Is it true that:

If the intersection of the Sylow p-subgroups is trivial, then the intersection of their normalizers is equal to the intersection of their centralizers?

I half remember this being true for odd p, but I cannot find the reference. I have not found a counterexample for p=2 or p=3.

  • $\begingroup$ In ams.org/mathscinet-getitem?mr=55340 Baer (1953 TAMS) studies the intersection of the normalizers of all Sylow subgroups (allowing p to vary), and assuming the Fitting subgroup is trivial, that intersection (the hypercenter) will be the intersection of the centralizers of the Sylow subgroups. So I guess in some sense I'm looking for the single prime version of this. $\endgroup$ – Jack Schmidt Jan 22 '11 at 1:08

I think the answer is yes. Let $K$ be the intersection of the normalizers of the Sylow $p$-subgroups of $G$, and $P$ any Sylow $p$-subgroup. Then $K$ is a normal subgroup of $G$, so $[K,P] \le K \cap P$. If $K \cap P$ is nontrivial, then a nonidentity element $g$ has order a power of $p$ and normalizes all Sylow $p$-subgroups, so it must lie in all Sylow $p$-subgroups, contradicting your assumption. So $[K,P] = 1$ and hence $K$ centralizes all Sylow $p$-subgroups.

  • $\begingroup$ Thanks! I was trying to prove K = Op(G)*Core(Centralizer(G,P)), which is obviously false when G is non-abelian of order 6. Your [K,P] ≤ Op(G) makes much more sense and proves it nicely. $\endgroup$ – Jack Schmidt Jan 23 '11 at 16:46
  • 2
    $\begingroup$ Alternatively, each Sylow $p$-normalizer has a normal Sylow $p$-subgroup, so $K$ certainly does. Since $K \lhd G$, the unique Sylow $p$-subgroup of $K$ is normal in $G$, hence trivial. Now $K$ has order prime to $p$, and the same argument with $[K,P]$ works. (Same argument, slighly recast). $\endgroup$ – Geoff Robinson Jul 10 '11 at 9:21

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.