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(Edit: The wikipedia article is correct. I messed up the notions normalizer and normal closure.)

Please take a look at the wikipedia article on Sylow's theorems here, more precisely, at the last bullet of Theorem 3.

It says that $n_p=\lvert G:N_G(P)\rvert$, where $P$ is a $p$-Sylow group in $G$, $N_G$ denotes the normalizer and $n_p$ is the number of $p$-Sylow groups in $G$.

Isn't this false? If $P$ is already normal, then we know that $n_p=1$ and $N_G(P)=P$. But $G:P$ is $m$ and not $1$ (where $\lvert G\rvert=p^n\cdot m$).

I would guess that $\frac{m}{n_p}=\lvert G:N_G(P)\rvert$ might be correct. What do you say?

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1 Answer 1

up vote 6 down vote accepted

If P is normal in G, then NG(P) = G, and [G:G]=1 as claimed. More generally, NG(P) is the stabilizer of P under the conjugation action of G, and the size of the orbit (the number of Sylows) is the index [G:NG(P)] of the stabilizer, by the Orbit-Stabilizer theorem.

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Oh, its the stabilizer? In the article it says normalizer. –  Rasmus May 6 '11 at 21:53
    
The normalizer of a set S is the (it turns out) group of elements g such that $gSg^{-1}\subset S$. Evidently this is the same as the stabilizer of the set S under the group action of conjugation (which that conjugation is a group action is what the proof that the normalizer is a subgroup boils down to). –  Vladimir Sotirov May 6 '11 at 22:01
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@Vladimir, Jack: You are right. I messed up the notions normalizer and normal closure. Thank you for your help! –  Rasmus May 6 '11 at 22:15
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What does completely independent mean precisely? –  Rasmus May 7 '11 at 8:36
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I see. Thank you very much for the explanation. –  Rasmus May 7 '11 at 16:47

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