Timeline for Find a surface or 3-manifold whose fundamental group is $(\mathbb{Z}/n\mathbb{Z}) \rtimes (\mathbb{Z}/2\mathbb{Z})$
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Sep 10, 2018 at 16:35 | answer | added | Neil Hoffman | timeline score: 2 | |
| Aug 24, 2018 at 16:08 | vote | accept | john mangual | ||
| Aug 24, 2018 at 14:56 | comment | added | Neil Hoffman | Thurston's book "Three-dimensional Geometry and Topology" (not the notes) also has a fairly nice, self-contained discussion of 3-manifolds with finite fundamental group in the elliptic manifolds chapter. More generally, I have always found his flow charts at the end of the book to be really useful laying bare the relations between certain group properties and the geometric classification of 3-manifolds. | |
| S Aug 2, 2018 at 8:57 | history | suggested | user21230 |
Adding tag 3-manifolds, because this is question for 3-manifolds (or surfaces, but surfaces can only have Z_2 fundamental group)
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| Aug 2, 2018 at 8:17 | comment | added | user21230 | I am happy that this question survived. Even it is easy for specialists, it contains quite interesting details as you can see from the answers and comments: 1) is dihedral group subgroup of $S^3$; 2) what abelian groups can be fundamental groups of 3-manifolds. Such details are useful for persons like me who are learning 3-manifolds. This is good homework to do :) | |
| Aug 2, 2018 at 8:12 | review | Suggested edits | |||
| S Aug 2, 2018 at 8:57 | |||||
| Jul 29, 2018 at 11:05 | comment | added | LSpice | @HJRW, thanks; I thought something looked wrong there. | |
| Jul 29, 2018 at 10:46 | answer | added | HJRW | timeline score: 6 | |
| Jul 29, 2018 at 6:59 | comment | added | HJRW | @LSpice , regarding your second comment, I think you have the Galois correspondence the wrong way round: quotients are realised as deck groups, not as fundamental groups of covering spaces. | |
| Jul 28, 2018 at 22:21 | comment | added | Allen Hatcher | In the reference you cite it is shown that $\pi_1({\mathbb R}^3-K)$ is the group generated by two elements $a$ and $b$ subject to the relation $a^m=b^n$. This group is torsionfree (as is true for all knots, not just torus knots) and it has ${\mathbb Z}_m*{\mathbb Z}_n$ as the quotient group when the center, which is the infinite cyclic group generated by the element $a^m=b^n$, is factored out. | |
| Jul 28, 2018 at 17:06 | answer | added | Igor Rivin | timeline score: 2 | |
| Jul 28, 2018 at 15:50 | review | Close votes | |||
| Aug 1, 2018 at 6:22 | |||||
| Jul 28, 2018 at 15:36 | comment | added | LSpice | Also, since the semi-direct product is a quotient of the amalgamated product, you should be able just to take a suitable cover of your $\mathbb R^3 \setminus K$, no? | |
| Jul 28, 2018 at 15:35 | comment | added | LSpice | Regarding uniqueness, there is always the direct product $\mathbb Z/m\mathbb Z \times \mathbb Z/n\mathbb Z$, and sometimes that's the only one (if $m$ does not divide $\varphi(n)$). If we do have a non-Abelian semi-direct product, then it's unique for $n$ square-free at least, and I think in general. | |
| Jul 28, 2018 at 15:35 | comment | added | john mangual | @MarkSapir I've been out of school 5 years. | |
| Jul 28, 2018 at 15:34 | comment | added | user6976 | This looks like a homework. Voted to close. | |
| Jul 28, 2018 at 15:24 | history | edited | john mangual | CC BY-SA 4.0 |
added 4 characters in body
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| Jul 28, 2018 at 15:08 | history | asked | john mangual | CC BY-SA 4.0 |