My assignment is to prove the following proposition, and I'm unsure if my proof is correct: Let $P$ be a Sylow $p$-subgroup of $G$, and let $H$ be the normalizer of $P$ in $G$. Prove that the normalizer of $H$ in $G$ is $H$ itself (i.e. normalizers of Sylow $p$-subgroups are self-normalizing).

I argued this: $|P|=p^{\alpha}$, and $P$ is normal in $H$, since $P$ is a subgroup of the normalizer of $P$ in $G$. Since $P$ is a subgroup of $H$, $|H:P|= 1$ (case 1) or $m$ (case 2).

$H$ is a subgroup of $G$, so $|H|$ divides $|G|$, therefore we have $|H:P||G:H|= 1$ (following from case 1 from above), or $m$ (following from case 2 from above).

Now I analyzed each case:

1) $1*|G:H|=m \implies |H|=|G|/m \implies |H|=p^{\alpha} \implies H=P$, and we already know that the normalizer of $P$ in $G$ is $H$, so if $H=P$, the normalizer of $H$ in $G$ is $H$.

2) $m*|G:H|=m \implies |G:H|=1 \implies G=H$, so obviously the normalizer of $H$ in $G$ is the same as the normalizer of $H$ in $H$, which is obviously all of $H$.

Is this legitimate?

  • $\begingroup$ I've edited your post to include MathJax. Please verify that it still says what you intended. $\endgroup$ – user61527 Nov 24 '13 at 23:19
  • $\begingroup$ yes, thank you! $\endgroup$ – malxmusician212 Nov 24 '13 at 23:29
  • $\begingroup$ What's $m$? Was it defined in your assignment? $\endgroup$ – Ivan Loh Nov 24 '13 at 23:30
  • $\begingroup$ no, this is for any finite group. $\endgroup$ – malxmusician212 Nov 24 '13 at 23:31
  • $\begingroup$ This is not a correct proof. The normalizer of a Sylow subgroup does not have to be either the Sylow subgroup or the entire group as you have concluded. You're mistake is when you assume that $|H:P||G:H|$ is either $1$ or $m$. If $|H:P| = m$ and $|H:P||G:H| \neq 1$ then why do think we must have $|H:P||G:H| = m = |H:P|$? $\endgroup$ – Jim Nov 24 '13 at 23:31

In general: let $S \in Syl_p(G)$ and $H \leq G$, with $N_G(S) \subseteq H$, then $N_G(H)=H$. Moreover, $[G:H]\equiv 1$ mod $p$.

Proof for the first part it suffices to show that $N_G(H) \subseteq H$: observe that $S \in Syl_p(H)$ and take a $g \in N_G(H)$. Then $S^g \subseteq H$ and hence $S^g=S^h$ for some $h \in H$. That means $gh^{-1}\in N_G(S)$, so $g \in hN_G(S) \subseteq H$.

For the second part we use the fact that in general for a $p$-subgroup $P$ of $G$, it holds that $[G:P]\equiv[N_G(P):P]$ mod $p$ (this can be shown by letting $G$ act by right multiplication on the right cosets of $P$). Further, since $N_G(S) \subseteq H$, $N_H(S)=N_G(S) \cap H = N_G(S)$, so the number of Sylow $p$-subgroups of $G$ and $H$ are equal. But $[G:S]=[G:H][H:N_G(S)][N_G(S):S]$ and taking the equation mod $p$ and using the fact that the number of Sylow $p$-subgroups $\equiv 1$ mod $p$ yields the required result.


I mentioned in the comments why your proof is incorrect. Now here's a hint to get on track to finding the correct proof:

Hint: Note that $H$ is itself a group and $P$ is the unique (!) Sylow $p$-subgroup of $H$. So if $g \in G$ normalizes $H$, where does conjugation by $g$ send $P$?

  • $\begingroup$ conjugation by g sends P to itself (since P is the unique Sylow p-subgroup in H, it is normal in H, and g normalizes everything in H), so that implies that P is normal in G, but how does that help to prove that the normalizer of H in G is H? I don't see the connection... $\endgroup$ – malxmusician212 Nov 24 '13 at 23:51
  • $\begingroup$ @MalcolmLazarow "so that implies that $P$ is normal in $G$": It does not. Recall that $g$ was not arbitrary; it was chosen to be in the normalizer of $H$. So in fact you have shown that $P$ is normal in the normalizer of $H$. What does that tell you? $\endgroup$ – Ivan Loh Nov 25 '13 at 0:04
  • $\begingroup$ @MalcolmLazarow Futhermore, strictly speaking, your argument for the fact that conjugation by $g$ sends $P$ to itself is problematic: You should be using the fact that $P$ is the unique p-subgroup of $H$ directly, and not just that $P$ is normal in $H$. It is not true in general that $g$ sends a normal subgroup of $H$ to itself; here one may show that $P$ gets sent to itself precisely by directly using the fact that $P$ is the unique Sylow p-subgroup of $H$. $\endgroup$ – Ivan Loh Nov 25 '13 at 0:13
  • $\begingroup$ I think I'm going in circles and doing something really stupid that I can't catch... So P is normal to H and we now know that P is normal to the normalizer of H in G. So we can say that for all h in H, hPh(-1)=P and for all g in the normalizer of H in G gPg(-1)=P, so we can then say that ghPh(-1)g(-1), which implies that the normalizer of H in G is normal to H? Since H is normal to its normalizer and its normalizer is normal to H, can we conclude that H=its normalizer? Am I totally screwing up definitions right now or is that on the right track? $\endgroup$ – malxmusician212 Nov 25 '13 at 1:17
  • $\begingroup$ nevermind, that's completely incorrect, i'm totally lost... $\endgroup$ – malxmusician212 Nov 25 '13 at 1:21

Here is another proof. Let N denote the normalizer. Let P be a sylow subgroup of G.

Thm: N(N(P)) subgroup N(P)


  1. P sylow in G ==> P sylow in N(P)
  2. P sylow and normal in N(P) ==> P char in N(P)
  3. P char in N(P) and N(P) normal in N(N(P)) ==> P normal in N(N(P))
  4. P normal in N(N(P)) ==> N(N(P)) subgroup N(P)

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