Timeline for If $A$, $B$ are abelian groups such that $\mathrm{Hom}(A, G) \cong \mathrm{Hom}(B, G)$ for all abelian groups $G$, must $A$ and $B$ be isomorphic?
Current License: CC BY-SA 4.0
14 events
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| Mar 5, 2023 at 0:55 | history | edited | LSpice | CC BY-SA 4.0 |
Typo, while this is on the front page
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| Mar 4, 2023 at 19:18 | answer | added | Jeremy Rickard | timeline score: 30 | |
| Jul 4, 2020 at 23:44 | comment | added | Ian Agol | @CarlosEsparza that’s right, the topology on Hom(A,G) is not unique. Somehow this is dual to the fact that Hom(Z,A) determines A. | |
| Jul 4, 2020 at 22:26 | comment | added | Carlos Esparza | @IanAgol very interesting... although I suppose that since $S^1$ has noncontinuous group automorphisms it is not possible to get the topology of $\mathrm{Hom}(A, G)$ just from the group structure. | |
| S Jul 2, 2020 at 11:03 | history | bounty ended | CommunityBot | ||
| S Jul 2, 2020 at 11:03 | history | notice removed | CommunityBot | ||
| Jun 26, 2020 at 18:02 | comment | added | Ian Agol | If $G=U(1)$ is a topological group, then this follows from Pontryagin duality (with $G=S^1=U(1)$, and the discrete topology on $A$ and $B$), where one remembers the topology on $Hom(A,G)$. en.wikipedia.org/wiki/… | |
| S Jun 24, 2020 at 9:06 | history | bounty started | Carlos Esparza | ||
| S Jun 24, 2020 at 9:06 | history | notice added | Carlos Esparza | Draw attention | |
| Jun 17, 2020 at 8:24 | comment | added | YCor | Ah, thanks indeed, I should have checked before. So the question has a positive answer when $A$ is an elementary abelian $p$-group for some prime $p$ (i.e., if for every $B$ the question has a positive answer for this given $A$); we can say that $A$ is "recognizable". | |
| Jun 16, 2020 at 22:35 | comment | added | Carlos Esparza | @YCor I think Eric Wofsey commented on mathSE (on a deleted answer to the linked question) that there always is a cardinal $\alpha$ such that $\alpha^\kappa \neq \alpha^\lambda$ for $\kappa \neq \lambda$ | |
| Jun 16, 2020 at 22:33 | comment | added | YCor | If $A=C_2^{(\aleph_0)}$ and $B=C_2^{(\aleph_1)}$, the these are isomorphic for every $G$ iff $\alpha^{\aleph_0}=\alpha^{\aleph_1}$ for all cardinal $\alpha$. This is false under CH (taking $\alpha=2$) but I don't know if it's consistent in general (=in ZFC). In case that yields a counterexample, one might restrict to $A,B$ countable, or assume GCH. | |
| Jun 16, 2020 at 21:19 | comment | added | Carlos Esparza | I 'migrated' this question from math.stackexchange (I deleted the question there) since it had already been asked | |
| Jun 16, 2020 at 21:17 | history | asked | Carlos Esparza | CC BY-SA 4.0 |