9
$\begingroup$

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 $\vert G \vert$.)

$\endgroup$
12
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Many thanks for this answer. $\endgroup$ – Panurge Dec 7 '17 at 11:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.