Editing Conjugacy class (section)

===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>