Timeline for Does the poset of free factors of a free group form a lattice?
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8 events
| when toggle format | what | by | license | comment | |
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| Mar 26, 2013 at 12:27 | comment | added | HenrikRüping | Thus it is $F(a,b)$ and hence $L=F(a,b,c)$. | |
| Mar 26, 2013 at 12:27 | comment | added | HenrikRüping | One difference to the free Abelian case is the following: In the free abelian case the rank is additive, i.e. the sum of the ranks of the greatest lower bound and the least upper bound of two elements equals the sum of the ranks of those elements. This does not hold in this case. The elements $c$ and $[a,b]c$ generate free factors of the free group generated by $a,b,c$. Their intersection is trivial. Their least upper bound $L$ contains $[a,b]$. Thus $L\cap F(a,b)$ is a free factor that contains $[a,b]$. But this commutator is not contained in rank $1$ free factor. | |
| Mar 26, 2013 at 8:03 | vote | accept | HenrikRüping | ||
| Mar 25, 2013 at 22:17 | comment | added | Lee Mosher | The Burns-Chau-Solitar counterexample is in a free group of countably infinite rank. For finite rank, everything works; I've amended my answer. | |
| Mar 25, 2013 at 21:51 | comment | added | Arturo Magidin | The intersection of any finite family of free factors in a free group is again a free factor, though there are infinite families for which this does not hold (see On the intersections of free factors of a free group, by Burns, Chau, and Solitar, Proc. Amer. Math. Soc. 64 (1977) no 1, 43-44. Also, the intersection of two retracts of a free group is a retract (Bergman, G.M., Supports of derivations, free factorizations, and ranks of fixed subgroups in free groups, Trans. Amer. Math. Soc. 351 (1999) no 4, 1531-1550. As has been mentioned, join seems harder. | |
| Mar 25, 2013 at 16:33 | answer | added | Lee Mosher | timeline score: 8 | |
| Mar 25, 2013 at 15:10 | vote | accept | HenrikRüping | ||
| Mar 25, 2013 at 21:35 | |||||
| Mar 25, 2013 at 14:06 | history | asked | HenrikRüping | CC BY-SA 3.0 |