Editing Conjugacy class (section)

==Conjugacy class equation==

If <math>G</math> is a [[finite group]], then for any group element <math>a,</math> the elements in the conjugacy class of <math>a</math> are in one-to-one correspondence with [[coset]]s of the [[centralizer]] <math>\operatorname{C}_G(a).</math> This can be seen by observing that any two elements <math>b</math> and <math>c</math> belonging to the same coset (and hence, <math>b = cz</math> for some <math>z</math> in the centralizer <math>\operatorname{C}_G(a)</math>) give rise to the same element when conjugating <math>a</math>: 
<math display="block">bab^{-1} = cza(cz)^{-1} = czaz^{-1}c^{-1} = cazz^{-1}c^{-1} = cac^{-1}.</math>
That can also be seen from the [[orbit-stabilizer theorem]], when considering the group as acting on itself through conjugation, so that orbits are conjugacy classes and stabilizer subgroups are centralizers. The converse holds as well.

Thus the number of elements in the conjugacy class of <math>a</math> is the [[Index of a subgroup|index]] <math>\left[ G : \operatorname{C}_G(a)\right]</math> of the centralizer <math>\operatorname{C}_G(a)</math> in <math>G</math>; hence the size of each conjugacy class divides the order of the group.

Furthermore, if we choose a single representative element <math>x_i</math> from every conjugacy class, we infer from the disjointness of the conjugacy classes that 
<math display="block">|G| = \sum_i \left[ G : \operatorname{C}_G(x_i)\right],</math> 
where <math>\operatorname{C}_G(x_i)</math> is the centralizer of the element <math>x_i.</math> Observing that each element of the center <math>\operatorname{Z}(G)</math> forms a conjugacy class containing just itself gives rise to the '''class equation''':<ref>Grillet (2007), [{{Google books|plainurl=y|id=LJtyhu8-xYwC|page=57|text=The Class Equation}} p. 57]</ref>
<math display="block">|G| = |{\operatorname{Z}(G)}| + \sum_i \left[G : \operatorname{C}_G(x_i)\right],</math>
where the sum is over a representative element from each conjugacy class that is not in the center.

Knowledge of the divisors of the group order <math>|G|</math> can often be used to gain information about the order of the center or of the conjugacy classes.

===Example===

Consider a finite [[p-group|<math>p</math>-group]] <math>G</math> (that is, a group with order <math>p^n,</math> where <math>p</math> is a [[prime number]] and <math>n > 0</math>). We are going to prove that {{em|every finite <math>p</math>-group has a non-[[Trivial (mathematics)|trivial]] center}}.

Since the order of any conjugacy class of <math>G</math> must divide the order of <math>G,</math> it follows that each conjugacy class <math>H_i</math> that is not in the center also has order some power of <math>p^{k_i},</math> where <math>0 < k_i < n.</math> But then the class equation requires that <math display="inline">|G| = p^n = |{\operatorname{Z}(G)}| + \sum_i p^{k_i}.</math> From this we see that <math>p</math> must divide <math>|{\operatorname{Z}(G)}|,</math> so <math>|\operatorname{Z}(G)| > 1.</math>

In particular, when <math>n = 2,</math> then <math>G</math> is an abelian group since any non-trivial group element is of order <math>p</math> or <math>p^2.</math> If some element <math>a</math> of <math>G</math> is of order <math>p^2,</math> then <math>G</math> is isomorphic to the [[cyclic group]] of order <math>p^2,</math> hence abelian. On the other hand, if every non-trivial element in <math>G</math> is of order <math>p,</math> hence by the conclusion above <math>|\operatorname{Z}(G)| > 1,</math> then <math>|\operatorname{Z}(G)| = p > 1</math> or <math>p^2.</math> We only need to consider the case when <math>|\operatorname{Z}(G)| = p > 1,</math> then there is an element <math>b</math> of <math>G</math> which is not in the center of <math>G.</math> Note that <math>\operatorname{C}_G(b)</math> includes <math>b</math> and the center which does not contain <math>b</math> but at least <math>p</math> elements.  Hence the order of <math>\operatorname{C}_G(b)</math> is strictly larger than <math>p,</math> therefore <math>\left|\operatorname{C}_G(b)\right| = p^2,</math> therefore <math>b</math> is an element of the center of <math>G,</math> a contradiction. Hence <math>G</math> is abelian and in fact isomorphic to the direct product of two cyclic groups each of order <math>p.</math>