Timeline for relatively free groups in $Var(S_3)$
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 23, 2018 at 6:40 | vote | accept | Sh.M1972 | ||
| Sep 16, 2018 at 11:50 | comment | added | Gro-Tsen | For those who, like me, had no idea what "variety" means in this context or what a "relatively free" group is, see encyclopediaofmath.org/index.php/Variety_of_groups and encyclopediaofmath.org/index.php/Free_group — I think people asking a question should make the minimal effort of providing such links or explanations. | |
| Sep 16, 2018 at 11:21 | answer | added | YCor | timeline score: 5 | |
| Sep 16, 2018 at 8:07 | answer | added | Keith Kearnes | timeline score: 4 | |
| Sep 11, 2018 at 17:38 | answer | added | R. Keith Dennis | timeline score: 2 | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Jan 9, 2014 at 13:01 | vote | accept | Sh.M1972 | ||
| Sep 23, 2018 at 6:40 | |||||
| Jan 9, 2014 at 6:45 | comment | added | Sh.M1972 | What do you think about the answer of Anton Klyachko? It seems interesting but I can't find any reference. Some thing is also missing in his answer. | |
| Jan 9, 2014 at 6:29 | comment | added | The Masked Avenger | Likely the next candidate for the free 2-algebra is $S_3^2 \times C_3^2$, or some similar product. | |
| Jan 9, 2014 at 5:04 | comment | added | Arturo Magidin | Fair enough, but it must also be $2$-generated, so it cannot be $S_3\times S_3\times C_6$, which cannot be generated by $3$ elements. So it cannot be of the form $S_3^n\times A$ for some abelian $A$. It must involve some different nonabelian group. Of course, $V$ is finitely based by Powell-Oats, so in principle you would be able to write down a basic set of laws and figure it out explicitly. | |
| Jan 9, 2014 at 4:52 | comment | added | Sh.M1972 | @:Arturo Magidin: $F_V(x_1, \ldots, x_n)$ most be the largest $n$-generator element of $V$. So, $F_V(x,y)$ is not $S_3\times C_6$, since we have also $C_6\times C_6\in V$ which is not a quotient of $S_3\times C_6$. | |
| Jan 9, 2014 at 3:30 | comment | added | Arturo Magidin | I think that your second question is in general difficult except for special cases, such as critical groups; for $G$ a finite nonabelian simple group, for example, it is not hard to show that the finitely generated groups in $\mathrm{Var}(G)$ is of the form $G^n\times K$, where $K\in\mathrm{Var}((\mathbf{HS}-1)(G))$ (see for example the proof of Lemma 3.2 in Sheila Oates's "Identical relations in groups", J. London Math. Soc. 38 (1963), 71-78). $S_3$ is critical (since the proper subfactors are all abelian) so perhaps something similar can be done. I expect $F_V(x,y)\cong S_3\times C_6$. | |
| Jan 8, 2014 at 23:26 | answer | added | Anton Klyachko | timeline score: 5 | |
| Jan 8, 2014 at 13:44 | comment | added | Sh.M1972 | In the first look it seems as a trivial question. If $V=Var(S_3)$, then clearly we have $F_V(x)=C_6$ the cyclic group of order 6. It maybe also true that $F_V(x, y)=S_3\times S_3\times C_6$. Now, what about $F_V(x_1, \ldots, x_n)$? If $G$ is and arbitrary group and $V=Var(G)$, then how we can express $F_V(x_1, \ldots, x_n)$ in terms of $G$ and known groups? | |
| Jan 7, 2014 at 21:22 | history | asked | Sh.M1972 | CC BY-SA 3.0 |