TITLE: On the subset $nil(G):=\{x\in G \mid \langle x,y \rangle \text{ is nilpotent for all } y \in G\}$ of a group $G$.
QUESTION [6 upvotes]: 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.

REPLY [5 votes]: 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 $Z(G)=1$.