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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.

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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. –  Jack Schmidt Jan 22 '11 at 1:08

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up vote 6 down vote accepted

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.

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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. –  Jack Schmidt Jan 23 '11 at 16:46
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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). –  Geoff Robinson Jul 10 '11 at 9:21

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