Timeline for Normal subgroups of an extension of the Higman group
Current License: CC BY-SA 3.0
23 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| S Nov 5, 2015 at 19:51 | history | bounty ended | H A Helfgott | ||
| S Nov 5, 2015 at 19:51 | history | notice removed | H A Helfgott | ||
| Nov 1, 2015 at 15:01 | history | edited | H A Helfgott | CC BY-SA 3.0 |
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| Oct 30, 2015 at 1:49 | answer | added | andrew | timeline score: 0 | |
| Oct 29, 2015 at 20:06 | history | edited | H A Helfgott | CC BY-SA 3.0 |
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| S Oct 29, 2015 at 19:57 | history | bounty started | H A Helfgott | ||
| S Oct 29, 2015 at 19:57 | history | notice added | H A Helfgott | Current answers are outdated | |
| Oct 26, 2015 at 12:29 | history | edited | H A Helfgott | CC BY-SA 3.0 |
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| Oct 26, 2015 at 6:19 | answer | added | andrew | timeline score: 4 | |
| Oct 24, 2015 at 15:07 | comment | added | H A Helfgott | No, I don't think so; conjugate $t^2$ by $a$, say. | |
| Oct 24, 2015 at 13:03 | comment | added | მამუკა ჯიბლაძე | But is $2\mathbb Z/4\mathbb Z$ normal itself? | |
| Oct 24, 2015 at 11:07 | history | edited | H A Helfgott | CC BY-SA 3.0 |
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| Oct 24, 2015 at 10:56 | comment | added | H A Helfgott | Thanks! Well, neither $\{e\}$ nor a group containing $H_4$, then. (I suspect your normal closure does coincide with $2\math{Z}/4\mathbb{Z} | |
| Oct 24, 2015 at 10:56 | history | edited | H A Helfgott | CC BY-SA 3.0 |
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| Oct 23, 2015 at 23:10 | answer | added | YCor | timeline score: 11 | |
| Oct 23, 2015 at 22:14 | history | edited | YCor | CC BY-SA 3.0 |
fixed symbol of semidirect product (triangle sounds like "normal in")
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| Oct 23, 2015 at 21:00 | answer | added | andrew | timeline score: 8 | |
| Oct 23, 2015 at 20:46 | comment | added | მამუკა ჯიბლაძე | Sorry for confusion in comments, let me start over again. The kernel of the composite homomorphism $G\twoheadrightarrow\mathbb Z/4\mathbb Z\twoheadrightarrow\mathbb Z/2\mathbb Z$ is the normal subgroup $2\mathbb Z/4\mathbb Z\ltimes H_4$ containing $t^2$, so it contains the normal closure of the latter. Whether it coincides with that normal closure I don't see, but in any case the normal closure of $t^2$ is neither $\{e\}$ nor $H_4$ nor $G$. | |
| Oct 23, 2015 at 16:01 | comment | added | H A Helfgott | Yes, I mean a semi-direct product. I still don't see quite how that (very general) article helps. And yes, I exclude $k_1=k_2=k_3=0$, but the normal closure of that is all of $G$, simply because the Higman-like group with 3 instead of 4 in the definition is trivial. | |
| Oct 23, 2015 at 9:21 | comment | added | Mark Grant | Are you using the notation $H\triangleright K$ to denote the semi-direct product of $H$ acting on $K$? If so, then the following paper of Usenko might be relevant: link.springer.com/article/10.1007%2FBF01058705 | |
| Oct 23, 2015 at 8:13 | comment | added | მამუკა ჯიბლაძე | Presumably you exclude $k_1=k_2=k_3=0$? | |
| Oct 22, 2015 at 22:24 | history | asked | H A Helfgott | CC BY-SA 3.0 |