Timeline for Is every group a subgroup of a centerless group with the same prime order elements?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 3, 2019 at 19:06 | comment | added | Wojciech Aleksander Wołoszyn | Oh, I see. It didn't occur to me that I can easily use the free product. Thanks! | |
| Mar 3, 2019 at 18:51 | comment | added | YCor | 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. | |
| Mar 3, 2019 at 18:50 | comment | added | Wojciech Aleksander Wołoszyn | Thank you very much for your comments. I corrected the question appropriately. | |
| Mar 3, 2019 at 18:48 | history | edited | Wojciech Aleksander Wołoszyn | CC BY-SA 4.0 |
modified question due to comments
|
| Mar 3, 2019 at 16:45 | review | Close votes | |||
| Mar 10, 2019 at 3:10 | |||||
| Mar 3, 2019 at 16:14 | comment | added | YCor | (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. | |
| Mar 3, 2019 at 16:08 | comment | added | YCor | 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. | |
| Mar 3, 2019 at 16:03 | comment | added | Derek Holt | Well if $|G|=2$ then $G \unlhd H$ and therefore $G \le Z(H)$. | |
| Mar 3, 2019 at 16:00 | review | First posts | |||
| Mar 3, 2019 at 17:06 | |||||
| Mar 3, 2019 at 15:58 | history | asked | Wojciech Aleksander Wołoszyn | CC BY-SA 4.0 |