Timeline for Are groups determined by their morphisms from solvable groups?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 18, 2021 at 20:30 | vote | accept | Chris H | ||
| Oct 18, 2021 at 19:48 | comment | added | YCor | By the way there's no need of fancy groups to get the result. Take all infinite non-cyclic countable group in which every solvable subgroup is infinite cyclic and in which maximal abelian subgroups have pairwise trivial intersection. The same argument works for such groups. (They have automatically infinitely many maximal abelian subgroups, but assume it if necessary.) Then many nonisomorphic groups have this property, e.g., free groups of different rank between $2$ and $\omega$, surface groups, etc. | |
| Oct 18, 2021 at 19:23 | comment | added | Benjamin Steinberg | @TimCampion, I think our comments crossed. I was referring to you first comment when I wrote mine but your second comment appeared before I finished typing | |
| Oct 18, 2021 at 18:53 | comment | added | Tim Campion | @BenjaminSteinberg Sorry, I agree: hypoabelian groups are dense among all groups, by virtue of containing the free groups. I meant that if you allow hypoabelian $A$, then you get a trivial positive answer to this modified version of the OP's original question -- which is a negative answer to the question in my comment. | |
| Oct 18, 2021 at 18:51 | comment | added | Benjamin Steinberg | @TimCampion Free groups are hypoabelian so this seems hard to believe | |
| Oct 18, 2021 at 18:49 | comment | added | Tim Campion | oh wait -- I see, free groups are hypoabelian. So if you allow $A$ to be hypoabelian, then trivially you get a positive answer to the question. | |
| Oct 18, 2021 at 18:44 | comment | added | Tim Campion | Is it also true that if $G,G'$ are Tarski monsters of exponent $p$, then $Hom(A,G) \cong Hom(A,G')$ (naturally in $A$) for each hypoabelian group $A$? It seems to me that if we're considering infinite groups, then we really ought to generalize solvability to hypoabelianness... | |
| Oct 18, 2021 at 16:33 | comment | added | YCor | PS: this is natural in the above categorical sense, with respect to $A$ (natural transformation from $h_G$ to $h_{G'}$, using OP's notation), although not that natural in the intuitive sense since it depends on a choice of bijection between the set of cyclic subgroups of $G$ and $G'$. | |
| Oct 18, 2021 at 7:33 | comment | added | YCor | You didn't mention naturality, but indeed it's natural. Namely, for a Tarski monster of prime exponent $p$, or more generally any infinite countable group $G$ in which every nontrivial solvable subgroup has order $p$, $\mathrm{Hom}(A,G)$ is naturally in bijection $u_A$ with $\mathrm{Hom}(A,C_p)\times\mathbf{N}$, natural meaning that for every group homomorphism $f:A\to B$ inducing by composition $f^*_{H}:\mathrm{Hom}(B,H)\to\mathrm{Hom}(A,H)$, we have $$u_A\circ f^*_G=(f^*_{C_p}\times\mathrm{id}_{\mathbf{N}})\circ u_B.$$ | |
| Oct 18, 2021 at 2:32 | comment | added | markvs | @ChrisH: I am not sure that for finite groups there is an example like that, but I do not know finite simple groups well enough. Basically you need two non-solvable groups with the same solvable subgroups. | |
| Oct 18, 2021 at 2:31 | comment | added | Chris H | This is great, thanks. I would like to leave the question up for a little while to see if any affirmative results are known, if nothing, then I'll accept this answer. | |
| Oct 18, 2021 at 0:56 | history | answered | markvs | CC BY-SA 4.0 |