Does $p \mid \frac{|{\rm{Stab}}(Q)|}{|\bigcap_{P\in {\rm{Syl}}_p(G)}{\rm{Stab}}(P)|}, \space\forall Q \in \operatorname{Syl}_p(G)$?

Question

Ancillary to the main problem I'm trying to solve, I've come up to a partial result whose generalization would read as follows:

Let $G$ be a finite group, $p$ a prime divisor of $|G|$ and $\operatorname{Syl}_p(G)$ the set of the Sylow $p$-subgroups of $G$. Assume further that $|\operatorname{Syl}_p(G)|>1$. With reference to the transitive action of $G$ by conjugacy on $\operatorname{Syl}_p(G)$, the following holds:

$$p \mid \frac{|{\rm{Stab}}(Q)|}{|\bigcap_{P\in {\rm{Syl}}_p(G)}{\rm{Stab}}(P)|}, \space\forall Q \in \operatorname{Syl}_p(G) \tag 1$$

So far I couldn't prove it nor find a counterexample.

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Just for the records, I'm using $(1)$ to prove that, if $G$ has eight $7$-Sylow subgroups, then $G$ has a normal subgroup $N$ such that $56$ divides $[G:N]$.

Answer 1

Let $|G| = p^km$ withe $p$ not dividing $m$. Since $Q \in {\rm Syl}_p(G)$, we have $p^k$ divides $|{\rm Stab}(Q)| = |N_G(Q)|$.

If the result was false then $p^k$ would also divide $|R|$ where $R$ is the intersection of the normalizers of all Sylow $p$-subgroups of $G$. But then $R$ would contain a Sylow $p$-subgroup, say $P$, of $G$, and then you would have $P$ normalizing all Sylow $p$-subgroups, which is impossible because one Sylow $p$-subgroup cannot normalize a different Sylow $p$-subgroup.

Answer 2

I think this is correct. If I understand you well: $Stab(Q)$ is the stabilizer subgroup of $Q$ for the conjugation by $Q$ on $Syl_p(G)$, hence $Stab(Q)=Q$. And in general for $P \in Syl_p(G)$, one has $Stab(P)=N_Q(P)=P \cap Q$. Hence your quotient equals $$\frac{|Q|}{|\bigcap_{P \in Syl_p(G)} Stab(P)|}=\frac{|Q|}{|\bigcap_{P \in Syl_p(G)} (P \cap Q)|}=\frac{|Q|}{|O_p(G) \cap Q|}.$$ And this quotient equals $1$ if and only if $Q=O_p(G)$, that is, when $Q$ is normal, contradicting $|Syl_p(G)| \gt 1$.