2
$\begingroup$

Let $G$ be a radical group (a group having a normal series with locally nilpotent factors) and $H$ its Hirsch Plotkin racial (i.e the locally nilpotent radical of $G$). It is well known that $C_G(H)\leq H$, just as happens for the Fitting subgroup of a hyperabelian group.

Moving towards an extention for the concept of centralizer of an element, for every $x$ in $G$ we can define the nilpotentizer of $x$ as the set $Nil_G(x)=\{g\in G|\langle x,g\rangle \text{ is nilpotent}\}$. Such set is not in general a subgroup as one can see, for example, by looking at the nilpotentizer of $(12)(34)$ inside $S_4$. In general, given a subgroup $H$ of $G$, one can define the nilpotentizer of $H$ in $G$ in an obvious way as $Nil_G(H)=\bigcap\limits_{x\in H} Nil_G(x)$.

If we moreover consider $N_G(G)$, it is easily seen to include the hypercenter of $G$ and to be included in the set of right Engel elements of $G$, so $N_G(G)$ is a subgroup and lies inside the Hirsch-Plotkin radical in case $G$ is finitely generated and soluble by a nice result by Brookes.

Now my question is the following. Let $G$ be a finitely generated poly-(locally nilpotent) group and let $H$ be its Hirsch-Plotkin radical. Is it true that $Nil_G(H)\leq H$? In other words, is it possible to find, under these hypotheses, an element $x$ of $G$ such that $H\langle x\rangle$ is still locally nilpotent?

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.