Given a group $G$ with $p+1$ Sylow $p$-subgroups, I've deduced that $R = P \cap P'$, where $P, P'$ are Sylow $p$-subgroups, has index $p$ in each of $P, P'$; and that all $p+1$ Sylow $p$-subgroups of $G$ are in the normalizer $N_G(R)$.

I want to see that $R$ is exactly the intersection of all the Sylow $p$-subgroups, and have been struggling with this-- in any case I cannot see how precisely to utilize these two facts in proving this. Any help would be appreciated.

EDIT:(If anyone could explain the hint below a little further, I would be grateful.)


1 Answer 1


Proposition Let $P$ be a $p$-subgroup of a group $G$, and $S \in Syl_p(G)$, then $$ S \subseteq N_G(P) \text{ iff } P \unlhd S.$$ Proof Assume $S \subseteq N_G(P)$. Observe that apparently $S \in Syl_p(N_G(P))$. Of course, $P$ is a $p$-subgroup of $N_G(P)$ and must lie in some Sylow $p$-subgroup of $N_G(P)$, that is $P \subseteq S^x$, for some $x \in N_G(P)$. Since $P$ is also normal in $N_G(P)$, it follows that $P=P^{x^{-1}} \subseteq S$. Conversely, if $P \unlhd S$, then of course $S \subseteq N_G(P)$.

Now apply this to your situation: you already observed that all Sylow $p$-subgroups of $G$ are contained in $N_G(R)$. The proposition gives you that $R \unlhd P$ for all $P \in Syl_p(G)$, that is $R \subseteq \bigcap_{P \in Syl_p(G)}P=O_p(G)$. Obviously, $O_p(G) \subseteq R$. So in fact $R=O_p(G)$ and you are done.(It follows that $N_G(R)=G!$)

Remark Note that there is a famous theorem of Jerald Brodkey (see for instance the book of I.M. Isaacs, Finite Group Theory, 1.37), saying that if the Sylow $p$-subgroups are abelian, then the $O_p(G)$ can be realized as the intersection of a certain pair of different Sylow $p$-subgroups. In your case (with a constraint on the number of Sylow $p$-subgroups), we see that the $O_p(G)$ can be realized as the intersection of any pair of different Sylow $p$-subgroups!


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