## The normalizer of a maximal Sylow p-intersection contains more than one Sylow p-subgroup

Let $G$ be a group with more than one Sylow $p$-subgroup. Over all pairs of distinct Sylow $p$-subgroups let $P$ and $Q$ be chosen so that $|P \cap Q|$ is maximal. Show that $N_G(P \cap Q)$ has more than one Sylow $p$-subgroup and that any two distinct Sylow $p$-subgroups of $N_G(P \cap Q)$ intersect in $P \cap Q$. (Thus $|N_G(P \cap Q)|$ is divisible by $p|P \cap Q|$ and by some prime other than $p$. Note that Sylow $p$-subgroups of $N_G(P \cap Q)$ need not by Sylow in $G$.)

Note that $P \cap Q < N_P(P \cap Q)$ is proper since $P$ is a $p$-group. Since $N_P(P \cap Q) \leq N_G(P \cap Q)$, we have that $p|P \cap Q|$ divides $|N_G(P \cap Q)|$.

Suppose now that $N_G(P \cap Q)$ is a $p$-group. Then $N_G(P \cap Q) \leq R$ for some Sylow $p$-group $R$. If $R = P$, then $N_G(P \cap Q) \leq N_P(P \cap Q)$, and hence $N_G(P \cap Q) = N_P(P \cap Q)$. However, since $P \cap Q < N_Q(P \cap Q) \leq N_G(P \cap Q) \leq P$ and $P \cap Q < N_Q(P \cap Q) \leq Q$, we have a contradiction since there are elements in $P$ and $Q$ but not in $P \cap Q$ (namely, $N_Q(P \cap Q) \setminus P \cap Q$). Thus $R \neq P$. But then we have $P \cap Q < N_P(P \cap Q)$ $\leq N_G(P \cap Q)$ $\leq R$ and $P \cap Q < N_P(P \cap Q) \leq P$, so that $P \cap Q < N_P(P \cap Q) \leq P \cap R$. This implies that $P \cap Q$ is not (cardinality) maximal among the pairwise intersections of Sylow $p$-subgroups. Thus $N_G(P \cap Q)$ is not a $p$-group, so that its order is divisible by some prime other than $p$.

Suppose now that $N_G(P \cap Q)$ contains a unique Sylow $p$-subgroup, say $R$. Then $N_P(P \cap Q) \leq R$. Moreover, $R \leq P^\prime$ for some Sylow $p$-sugbroup $P^\prime$ of $G$. Since $P \cap Q < N_P(P \cap Q) \leq P \cap P^\prime$ and $P \cap Q$ is order maximal, we in fact have $R \leq P^\prime = P$. Similarly, $R \leq Q$. But then $P \cap Q \leq R \leq P \cap Q$, so that $P \cap Q$ is Sylow in $N_G(P \cap Q)$, a contradiction.

Let $A, B \in \mathsf{Syl}_p(N_G(P \cap Q))$. Since $P \cap Q$ is a $p$-group, $P \cap Q$ is contained in some Sylow $p$-group $R \leq N_G(P \cap Q)$. By Sylow’s Theorem, $R$ is conjugate to $A$ and $B$ in $N_G(P \cap Q)$, say by $x$ and $y$, respectively. Since $P \cap Q$ is normal in $N_G(P \cap Q)$, $P \cap Q \leq A, B$. Thus $P \cap Q \leq A \cap B$. Now $A \leq A^\prime$ and $B \leq B^\prime$ for some Sylow $p$-subgroups $A^\prime$ and $B^\prime$ of $G$. Suppose $A^\prime = B^\prime$. Then, since $A$ and $B$ are distinct, the subgroup $\langle A. B \rangle$ as generated in $A^\prime$ is a $p$-group, normalizes $P\cap Q$, and properly contains (for instance) $A$, which is a contradiction since $A$ is Sylow in $N_G(P \cap Q)$. Thus we have $A^\prime \neq B^\prime$. Moreover, $A \cap B \leq A^\prime \cap B^\prime$, and since $|P \cap Q|$ is maximal among the intersections of distinct Sylow $p$-subgroups and $G$ is finite, $P \cap Q = A \cap B$.

• mcoulont  On November 2, 2010 at 10:28 am

And if A’=B’ ?

• nbloomf  On November 2, 2010 at 10:45 am

Good question.

I’ll mark this “incomplete” until I have a chance to fix it. (Or until someone offers a fix.)

Thanks!

• nbloomf  On November 2, 2010 at 11:12 am

OK- I think it’s patched up now.

Let me know if you find another problem, and thanks for the heads up!

• mcoulont  On November 2, 2010 at 11:48 am

No, it looks ok.

• Lennon  On February 26, 2011 at 4:18 am

Can you please explain more clearly “why N_G(P n Q) is contained in P” in the 4th line ?thnx

• nbloomf  On February 26, 2011 at 7:52 am

That’s actually not necessarily true; fortunately it’s also not what I meant. 🙂 It now reads $P \cap Q < N_P(P \cap Q) \leq N_G(P \cap Q) \leq R$ and $P \cap Q < N_P(P \cap Q) \leq P$, so that $P \cap Q < N_P(P \cap Q) \leq P \cap R$.

Thanks for letting me know!

• Lennon  On February 26, 2011 at 9:43 am

thnx.(P n Q) is strictly contained in (P n R) but i think this is not a contradiction.it just implies that P=R.then we may possibly derive a contradiction but i couldnt find out what?

• nbloomf  On February 26, 2011 at 9:48 am

The contradiction is that $P \cap Q$ was chosen to have maximal cardinality among the pairwise intersections of Sylow $p$-subgroups in $G$. If $P \cap Q < P \cap R$, then there is a larger intersection.

• nbloomf  On February 26, 2011 at 9:49 am

Oops- I see what you mean now.

Hmm…

• nbloomf  On February 26, 2011 at 9:56 am

Okay, I think it’s fixed.

If $R = P$, then we get that $N_Q(P \cap Q)$ is contained in both $P$ and $Q$ but strictly contains $P \cap Q$, which is a contradiction.

Thanks for reading so closely! Let me know if you find another problem.