Timeline for A question about generating set of groups and epimorphism
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Feb 6, 2018 at 13:40 | vote | accept | MSMalekan | ||
| Mar 15, 2016 at 0:42 | answer | added | Simon Thomas | timeline score: 5 | |
| Feb 13, 2016 at 16:22 | answer | added | MSMalekan | timeline score: 1 | |
| Feb 10, 2016 at 9:35 | comment | added | YCor | ... I don't know what you checked, but these groups are definitely finitely generated and the proof is quite standard. | |
| Feb 9, 2016 at 21:17 | comment | added | YCor | @MeisamSoleimaniMalekan: yes, the center of a finitely generated group is not always finitely generated. | |
| Feb 9, 2016 at 18:25 | comment | added | MSMalekan | @YCor: 2.If your claim is right, then groups $\mathbb Z$ and $2\mathbb Z$ also satisfied the condition, and this is absurd. 3. In the above link, we also need that $d_{ii}$ belongs to $\langle t\rangle$ instead of belonging to $Z^\times=\mathbb Z^\times\langle t\rangle$. | |
| Feb 9, 2016 at 18:24 | comment | added | MSMalekan | @YCor: I checked the details of your example, and I found that: 1. The groups of your example are not f.g., for the central subgroup $\mathbb Z[t,t^{-1}]$ could not be generated by finite elements of the groups. | |
| Jan 12, 2016 at 10:37 | comment | added | YCor | I mean, you mod out the matrix group by the central subgroup consisting of matrices of the form $e_{14}(x)$ where $x$ ranges over those Laurent polynomials of the form $2nt^{-1}+P$, where $n$ ranges over $\mathbf{Z}$ and $P$ ranges over $\mathbf{Z}[t]$. | |
| Jan 12, 2016 at 8:28 | comment | added | MSMalekan | @YCor: I did not understand the "The intermediate quotient $2t^{-1}Z\oplus Z[t]$", I asked groupiest in our department and have not received that's meaning. Could you please make it clear? | |
| Jan 6, 2016 at 16:37 | comment | added | YCor | In my example $G$ and $H$ are solvable (hence amenable). They are not finitely presented but I can believe this can be arranged. On the other hand residually finite is hopeless since any groups answering your question are f.g. and non-Hopfian, hence not residually finite. | |
| Jan 6, 2016 at 15:20 | comment | added | MSMalekan | @YCor: Could you please make me known that if $G$ or $H$ is amenable, finitely presented and residualy finite? | |
| Jan 6, 2016 at 5:03 | comment | added | Alireza Abdollahi | @YCor: Could you please kindly write down your groups $G$ and $H$ in details. It is very interesting for me too. But I could not follow your quick point in the above. | |
| Dec 30, 2015 at 22:12 | comment | added | YCor | Your "that's mean" is not correct: "every generating set of these group is an image of a generating set" is always true (and remains true if you add "finite"). In words, your (new) assumption is "every finite generating subset of $H$ is a one-to-one image of a generating subset of $G$ and vice versa". Anyway my example fulfills this assumption [and even the stronger assumption that every lift of a generating subset of $H$ is a generating subset of $G$, and vice versa]. | |
| Dec 30, 2015 at 20:05 | history | edited | MSMalekan | CC BY-SA 3.0 |
added 261 characters in body
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| Dec 30, 2015 at 11:03 | comment | added | YCor | The condition "every generating subset is an image of an generating subset", for a group epimorphism, is empty, since the inverse image of any generating subset is generating. So it is enough to answer the first part of the question. Take Abels's group as in my answer here mathoverflow.net/questions/59209/hopfian-property, but with $F_p$ replaced by $Z=\mathbf{Z}$. Killing the central $Z[t]$ yields an isomorphic group. The intermediate quotient $2t^{-1}Z\oplus Z[t]$ yields an intermediate quotient which is not isomorphic to the original group with epimorphisms in both directions. | |
| Dec 30, 2015 at 9:51 | history | edited | MSMalekan |
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| Dec 30, 2015 at 8:43 | review | First posts | |||
| Dec 30, 2015 at 9:01 | |||||
| Dec 30, 2015 at 8:42 | history | asked | MSMalekan | CC BY-SA 3.0 |