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I was reading this proof:

https://proofwiki.org/wiki/Number_of_Conjugates_is_Number_of_Cosets_of_Centralizer

and I don't understand the very last part, why the order of the conjugacy class divides the order of the group. It says that it follows by Lagrange's but the conjugacy class is not necessarily a subgroup? I'm thinking that perhaps the index of the centralizer is exactly equal to the order of some subgroup of $G$, but I'm not sure. Can someone explain?

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By Lagrange the order of the centralizer divides the order of the group. The number of conjugacy classes is the number of cosets of the centralizer, which is the same as the index of the centralizer. This is a divisor of the order of the group because it is the order of the group divided by the order of the centralizer.

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  • $\begingroup$ I know, but lagrange's only says the order of the centralizer divides the order of the group. Why does this mean the number of cosets of the centralizer divides it as well? $\endgroup$ – b_pcakes Nov 7 '16 at 0:12
  • $\begingroup$ @b_pcakes Edited further explanation. Alan Wang's answer says the same, but with notation. $\endgroup$ – Servaes Nov 7 '16 at 0:13
  • $\begingroup$ Hm, this the case because the centralizer is a normal subgroup? $\endgroup$ – b_pcakes Nov 7 '16 at 0:14
  • $\begingroup$ No; in general if $d$ is a divisor of $n$, then $\tfrac{n}{d}$ is also a divisor of $n$, because $d\cdot\tfrac{n}{d}=n$. $\endgroup$ – Servaes Nov 7 '16 at 0:14
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    $\begingroup$ Oops nvm! It is $\endgroup$ – b_pcakes Nov 7 '16 at 0:19
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Since $|C_a|=[G:C_G(a)]$, if $G$ is a finite group, we have $$|C_a|=\frac{|G|}{|C_G(a)|}$$ $$|G|=|C_a||C_G(a)|$$ So obviously, $|C_a|$ divides $|G|$

Maybe the proof there wants to mean that the index of a subgroup $[G:H]$ divides $|G|$.

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