Timeline for When can we establish an isomorphism between two not-finitely presented groups?
Current License: CC BY-SA 3.0
17 events
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| Jan 27, 2016 at 18:49 | comment | added | LSpice | Sorry; I meant: given the conditions that you state on the ordered pair $(G, H)$, can one deduce the same conditions on the ordered pair $(H, G)$—or is it just that, in your context, usually one happens to know the conditions for both ordered pairs a priori? I don't mean to say that it can't be done, just that I don't see it immediately. | |
| Jan 27, 2016 at 6:49 | comment | added | MSMalekan | @LSpice: In my research problem, YES. | |
| Jan 26, 2016 at 20:23 | comment | added | LSpice | Is it obvious that this situation is really symmetric in $G$ and $H$? | |
| Jan 24, 2016 at 6:00 | comment | added | MSMalekan | @HJRW: What extra assumptions needed to make groups $G$ and $H$, at least, quotient of each other? In my problem, these two groups have the same finite class number, also, the described situation is symmetric, means the statement also remains true if we change the role of $G$ and $H$. | |
| Jan 21, 2016 at 22:38 | review | Close votes | |||
| Jan 22, 2016 at 10:49 | |||||
| Jan 21, 2016 at 19:58 | comment | added | HJRW | @YCor, I agree that what you describe is a contradiction to the question as literally read. But I presume the OP is looking for some non-trivial phenomenon in this vicinity, and so I thought an example of that form might be helpful to him. It was just a suggestion. | |
| Jan 21, 2016 at 13:21 | comment | added | YCor | @HJRW: at the moment, my first comment yields trivial instances where $G$ is not a quotient of $H$ (e.g., when $H$ is a quotient of $G$ with $H$ finite and $G$ infinite, or $H$ abelian and $G$ non abelian), and this is also in contradiction with your interpretation of the question. | |
| Jan 21, 2016 at 12:06 | comment | added | HJRW | @YCor, it seems to me that the OP would benefit from an example of a group G which is a limit of groups H, but such that H is not a quotient of G. I'm under the impression that such a (necessarily infinitely presented) G should exist, but don't know a construction. Do you know one? | |
| Jan 21, 2016 at 10:18 | comment | added | MSMalekan | @YCor: Yes, I mean $\ker(\text{Free}(\mathfrak g)\rightarrow H)$ contains $\mathfrak S$. My apologizes for not following the grammar! | |
| Jan 21, 2016 at 10:06 | comment | added | YCor | "which its kernel contains $S$" is not correct English. Do you mean "whose kernel contains $S$"? | |
| Jan 21, 2016 at 10:05 | comment | added | MSMalekan | @HJRW: I have two not-finitely presented group $G$ and $H$, I can finitely manage the relations, that's mean: For every finite subset $\mathfrak S$ of $\ker(\text{Free}(\mathfrak g)\rightarrow G)$, there is an epimorphism $\text{Free}(\mathfrak g)\rightarrow H$, which its kernel contains $\mathfrak S$. I want to know, what we can say then about $G$ and $H$? | |
| Jan 21, 2016 at 9:57 | comment | added | MSMalekan | @YCor: The question is not about the existence of such groups. I want to know, if it is the case that $G$ and $H$ are always quotient of each other? | |
| Jan 20, 2016 at 23:53 | comment | added | YCor | @HJRW as far as I understand, every quotient $H$ of $G$ in which $\mathfrak{g}$ is mapped injectively works... | |
| Jan 20, 2016 at 22:44 | comment | added | HJRW | You may like to look up the Gromov--Grigorchuk space of marked groups. If I've understood you correctly, your condition means that $G$ is in the closure of the subspace of groups isomorphic to $H$. | |
| Jan 20, 2016 at 19:06 | comment | added | MSMalekan | @LSpice: Yes, it is exactly all of $\ker(\text{Free}(\mathfrak g)\rightarrow G)$. | |
| Jan 20, 2016 at 18:32 | comment | added | LSpice | In what set does $\mathfrak R$ live? Is it another name for $\ker(\operatorname{Free}(\mathfrak g) \to G)$? | |
| Jan 20, 2016 at 18:27 | history | asked | MSMalekan | CC BY-SA 3.0 |