# A strengthening of a Sylow theorem?

Let $G$ be a finite group, let $p$ be a prime number. It is a well known theorem of Sylow that the number of Sylow $p$-subgroups of $G$ divides $\vert G \vert$.

I wonder if the following strenghtening is true :

Statement 1. Let $G$ be a finite group, let $p$ be a prime number, let $G_{0}$ be a $p$-subgroup of $G$. Then the number of Sylow $p$-subgroups of $G$ containing $G_{0}$ divides $\vert G \vert$.

If I'm not wrong, the following special case is true :

Theorem. Let $G$ be a finite group, let $p$ be a prime number, let $G_{0}$ be a $p$-subgroup of $G$. Assume that $G_{0}$ is normal in every Sylow $p$-subgroup of $G$ containing $G_{0}$. Then the number of Sylow $p$-subgroups of $G$ containing $G_{0}$ divides $\vert G \vert$.

Proof. Let $P_{1}, \ldots , P_{r}$ be the distinct Sylow $p$-subgroups of $G$ containing $G_{0}$. We have to prove that $r$ divides $\vert G \vert$. By hypothesis, $P_{1}, \ldots , P_{r}$ normalize $G_{0}$, thus $G_{0}$ is normal in the subgroup $P = <P_{1}, \ldots , P_{r}>$ of $G$ generated by $P_{1}, \ldots , P_{r}$, thus $G_{0}$ is contained in every Sylow $p$-subgroup of $P$. Since the Sylow $p$-subgroups of $P$ are clearly Sylow $p$-subgroups of $G$, we have proved that every Sylow $p$-subgroup of $P$ is a Sylow $p$-subgroup of $G$ containing $G_{0}$. In other words, every Sylow $p$-subgroup of $P$ is a $P_{i}$. Reciprocally, every $P_{i}$ is a Sylow $p$-subgroup of $P$, thus $r$ is the number of Sylow $p$-subgroups of $P$, thus (Sylow theorem) $r$ divides $\vert P \vert$ and thus divides $\vert G \vert$.

Do you know if the normality hypothesis can be cancelled in the above theorem ? And perhaps, do you know a reference to the literature concerning this matter ? Thanks in advance.

(By the way, I think that the following is true : Let $G$ be a finite group, let $p$ be a prime number, let $G_{0}$ be a $p$-subgroup of $G$; then the number of Sylow $p$-subgroups of $G$ containing $G_{0}$ is $\equiv 1 \pmod{p}$. Thus Statement 1, if true, can be formulated in the more precise manner : Let $G$ be a finite group, let $p$ be a prime number, let $G_{0}$ be a $p$-subgroup of $G$. Then the number of Sylow $p$-subgroups of $G$ containing $G_{0}$ divides $\vert G \vert / p^{n}$, where $p^{n}$ denotes the greatest power of $p$ dividing $\vert G \vert$.)

It's not true in general. For example, the number of Sylow $2$-subgroups that contain a subgroup $S$ of order $2$ (there is only one up to conjugacy) in the simple group ${\rm PSL}(2,7)$ of order $168$ is $5$. $S$ is normal (and central in) in one of these, and not normal in the others.