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In order to prove this, I did the following:

first, I showed that conjugacy forms an equivalence relation, then I can find its conjugacy classes. I understand how to form a conjugacy class, given a group. So, the index of the normalizer of $\{x\}$ in $G$ is the number of cosets of $N_x$ in $G$, right?

The set $N_x$ should be $\{a: ax = xa\}$, right? How does this set look? I need to find it, so I can take its quotient with $G$ and count how many of them exist. But how to do it? I'm lost.

UPDATE:

I think I got it:

$a\in N_G(H)b \iff ab^{-1} \in N_G(H) \iff ab^{-1}H = Hab^{-1} \iff b^{-1}H = a^{-1}Hab^{-1} \iff b^{-1}Hb = a^{-1}Ha$

So, two conjugates are equal $\iff $ their elements are in the same coset of $N_G(H)$

Thus, there is an explicit bijection between the set of different conjugates and the normalizer, so there are $[G:N_G(H)]$ different conjugates of $H$.

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2 Answers 2

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What I like to do is to set up an explicit bijection between the set of conjugates of $x$ and the set of cosets of $N_x$.

Now the set of conjugates of $x$ looks like $ \{y^{-1}xy \ | \ y \in G\}$. The obvious mapping,

$$ y^{-1}xy \; \mapsto \; \text{right coset of $N_x$ for $y$} $$

works just fine.Show that this map is one-to-one which is a fun little exercise. The fact that it is onto is plain.

Then you can argue that the domain and the range of the bijective map above must have the same number of elements as long as they are finite. The number of elements in the range of the bijection is the index of $N_x$in $G$.

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  • $\begingroup$ how would $\text{right coset of $N_x$ for $y$}$ look like? is it $N_x y$? I'm having trouble visualizing it. Could you help me? $\endgroup$ Aug 17, 2015 at 16:27
  • $\begingroup$ is it for all $y$ or a particular $y$? $\endgroup$ Aug 17, 2015 at 16:27
  • $\begingroup$ Yes, well the right coset of $N_x$ for $y$ is the set $\{ny \ \mid \ n \in N_x \}$ as usual. I used words since $N_x y$ looks ugly. Every conjugate of $x$ can be identified as $y^{-1}xy$ for some $y \in G$. Map this particular conjugate to the corresponding right coset involving $y$. Of course this $y$ here need not be unique. You can totally have $y^{-1}xy = z^{-1}xz$ with $y \neq z$. So you need to also prove that this function is well-defined. $\endgroup$
    – Ishfaaq
    Aug 17, 2015 at 16:30
  • $\begingroup$ What about: $\phi (y_1^{-1}xy_2)=\phi (y_2^{-1}xy_2)\implies N_xy_1 = N_xy_2$? I need to end up with $y_1^{-1}xy_2=y_2^{-1}xy_2$ rigth? I know that the set $N_x$ is the set of all elements of $G$ that commute with $x$, so it can be used to prove the result? Also, why this mapping is obvious? $\endgroup$ Aug 21, 2015 at 21:32
  • $\begingroup$ Í think I'm understanding this question, but why would somebody ask that? Is there a reason for this? It seems like a pretty unintuitive result for me. $\endgroup$ Aug 21, 2015 at 21:33
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If you're familiar with the orbit stabilizer theorem, this result is a quick application. (If you aren't familiar with this result, it still might be useful to know that this problem is a special instance of a more general phenomenon!)

The theorem states that if $G$ is a finite group acting on a set $S$, and $O$ is the orbit of some $x \in S$, then

$$|O| = [G: \operatorname{Stab}_x]. \,\,\,\,\,\, (*)$$

That is, the size of the orbit is the index of the stabilizer.

In your case, consider the action of $G$ on itself by conjugation. Assuming $(*)$, do you see how to finish?

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  • $\begingroup$ I'll certainly unerstand it by the orbit stabilizer problem, but I needed a general proof without this theorem $\endgroup$ Aug 22, 2015 at 0:49
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    $\begingroup$ @GuerlandoOCs Your comment does not make sense to me. Since what you're trying to prove is a special case of orbit-stabilizer, you're talking about a specific proof essentially of the theorem, that's about as opposite of "a general proof without this theorem" as you can get. $\endgroup$
    – anon
    Aug 22, 2015 at 5:51

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