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Let $G$ be a group and $nil(G):=\{x\in G \mid \langle x,y \rangle \text{ is nilpotent for all } y \in G\}$. Is $nil(G)$ always a subgroup of $G$?

Many thanks for any help.

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  • $\begingroup$ Why do you expect this to be the case? $\endgroup$ – Tobias Kildetoft Nov 29 '13 at 10:24
  • $\begingroup$ I have found that in many cases it is true and equal to the hypercenter of G. For example when G is a finitely generated solvable group. $\endgroup$ – D. N. Nov 29 '13 at 10:29
  • $\begingroup$ Hmm, so whenever is is a subgroup, it is automatically normal. Am I right in thinking that the quotient will have trivial Fitting subgroup in that case? $\endgroup$ – Tobias Kildetoft Nov 29 '13 at 10:45
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    $\begingroup$ That doesn't sound right. For $G=S_3$, ${\rm nil}(G)=1$, but the Fitting subgroup is not trivial. $\endgroup$ – Derek Holt Nov 29 '13 at 11:29
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    $\begingroup$ So a closely related question would be does $Z(G)=1 $ imply ${\rm nil}(G)=1$? $\endgroup$ – Derek Holt Nov 29 '13 at 11:53
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The stronger conjecture is that ${\rm nil}(G)$ is the hypercentre $Z^\infty(G)$ of $G$. I think I can prove that when $G$ is finite. I am not sure about the general case.

Assume $G$ is finite. It is easy to see that $Z^\infty(G) \le {\rm nil}(G)$. To prove the reverse inclusion we can factor our $Z^\infty(G)$, so it is enough to prove that $Z(G)=1$ implies ${\rm nil}(G)=1$.

So assume $G$ is finite with $Z(G)=1$, and there exists $1 \ne x \in {\rm nil}(G)$. Clearly any power of $x$ is in ${\rm nil}(G)$, so we can can assume that $x$ has prime order $p$. Now $\langle x,y \rangle$ nilpotent for all $y \in G$ implies that $x $ commutes with all elements of $p'$-order. So $x \in C_G(O^{p'}(G))$, where $O^{p'}(G)$ is the subgroup generated by all elements of $p'$-order. So, if $x \in P \in {\rm Syl}_p(G)$, then $P \cap C_G(O^{p'}(G))$ is a nontrivial normal subgroup of $P$, so it intersects $Z(P)$ nontrivially. But $Z(P) \cap C_G(O^{p'}(G)) \le Z(G)$, contradicting $Z(G)=1$.

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  • $\begingroup$ Thank you so much for the answer. $\endgroup$ – D. N. Nov 29 '13 at 13:49
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    $\begingroup$ Actually, this holds for any group satisfying the max condition for subgroups. In such a group $G$ the set of right Engel elements coincides with the hypercenter of $G$, which equals $Z_n(G)$ for some finite $n$ (as the upper central series is finite). Of course one should observe that $\nil(G) =R(G)$. $\endgroup$ – Yassine Guerboussa Nov 29 '13 at 17:29

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