Are groups determined by their morphisms from solvable groups?

$$\newcommand{\Grp}{\mathrm{Grp}}$$Consider the category of groups $$\Grp$$, and within it we have the solvable groups $$S$$. Then any group $$G$$ determines the functor from solvable groups: $$h_G:=\text{hom}_{\Grp}(\_,G):S^{\mathrm{op}}\rightarrow \text{Set}$$

Does this functor determine the group $$G$$? More concretely, if we have a natural bijection $$\text{hom}_{\Grp}(A,G)\cong \text{hom}_{\Grp}(A,G')$$ for all solvable groups $$A$$, must this be induced by an isomorphism $$G\rightarrow G'$$?

Are there any circumstances/more restrictive hypotheses where the answer to this is known to be affirmative, for conceptual reasons?

By other circumstance, I mean imposing finiteness conditions, or other "well behavior conditions", or looking at group objects in a more exotic category, etc. I mean conceptual reasons in the sense of being independent of a classification result of all the objects involved, it wouldn't surprise me if this result were true for finite groups, but only verifiable by induction/case checking for the simple groups. Though interesting, I am more interested in any setting where we have a well understood reason for this to hold, or counterexamples/obstacles to its potential truth.

• Sylow subgroups are solvable, and finite groups are in some sense "controlled" by their Sylow subgroups, right? So it seems reasonable to guess that maybe you get a positive answer for finite groups. In any event, if you keep track of the whole groupoid of homomorphisms from each solvable group to $G$, then you probably get much closer to a positive answer -- but maybe that's cheating? Oct 18 '21 at 18:41
• By the way the category-theoretic term is dense. That is, you're asking whether the solvable groups are dense among all groups, or among some restricted subcategory thereof. Oct 18 '21 at 18:44

Let $$G, G'$$ be two non-isomorphic Tarski monsters of prime exponent $$p$$ or two non-isomorphic torsion-free Tarski monsters. Then for every solvable group $$A$$, $$\mathbb{hom}(A,G)\cong \mathbb{hom}(A,G')$$.

• 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 '21 at 2:31
• @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 '21 at 2:32
• 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 $$u_A\circ f^*_G=(f^*_{C_p}\times\mathrm{id}_{\mathbf{N}})\circ u_B.$$
– YCor
Oct 18 '21 at 7:33
• 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 $G'$.
– YCor
Oct 18 '21 at 16:33
• 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.
– YCor
Oct 18 '21 at 19:48