Timeline for Analogues of the dihedral group
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 15, 2014 at 18:29 | comment | added | Steve D | Yes you are right, I had overlooked the case where the F of the quotient was not coming from the whole commutator. Indeed $F_2$ itself subjects onto $C_2\ast C_3$! | |
| Nov 14, 2014 at 22:12 | comment | added | HJRW | @SteveD, that's a nice idea. You're right that $F_2/K$ can't be free, but it could still be virtually free, so I don't see how to finish the argument... | |
| Nov 14, 2014 at 15:05 | comment | added | Steve D | I think an easier argument gives something stronger about $C_2\ast C_3$. Namely, if $K$ is the (non-trivial) kernel of some surjection, then either $K$ has torsion (and thus the quotient is abelian), or $K$ is torsion free. But every torsion-free subgroup of $C_2\ast C_3$ is contained in the commutator ($F_2$), and there are no non-trivial quotients of $F_2$ which are both non-abelian and free. | |
| Jan 4, 2012 at 15:38 | vote | accept | HenrikRüping | ||
| Jan 4, 2012 at 13:48 | comment | added | Derek Holt | You are right that care is needed! It is not hard to construct examples with $d(A*_CB) < d(A),d(B)$ with $A$ and $B$ finite, which surprised me! I can see how to do that with $d(A*_CB) \le 4$ and $d(A),d(B)$ arbitrarily large. I expect it is possible with $d(A*_CB) = 2$, but I haven't managed that yet! | |
| Jan 4, 2012 at 10:37 | comment | added | HJRW | 'There must be known results about minimal generator numbers of free products with amalgamation in terms of the generator numbers of the factors'. Some care is needed here! For example, the free group of rank two admits very complicated graph-of-group decompositions. See, for instance, the examples on page 454 of Bestvina and Feighn's Inventiones paper 'Bounding the complexity of simpliciai group actions on trees'. | |
| Jan 4, 2012 at 9:15 | comment | added | Derek Holt | Let $G := F \rtimes H = f(\Gamma)$. Since $G$ is virtually free, it is a fundamental group of a graph of groups with finite vertex groups. But (like $\Gamma$) $G$ has finite abelianization, so $G$ can be defined by free products with amalgamation of a finite number of finite groups. There must be known results about minimal generator numbers of free products with amalgamation in terms of the generator numbers of the factors. I would expect the fact that $G$ is 2-generated to imply that it is a free product with amalgamation of two cyclic groups, which should lead rapidly to a contradiction. | |
| Jan 3, 2012 at 19:39 | history | answered | HJRW | CC BY-SA 3.0 |