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I'm trying to prove that the intersection of the normalizers of the Sylow subgroups of a [finite] group $G$ is equal to its hypercenter, i.e., $$Z_\infty(G)=\bigcap\limits_{S\in Syl(G)}\textbf{N}_G(S)$$

It is really easy if we use the theorem 3.v from the Baer's paper Group elements of prime power index. I wonder if there is a more direct way to prove this fact. I also tried to prove the next fact:

The hypercenter is the intersection of the maximal nilpotent subgroups.

I'm really stuck on these theorems provided that I can't figure out a satisfactory direct proof of them.

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  • $\begingroup$ Is $G$ a finite group? $\endgroup$ – Derek Holt Jul 28 '15 at 18:08
  • $\begingroup$ Oh, yes, It's important to asume finiteness in order to ensure the existence of the hypercenter. $\endgroup$ – Jose Paternina Jul 28 '15 at 18:13
  • $\begingroup$ I think I could prove this, but I don't particularly want to write down a proof, given that you know how to prove it already. Let $H$ be the hypercentre and $I$ the intersection of the Sylow normalizers. It is not too hard to show that $H \le I$ and to prove $I \le H$ we can assume that $Z(G)=1$. So the hardet part is to prove that $Z(G)=1 \Rightarrow I=1$. For that I would assume not, let $N \le I$ with $N$ minimal normal in $G$, so $N$ is an elementary abelian $p$-group, show that there exists a $p'$-element normalizing bu not centralizing $N$ and use that to contradict $N \le I$. $\endgroup$ – Derek Holt Jul 29 '15 at 11:42

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