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I want to know that whether follow equality holds: $ |N_G(C_G(a)):C_G(a)|=|a^G\cap C_G(a)|.$ It is easy to see that the left hand is no more than the right hand. I think this equality does not hold, but I can't find a counter example.

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 I think this question would be better phrased: "If $b$ is conjugate to $a$ and in $C_G(a)$, then there exists $c\in N_G(C_G(a))$ such that $b=cac^{-1}$." – Will Sawin May 16 2012 at 5:40
@Will Sawin, your statement is weaker than the question. because you gave a map from one set to another. However, the question is only asking the equal cardinality statement. – i707107 May 16 2012 at 6:18
Since the map is always an injection, equal cardinality is the same as surjectivity. – Will Sawin May 16 2012 at 6:26
You might be interested in this thread - mathoverflow.net/questions/96652/… - where there are some questions concerning the quantity $|a^G\cap C_G(a)|$ . – Nick Gill May 16 2012 at 10:47
Thanks, Nick. I am very interested in the problem discussed in this thread. – Wei Zhou May 16 2012 at 12:49

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Take $G$ to be a symmetric group of degree at least 5, and $a$ a transposition.

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Thanks. You are right. I have considered this group, but only considered the elements of order 3 and 5, where the equality holds. – Wei Zhou May 16 2012 at 5:53
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Write $C$ for $C_G(a)$ and $N$ for $N_G(C_G(a))$. Although, as Jeremy has demonstrated, the equality $|N:C| = |a^G\cap C|$ does not hold in general, it seems to me that there is a chance of characterising when it is true.

Consider, for instance, the situation when $C$ is a TI-subgroup. That is to say, $|C\cap C^g|$ is equal to ${1}$ or $G$ for every $g\in G$. I think it's fairly clear that in this case one does indeed have $$|N:C| = |a^G\cap C|.$$

It seems to me that something like a converse might also be true. I don't have the wherewithal to work this through right now but I will try and return to it in due course...

Edit: As Mark has pointed out in comments below, the TI-condition is too strong if we want to characterise when $$|N:C| = |a^G\cap C|.$$ The condition that we want is this: $$x\in G\backslash N \Longrightarrow h^x \not\in H.$$ The contrapositive might be clearer: $$h^x \in H \Longrightarrow x\in N.$$

Clearly the TI-condition implies this, but it also accounts for the example given by Mark... And it's quite easy to see that it precisely characterizes when $|N:C| = |a^G\cap C|.$

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 Hi Nick. Maybe there is some sort of converse, but it will definitely need some extra hypotheses. For example, take $G = S_6$ and let $a = (123456) \in S_6$. Then $C = \left< a \right>$ and $N$ is the dihedral group of order $12$, so $|N : C| = 2$ and $a^G \cap C = \lbrace a,a^{-1} \rbrace$. But $C$ is not a trivial intersection subgroup of $G$ because both $C^{(23)(56)} = \left< (132465) \right>$ and $C$ contain $(14)(25)(36)$. – Mark Wildon May 16 2012 at 14:54
I think that the hypotheses that $C_G(a)=\langle a \rangle$ also implies that equality holds – Wei Zhou May 16 2012 at 15:08
Wei Zhou, you are right, and the condition in the edit above confirms this. – Nick Gill May 18 2012 at 13:16
Nick, do you know any groups satisfying the equality for every $h \in G$, except Dedekind groups? – Wei Zhou May 19 2012 at 0:58
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 I could be wrong but it seems to me that $G=C_p\rtimes C_{p-1}$ would satisfy the equality for all $g\in G$. In fact Frobenius groups are probably a good source of examples in general - one would just need to be sure that elements in the kernel satisfy the equality. Taking the kernel to be abelian would be enough (as for $G=C_p\rtimes C_{p-1}$) but there may well be other examples... – Nick Gill May 19 2012 at 9:21
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