Let $G$ be a finite group having exactly $n$ Sylow $p$-subgroups for some prime $p$. Show
that there exists a subgroup $H$ of the symmetric group $S_n$ such that $H$ also has exactly
$n$ Sylow $p$-subgroups.
My attempt: Say $G$ has only $1$ Sylow $p$-subgroups for some $p$ ($n=1$). Then, $S_n=S_1=\{e\}$ so it has no subgroup. Couldn't it be a counterexample? Otherwise, I should look into the case $n>p$, because $n\equiv 1\ (mod\ p)\implies n=1\ or\ n>p$ by the 2nd Sylow theorem.
In that case ($n>p$), $p$ devides $|S_n|=n!$, so $S_n$ has Sylow $p$-subgroups by the 1st Sylow theorem. Say $P_1,P_2,…,P_k$ are them. Now by the 3rd Sylow theorem, $k=(S_n:N_{S_n}(P_i))$ for $i=1,2,…,k.$ And from here, I want to show that $k=n$ and $N_{S_n}(P_i) = N_{S_n}(P_j)$ for $i,j=1,2,…,k$ and conclude $\exists H=N_{S_n}(P_1)$, it seems gone wrong.
So I have two main troubles here: Can't discard the case $n=1$, and can't go further beyond $p|n!$ in the $n>p$ case (I guess my direction – trying to show $H=N_{S_n}(P_1)$ is just wrong?).
Any hints or suggestions would be appreciated.
Best Answer
Let $P \in Syl_p(G)$ and hence $|G:N_G(P)|=n$. The $n!$-Theorem ($G$ acting on the right cosets of in this case $N_G(P)$ by right multiplication, see M. Isaacs, Finite Group Theory Theorem (1.1) for example) implies that there is a homomorphism $\bar{} : G \rightarrow S_n$, with kernel $K=core_G(N_G(P))$. We know that $\overline{PK}=PK/K$ is a Sylow $p$-subgroup of $\bar{G}$ a subgroup of $S_n$.
Observe that there is a one-to-one correspondence between the subgroups of $\overline{G}$ and the the subgroups of $G$ containing $K$. This implies that $\overline{N_G(PK)}=N_{\overline{G}}(\overline{PK})$ and in particular $|\overline{G}:N_{\overline{G}}(\overline{PK})|=|G:N_G(PK)|$.
Finally we will show that in fact $N_G(PK)=N_G(P)$ and then we are done. Since $K \lhd G$ it is easy to see that $N_G(P) \subseteq N_G(PK)$. But then, since $PK \lhd N_G(PK)$ and $P \in Syl_p(PK)$, the Frattini argument gives $N_G(PK)=N_{N_G(PK)}(P)PK=(N_G(P) \cap N_G(PK))PK=N_G(P)PK=N_G(P)$.