bio  website  dpmms.cam.ac.uk/~hjrw2 

location  Cambridge  
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visits  member for  5 years, 1 month 
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8h

comment 
Is a free group a product of f.g subgroups of infinite index?
@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. 
8h

comment 
Is a free group a product of f.g subgroups of infinite index?
@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 $a=b$. 
12h

comment 
Is a free group a product of f.g subgroups of infinite index?
@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 $X$. 
14h

comment 
Is a free group a product of f.g subgroups of infinite index?
@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. 
14h

revised 
Is a free group a product of f.g subgroups of infinite index?
Corrected final paragraph. 
1d

comment 
Is a free group a product of f.g subgroups of infinite index?
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. 
1d

comment 
Is a free group a product of f.g subgroups of infinite index?
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. 
1d

answered  Is a free group a product of f.g subgroups of infinite index? 
1d

comment 
Is a free group a product of f.g subgroups of infinite index?
I'm not sure about the votes to close  this seems a reasonable question to me. Anyway, the answer is 'no'. I'll try to write down an answer when I have time. 
2d

comment 
Free abelian subgroups and distorsion
That should be undistorted. Anyway, mapping tori of suitable automorphisms of $\mathbb{Z}^{n+1}$ should give examples for any $n$. 
Dec 16 
comment 
Free abelian subgroups and distorsion
What about lattices in Sol? I think their cyclic subgroups are in distorted. 
Dec 14 
comment 
Approximating Lie groups by finite groups
Presumably @BenMcKay meant 'nonabelian' (in which case this statement is true). 
Dec 4 
comment 
Is the free abstract group residually of rank d > 2?
Sorry, to be precise, your question asks whether $F$ is residuallu $\mathcal{F}_d$ for some $d$ (I think). Anyway, in particular, a positive answer for $d=2$ would suffice. 
Dec 4 
comment 
Is the free abstract group residually of rank d > 2?
@Pablo, your question asks whether $F$ is residually $\mathcal{F}_2$. My answer shows that $F_n$ (for any $n$, and hence $F$) is residually $F_2$. Since 'residually' is idempotent, it suffices to prove that $F_2$ is residually $\mathcal{F}_2$. Note that $F_2$ is known to be residually many classes of finite groups; what are some interesting examples of groups in $\mathcal{F}_2$? 
Dec 4 
comment 
Is the free abstract group residually of rank d > 2?
@Pablo  I'm sorry, you're right  I hadn't noticed this requirement. I guess my answer shows that you can take $n=d=2$. 
Dec 4 
comment 
Is the free abstract group residually of rank d > 2?
@AntonKlyachko  good point! Because $F_2$ is residually finite. 
Dec 4 
answered  Is the free abstract group residually of rank d > 2? 
Dec 4 
comment 
Is there a nontrivial profinite word which is trivial in any group with at most d generators?
You don't need to appeal to the classification for the discrete case. It follows because $F_{n+1}$ is (fully) residually $F_n$, for any $n>1$. 
Dec 3 
comment 
Is there a nontrivial profinite word which is trivial in any group with at most d generators?
The answer to the analogous question in the discrete setting is 'yes'. For every finite subset $X\subseteq F_\infty$, there is a 2generator finite group $Q$ and an epimorphism $q:F_\infty\to Q$ so that $q$ is injective on $X$. The profinite version may follow, but it's not completely clear to me right now. 
Nov 29 
comment 
Action of the homotopy braid groups on reduced free groups
The easiest way to see the action of $B_n$ on $F_n$ is topologically, thinking of $B_n$ as the mapping class group of the $n$punctured disc (with a base point at infinity). Is there some corresponding interpretation of the homotopy braid group? 