Timeline for Groups with no bounds on the size of abelian subgroups without infinite ones
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 16, 2011 at 17:59 | comment | added | user6976 | Grigorchuk has many groups. | |
| Nov 16, 2011 at 16:05 | comment | added | Ashot Minasyan | @Mark: any infinite $2$-group contains an infinite abelian subgroup (Held D., "On abelian subgroups of infinite $2$-groups", Acta Sci. Math. (Szeged) 1966, v 27, 97-98). This applies to Grigorchuk's first group, I think. | |
| Nov 16, 2011 at 15:52 | comment | added | Ashot Minasyan | @Mark: thanks for the link! I am quite surprised by this result. | |
| Nov 16, 2011 at 15:24 | comment | added | user6976 | See: jlms.oxfordjournals.org/content/s1-39/1/235.extract | |
| Nov 16, 2011 at 15:08 | comment | added | user6976 | @Ashot: Locally finite groups tend to have infinite Abelian subgroups. All answers so far are lacunary hyperbolic groups. This is almost certainly not the only way to construct an example. One can probably take the "lacunary large" construction of Olshanskii and Osin (impose relations $w^{n_w}$ keeping the group large), Golod-Shafarevich groups may not contain infinite Abelian subgroups. Also I do not know if Grigorchuk group contains an infinite Abelian subgroup (that must be known because the centralizers of elements in Grigorchuk groups are known). | |
| Nov 16, 2011 at 11:07 | comment | added | Ashot Minasyan | I think that there might also be infinitely generated examples, which should be easier to construct. Something like an iterated semidirect product $((A1\rtimes A2)\rtimes A3)\dots$ of finite groups $A_i$, such that $A_n$ acts on the semidirect product $B_{n−1}$, before it, so that the induced action on the conjugacy classes of $B_{n−1}$ is free. However, I do not know how to construct such a sequence... | |
| Nov 15, 2011 at 21:51 | comment | added | user6976 | @Denis: Yes, that is the oldest example. Also one does not need to refer to a proof, only to the result. | |
| Nov 15, 2011 at 20:26 | comment | added | Denis Osin | Tarski Monsters originally constructed by Olshanskii satisfy the requirements. All abelian subgroups there are cyclic, but the exponent is unbounded. The construction can be found in the book, of course. | |
| Nov 15, 2011 at 19:58 | history | answered | user6976 | CC BY-SA 3.0 |