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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?

1 Answers 1

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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*.2011-05-06
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    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).2011-05-06
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    @Vladimir, Jack: You are right. I messed up the notions *normalizer* and *normal closure*. Thank you for your help!2011-05-06
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    The normal closure of a Sylow p-subgroup is very important, and called $O^{p'}(G)$. It is the subgroup generated by all elements whose order is a power of p, and is the unique smallest normal subgroup whose quotient has no elements of order p (such quotient groups are called "p-perfect"). The index $[G:O^{p'}(G)]$ is actually *completely* independent of $n_p$, which is kind of freaky.2011-05-07
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    What does completely independent mean precisely?2011-05-07
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    @Rasmus: For any finite group H with Sylow p-subgroup P, and any positive integer k coprime to p, there is a group G containing $O^{p'}(H)$ such that $[G:N_G(P)] = [H:N_H(P)]$, so that $n_p(G) = n_p(H)$, and such that $[G:O^{p'}(G)] = k$, so that the index is chosen arbitrarily.2011-05-07
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    For instance, take p=3, H the symmetric group of degree 4, k=47. $[H:N_H(P)]=4$ and H contains 4 Sylow 3-subgroups, <123>, <124>, <134>, and <234>. These generate $O^{3'}(H) = A_4$, the alternating group of degree 4. We can take $G = A_4 \times C_{47}$ to be the direct product of $O^{3'}(H)$ with a cyclic group of order k. Then $O^{3'}(G) = A_4 \times 1$ and $[G:O^{3'}(G)] = 47$ and $[G:N_G(P)]=4$.2011-05-07
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    I see. Thank you very much for the explanation.2011-05-07