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.

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Comments

  • 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.

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