Timeline for Is every abelian group a colimit of copies of Z?
Current License: CC BY-SA 3.0
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| May 6, 2015 at 21:41 | history | edited | YCor | CC BY-SA 3.0 |
Expanded 2nd and 3rd paragraphs
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| May 6, 2015 at 17:19 | comment | added | Qiaochu Yuan | @YCor: in the second paragraph you claim without proof that torsion-free implies locally cyclic. You could be a lot more explicit here. I recognize that the argument isn't hard but locally cyclic is an unfamiliar condition to me and e.g. I don't know what equivalences between equivalent definitions of it you might implicitly be using. | |
| May 6, 2015 at 15:14 | comment | added | Eric Wofsey | @TimCampion: That group is not torsion-free because $2(x-y)=0$. | |
| May 6, 2015 at 14:21 | comment | added | Tim Campion | Hm. Then using the same idea, isn't the group $\langle x,y,z \mid 2x=z,2y=z \rangle$ torsion-free, simply-presented (with a connected graph of relations), but not locally cyclic? | |
| May 6, 2015 at 14:03 | comment | added | YCor | @TimCampion: not right. Try with the abelian group presentation $\langle x,y,z,v,w\mid 2x=z,2y=z,2z=w,v=w,v=2w\rangle$ (in which $v,w$ are trivial, reducing to the presentation $\langle x,y,z,v,w\mid 2x=z,2y=z,2z=0\rangle$ of $C_2\times C_4$, where $C_i$ is cyclic of order $i$) | |
| May 6, 2015 at 12:58 | comment | added | Tim Campion | I don't see the need for torsion-freeness, either: if $G$ is simply presented, and $A$ is a finitely-generated subgroup, then write the generators in the form $\sum n_{ij}e_j$, where the $e_j$ appear in the simple presentation of $A$. Then the $e_j$ generate a finitely-generated subgroup $E$ of $G$ containing $A$; it's pretty clear that $E$ is cyclic, so the subgroup $A$ is also cyclic. Right? | |
| May 6, 2015 at 7:03 | comment | added | Eric Wofsey | @QiaochuYuan: Perhaps an easier way to think about what happens in the torsion-free case once you've reduced to a single equivalence class is to tensor everything with $\mathbb{Q}$. Doing this turns your diagram into a commuting diagram where every object is $\mathbb{Q}$ and every map is an isomorphism, so the colimit is clearly $\mathbb{Q}$. Since your group was torsion-free, it injects into its tensor product with $\mathbb{Q}$ and is hence a subgroup of $\mathbb{Q}$. | |
| May 6, 2015 at 6:31 | comment | added | YCor | @QiaochuYuan: how could I be more explicit? in the first paragraph I decomposed $G$ as a direct sum without using torsion-freeness and in the second paragraph I use torsion-freeness to show that components in this decomposition are locally cyclic. | |
| May 6, 2015 at 5:36 | comment | added | Qiaochu Yuan | Can you give just a few more details? Aside from "cyclic" probably being "locally cyclic" in the first paragraph, I'd like to confirm that I know how you're using the hypothesis that $G$ is torsion-free. | |
| May 5, 2015 at 22:38 | comment | added | YCor | Yes, every locally cyclic group is either isomorphic to a subgroup of $\mathbf{Q}$, or to a subgroup of $\mathbf{Q}/\mathbf{Z}$. (An arbitrary torsion-free locally cyclic module over a domain $A$ is isomorphic to an $A$-submodule of the field of fractions $K$ of $A$. If $A$ is a PID, the only other locally cyclic modules are the submodules of $K/A$.) | |
| May 5, 2015 at 20:30 | comment | added | Tim Campion | Should the second sentence be "Then $G$ is a direct sum of locally cyclic groups"? Also, it's interesting to learn that every torsion-free locally cyclic group is isomorphic to a subgroup of $\mathbb{Q}$. Is every locally cyclic group isomorphic to a subgroup of $\mathbb Q / \mathbb Z$? | |
| May 5, 2015 at 20:18 | history | answered | YCor | CC BY-SA 3.0 |