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bio website dpmms.cam.ac.uk/~hjrw2
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2d
comment What are your favorite instructional counterexamples?
@YCor - of course that is strictly correct. I was trying to give some rough idea of what it meant.
Aug
13
comment embedding of finite groups into product
The answer is surely still 'no'. You are asking if there is an infinite, finitely presented, simple group generated by torsion. Infinite finitely presented simple groups are not so easy to construct, but there is no reason why such a group should not exist.
Aug
9
comment Groups where word problem is solvable, but not quickly?
There are fairly standard constructions that should build fg groups for whom the word problem is equivalent to the membership problem for your favourite subset $S\subseteq \mathbb{N}$. For instance, I think the word problem in $\langle a,b\mid [a^{-n}ba^n,b]=1\Leftrightarrow n\in S\rangle$ can be seen to be at least as difficult as membership of $S$. (Of course, one needs to check that the word problem is actually solvable, but I don't think that's too difficult in this case.) For finitely presented examples, one could invoke Clapham's improvement of Higman's embedding theorem.
Aug
3
comment An amenable group containing a wreath product of itself
@ToddTrimble, that convention is not very firm. See, for instance, the Lamplighter Group: en.wikipedia.org/wiki/Lamplighter_group . In geometric and combinatorial group theory, where we are mostly interested in finitely generated groups, we almost always use direct sums.
Jul
24
comment Automorphisms of finite order in $Out(\widehat{F_2})$
Have you thought about $\mathrm{Out}(\widehat{\mathbb{Z}})$?
Jul
24
comment dense subgroup in pro v topology
Ah right, I misunderstood you. Yes, it is true in the pro finite topology.
Jul
24
comment dense subgroup in pro v topology
Unless I misunderstood your question, this fails trvially for the profinite topology, since any subgroup of finite index is closed.
Jul
21
revised Products of elliptic isometries
Added gr.group-theory tag.
Jul
20
revised Does there exist a homotopy equivalence from $\mathbb{C}P^{2n}$ to itself that reverses orientation?
Clarified title.
Jul
20
comment Products of elliptic isometries
As Richard's answer indicates, you seem to have misinterpreted $\mathrm{Fix}_\delta$ to mean a neighbourhood of the fixed point set, whereas it should actually be the approximate fixed point set.
Jul
14
comment Cycles covering the edges of the graph corresponding to the Van Kampen diagram of a presentation of a group
I too am confused. A van Kampen diagram is a geometric proof that one element $\gamma$ of the free group on the generators $x_1,\ldots,x_n$ maps to the trivial element in $G$. It does not tell you anything about the whole of $G$. Perhaps you meant 'presentation complex'?
Jul
13
comment When to postpone a proof?
It seems to me that this question should be community-wiki. I've flagged it as such.
Jul
8
comment Your favorite papers on geometric group theory
You already mentioned three of my favourites!
Jul
1
comment Hyperbolic knot complement groups and relative dimension
No problem. Nice theorem, by the way!
Jul
1
revised Hyperbolic knot complement groups and relative dimension
Minor corrections.
Jun
30
answered Hyperbolic knot complement groups and relative dimension
Jun
27
awarded  Good Answer
Jun
26
comment Classification of groups in which the centralizer of every non-identity element is cyclic
PS Ian, I know you know this. But I thought it worth setting the record straight.
Jun
26
comment Classification of groups in which the centralizer of every non-identity element is cyclic
In fact, Rips found (torsion-free) examples of finitely generated but infinitely presented (and hence not hyperbolic) subgroups of hyperbolic groups in the early '80s. I think Gromov may have asked whether every such finitely presented group is word-hyperbolic. A (torsion-free) counterexample was found by Noel Brady. I think the correct statement is now 'For torsion-free group of type $F_3$, your question is as difficult as Gromov's question.'
Jun
23
revised Abelianization of limit groups
Added Weidmann's proof that $G$ has rank four.