotient of $G$ with the centrilizer. $\endgroup$ – Permute Jun 20 '15 at 10:48
  • $\begingroup$ @Permute The size of an orbit is the index of the stabilizer. (See here.) In this case the action is conjugation, so the stabilizer is the centralizer. The size of the orbit of $x$ is thus $[G: C_G(x)]$. $\endgroup$ – Viktor Vaughn Jun 20 '15 at 17:03

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