Normalizers of Sylow $p$-subgroups have restricted order

Question

I saw an old post. It gives the following result.

Assume that $H$ is a subgroup of a finite group $G$, and that $P$ is a Sylow $p$-group of $H$. If $N_G(P) \subset H$ then $P$ is a Sylow p-subgroup of $G$.

My first question: I wonder if there should be “$m=k$” at the end of the proof. “$\alpha=s$” is the desired result and in fact we can only get $\alpha=s$ from $p^{\alpha-s}m\equiv k({\rm mod}~p)$. “$m=k$” gives $$|N_G(P)|=p^sk=p^{\alpha}m=|G|$$ and hence $N_G(P)=G$, which we don’t necessarily have.

My second question: If what I said in the first question is correct, then we still have $m\equiv k ({\rm mod}~p)$. If that is so, then the order of the normalizer of a Sylow $p$-subgroups is rather restricted. Is that right? Is there anything that I should note?

Any help is appreciated. Thank you!

Answer 1

Your first question: Yes, $m\equiv k\bmod p$, not $m=k$.

Second question: The order of a Sylow$p$-normalizer can be anything you want, in for example the group $P\times G$. If you mean inside a given group then yes, it has to be such that the index is congruent to $1$ modulo $p$. I'm not aware of any other serious restrictions, except things like Burnside's normal $p$-complement theorem. This has the corollary that if the Sylow $p$-subgroup is abelian and $N_G(P)=P$ then there is a normal $p'$-subgroup $K$ such that $G=KP$.

Answer 2

Different proof, using the fact that if $P$ is a $p$-subgroup (so not necessarily Sylow) of a group $G$, then $$|G:P| \equiv |N_G(P):P| \text{ mod } p.$$ Hint for a proof: let $P$ act by right multiplication on the right cosets of $P$ in $G$ and apply the Burnside/Frobenius/Cauchy orbit stabilizer lemma.

Now observe that in the situation you sketch, $N_H(P)=H \cap N_G(P)=N_G(P)$. Hence $$|G:P|=|G:H| \cdot |H:N_G(P)| \cdot |N_G(P):P|=\\|G:H| \cdot |H:N_H(P)| \cdot |N_G(P):P|= \\|G:H| \cdot n_p(H) \cdot |N_G(P):P| .$$ Note that $p \nmid |N_G(P):P|$, hence taking the above equation mod $p$ and applying the aforementioned fact, we get $|G:H| \equiv 1 \text{ mod } p$ and hence $|G:P|=|G:H| \cdot|H:P|$ is not divisible by $p$, that is $P$ is Sylow in $G$.

Now considering your second question: the order and hence the index of $N_G(P)$, with $P$ Sylow, is quite restricted in the following sense. If $N_G(P) \subseteq H \subseteq G$, then both $|G:H|$ and $|H:N_G(P)|$ must be $\equiv 1$ mod $p$. This implies for example that if a group has $21$ Sylow $5$-subgroups, then $N_G(P)$ is a maximal subgroup for each $P \in Syl_5(G)$.

In addition, one can prove the following (proof omitted).

Theorem Let $G$ be a finite group and $P \in Syl_p(G)$.
$(a)$ There does not exist a normal subgroup $N$ in $G$ with $N_G(P) \subseteq N \subsetneq G$.
$(b)$ There does not exist a nilpotent subgroup $H$ in $G$ with $N_G(P) \subsetneq H \subseteq G$.
$(c)$ There does not exist a subgroup $H$ in $G$ with $N_G(P) \lhd H \subseteq G$.