# $n$ Sylow $p$-subgroups of $G$ $\implies$ $\exists H<Sym_n$ s.t. $H$ has $n$ Sylow $p$-subgroups?

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.

• The case $n=1$ is not a counterexample, because the trivial subgroup is a Sylow $p$-subgroup of $S_1$ for all primes $p$. For the general case, let $S$ be the set of $n$ Sylow $p$-subgroups of $G$. Then $G$ acts on $S$ by conjugation, and the image of $G$ in ${\rm Sym}(S) \cong S_n$ defined by this action has exactly $n$ Sylow $p$-subgroups. – Derek Holt Feb 11 at 9:09

An equivalent action to the action given by @Nicky is the action of $$G$$ on the conjugates of $$P$$ for some Sylow $$p$$-subgroup of $$G$$ defined by conjugation in $$G$$ (so for $$x,y\in G$$ $$g\cdot xPx^{-1}=gxPx^{-1}g^{-1}$$). I'll skip the details as @Nicky's answers seems complete.
As there are $$n$$ Sylow subgroups of $$G$$ (precisely the conjugates of $$P$$), this defines a homomorphism $$f:G\to S_n$$.
Note that $$f(G)\cong G/K$$ where $$K=\ker(f)$$ and $$f(G)\le S_n$$. So $$f(H)$$ is a Sylow $$p$$-subgroup of $$f(G)$$. The action of $$f(G)$$ on the conjugates of $$f(H)$$ is equivalent to the action of $$G$$ on the conjugates of $$H$$ so $$f(G)$$, so $$f(H)$$ must have $$n$$ conjugates. That is $$f(G)$$ has $$n$$ Sylow $$p$$-subgroups.
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 $$N_G(PK)=N_{N_G(PK)}(P)PK=(N_G(P) \cap N_G(PK))PK=N_G(P)PK=N_G(P)$$.