2
$\begingroup$

Suppose $G$ is a group. Is $G$ a subgroup of some group $H$ such that:

  • $H$ is centerless;
  • If $h \in H$ is an element of prime order $p$, then there is also some $g \in G$ of order $p$.

In other words, does any group embed in a centerless group without introducing an element of a new prime order?

This is a modified version of my question; the previous version turned out to be trivial.

$\endgroup$
6
  • 3
    $\begingroup$ Well if $|G|=2$ then $G \unlhd H$ and therefore $G \le Z(H)$. $\endgroup$
    – Derek Holt
    Commented Mar 3, 2019 at 16:03
  • $\begingroup$ And if $G$ is a finite $p$-group (and $H$ is required to be finite), then $H$ has to be a finite $p$-group, and hence has a nontrivial center. $\endgroup$
    – YCor
    Commented Mar 3, 2019 at 16:08
  • 4
    $\begingroup$ (Not assuming any finiteness.) Let $L_G$ be the subgroup of $G$ generated by elements of prime order, and $A_G$ its automorphism group. Then the answer is yes iff ($\sharp$) $A_G$ fixes no nontrivial element of $L_G$. Indeed, Let $F_G$ be a nontrivial free group with a homomorphism onto $A_G$, and consider the amalgam $(F_G\ltimes L_G)\ast_{L_G}G$. It then has a trivial center as ($\sharp$) holds and has the same prime order elements. Conversely given $H\supset G$ with the same prime order elements, $L_G$ is normal in $H$, and hence if ($\sharp$) fails, we see that $H$ has a nontrivial center. $\endgroup$
    – YCor
    Commented Mar 3, 2019 at 16:14
  • $\begingroup$ Thank you very much for your comments. I corrected the question appropriately. $\endgroup$ Commented Mar 3, 2019 at 18:50
  • 1
    $\begingroup$ For the new question, the answer is yes: take $H$ to be the free product of $G$ with $\mathbf{Z}$ (which has trivial center, if $G\neq 1$). It even satisfies that every element of finite order in $H$ is conjugate to an element of $G$. Also, my first comment applies if you assume that $G$, $H$ are finite. $\endgroup$
    – YCor
    Commented Mar 3, 2019 at 18:51

0

You must log in to answer this question.

Browse other questions tagged .