Timeline for Is a free group a product of f.g subgroups of infinite index?
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 21, 2014 at 16:04 | comment | added | HJRW | @Pablo, yes, that's right. | |
| Dec 21, 2014 at 15:22 | comment | added | Pablo | @HJRW and your approach can also show that the product of finitely many subgroups does not contain a coset of a finite index subgroup, right? | |
| Dec 21, 2014 at 14:22 | comment | added | HJRW | @Pablo, mine does! | |
| Dec 21, 2014 at 10:17 | comment | added | Pablo | @BenjaminSteinberg does your argument also work for any finite number of subgroups and not just two (that is $H$ and $K$)? | |
| Dec 19, 2014 at 14:47 | vote | accept | Pablo | ||
| Dec 19, 2014 at 14:45 | comment | added | Benjamin Steinberg | Ok. Fair enough. | |
| Dec 19, 2014 at 14:08 | comment | added | HJRW | @BenjaminSteinberg, in fact, one doesn't need Marshall Hall's theorem at all. Geometrically, the point is that there are infinitely many different elevations of the the cover $X^H$ to the cover $X^K$ (in Wise's beautiful terminology). On the other hand, writing it down cleanly for a questioner who seems to have difficulties with the topological point of view is another matter. This seemed the best way to me. | |
| Dec 19, 2014 at 14:04 | comment | added | HJRW | @Pablo, algebraically, you could take $a$ to be a generator in the free splitting for $H$ not contained in $H$, and likewise for $b$ and $K$. You're quite right that it could be that $a=b$. | |
| Dec 19, 2014 at 13:55 | comment | added | Benjamin Steinberg | Why not use the slightly easier version of this argument suggested in my comment to the question? By Marshall hall you can assume that one of the subgroups H is a free factor. The covering space or Schreier graph associated to K has a finite core, the stallings graph. Outside this core you have infinitely many orbits of H so infinitely many double cosets. Both being free factors is not needed. | |
| Dec 19, 2014 at 13:37 | comment | added | Pablo | @HJRW Once $H$ and $K$ are free factors, what is your algebraic choice for $a$ and $b$? some basis elements? How do you know that all the double cosets are indeed distinct? It looks like it can even happen that $a = b$... | |
| Dec 19, 2014 at 10:08 | comment | added | Andreas Thom | Ok, I see. Why not explain this in the answer? | |
| Dec 19, 2014 at 9:59 | comment | added | HJRW | @AndreasThom, as I said above, this is obvious from a topological point of view. If $H$ is carried by an embedded subgraph $Y$ of $X$ and $X'$ is a cover of $X$, then $H\cap\pi_1X'$ is carried by the preimage of $Y$ in $X'$ which is, of course, embedded. So applying Marshall Hall's theorem twice, we obtain a cover in which both intersections are free factors. The corresponding cover is the graph $X$. | |
| Dec 19, 2014 at 8:56 | comment | added | Andreas Thom | @Henry, how do you find $X$? That is the point that Ashot already raised. Is it really true that there is some finite index subgroup $F'$ such that both $F'\cap H$ and $F'\cap K$ are free factors? | |
| Dec 19, 2014 at 8:33 | comment | added | HJRW | @AndreasThom, since there's already some discussion of how to modify the argument in comments, and you give no details of your critique, your comment is not very helpful. Anyway, I've now edited the answer to fix my small mistake. | |
| Dec 19, 2014 at 8:31 | history | edited | HJRW | CC BY-SA 3.0 |
Corrected final paragraph.
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| Dec 19, 2014 at 7:39 | comment | added | Andreas Thom | This argument is not correct. | |
| Dec 18, 2014 at 21:58 | comment | added | HJRW | Actually, sorry, you're right, I should be a little more careful. The question is whether, given two embedded subgraphs in a graph $X$, one can find a maximal tree that restricts to a maximal tree in each. But this issue isn't really important. One should just work in the graph $X$ (without contracting a maximal tree), and the same argument goes through. | |
| Dec 18, 2014 at 21:44 | comment | added | HJRW | Ashot, you need the observation that if $H$ is a free factor of $F$ then $H'$ is a free factor of $F'$. Topologically, this is obvious. | |
| Dec 18, 2014 at 21:31 | comment | added | Ashot Minasyan | How do you use M. Hall's theorem to find a single finite index subgroup $F' \leqslant F$ such that both $F' \cap H$ and $F' \cap K$ are free factors of $F'$? | |
| Dec 18, 2014 at 20:33 | history | answered | HJRW | CC BY-SA 3.0 |