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Apr
7
comment Can one construct a “Cayley diagram” that lacks only an inverse?
@RespawnedFluff Well, any monoid. But I think the problem is more subtle - in semigroups there are "left" Cayley graphs and there are "right" Cayley graphs, which are usually different. But here it is commutative so this doesn't matter. Also, I suspect some commutative monoids would have Cayley graphs which do not form part of a groups Cayley graph (so there is something else going on). I think this is why this example is nice - it is missing inverses, and nothing else. If we add in inverses we get a group/the Cayley graph of a group (note that there is a unique way of adding in the inverses).
Apr
2
comment Decomposition of a group whose Cayley graph is a tree
@Hanna Whereabout in Hatcher? Also, $\operatorname{PSL}_2(\mathbb{Z})$ acts on a tree (as does any free product - this is the Bass-Serre tree), but the action is not free. It is free on the edges thought (but every torsion element fixes some vertex).
Apr
1
comment Decomposition of a group whose Cayley graph is a tree
@Seirios Then there is something odd going on with, I suspect, a definition or a translation of a definition. A group $G$ has a tree for its Cayley graph only if $G$ is free (or, for people wanting to confuse people, you could make the Cayley graph of $F_n\ast C_2$ into a tree by writing the $C_2$ generator as an undirected edge). This is because a group acts freely on its Cayley graph, and if a group acts freely on a tree then it is a free group. Or use the covering space argument, above.
Apr
1
comment Decomposition of a group whose Cayley graph is a tree
Perhaps the OP meant quasi-isometric to a tree?
Apr
1
comment Decomposition of a group whose Cayley graph is a tree
@Qiaochu yes, true, but it still corresponds to a loop. If $w$ is an element of order three then it gives a loop in the Cayley graph. (As does any relator). Or, to put it another way, the presentation complex contains $2$-cells so the universal cover (the Cayley complex) cannot have a tree as its 1-skeleton. However, its 1-skeleton is the Cayley graph.
Mar
31
answered Can closure of quaternions under multiplication be shown with a cayley table?
Mar
31
revised Sort-of-simple non-Hopfian groups
added 24 characters in body
Mar
31
comment Decomposition of a group whose Cayley graph is a tree
I am probably missing something silly here, but how can a group with an element of order $3$ have a tree for its Cayley graph? Surely the element of order $3$ gives a loop of order $3$ in the graph?
Mar
28
awarded  Nice Answer
Mar
27
accepted Are isomorphisms always constructable?
Mar
26
comment Are isomorphisms always constructable?
@j.p. Yes, that is what I was envisaging. However, I would be surprised if how we were given the groups mattered (in the same way that a class of groups has soluble or insoluble isomorphism problem and this is independent of how the groups are given to us).
Mar
26
comment Are isomorphisms always constructable?
@user126154 No...$G=\langle S\rangle$ means that $S$ is a generating set for $G$, while $\langle G\mid\: \rangle$ (or something similar) would imply free.
Mar
26
comment Are isomorphisms always constructable?
@user126154 Yes, exactly. There is the added complication that merely generation is not enough, as it is possible that the groups might be non-Hopfian.
Mar
26
revised Are isomorphisms always constructable?
added 219 characters in body
Mar
26
comment Are isomorphisms always constructable?
@Bolt64 See wikipedia, or if you have access to a library find the book Combinatorial group theory by Magnus, Karrass and Solitar (or pick up any book on the same shelf).
Mar
26
asked Are isomorphisms always constructable?
Mar
23
comment Are there any conditions , such that 2 permutations in $S_4$ are commutative?
@Guldam Yes, left-first. If you want right-first then swap the $\tau$ and $\tau^{-1}$ in $\tau^{-1}(cycles)\tau$ line, and keep the next line the same.
Mar
11
comment Are all $\delta$-hyperbolic groups CAT(0)?
Incidentally, Calegari and Walker recently proved that Sapir's group contained a closed surface subgroup of genus 28. I think these two papers go some way to addressing the question asked (in a probabilistic, or probably hand-wavey, sense), but will have to think about it.
Mar
11
comment Are all $\delta$-hyperbolic groups CAT(0)?
I had just been reading over their paper before I read your comment! A freshly printed copy is sitting on my desk, and I was just checking stuff before taking it for a coffee... Incidentally, it is R. Kropholler - his father, P. Kropholler, works in the same area (although he is more topological, but he is credited for the translation of JSJ-decompositions from $3$-manifolds to group theory, as made famous by Sela et al.)
Feb
23
comment Can I recover a group by its homomorphisms?
In addition to @QiaochuYuan's comment, recent work of Bridson and Wilton essentially covers this topic. Specifically, this paper proves that, given two finitely presented, residually finite groups, it is undecidable whether their profinite completions are isomorphic. The "profinite completion" of a group is an object which encodes all its finite quotients. See also this related paper, which proves this it is undecidable whether a finitely presented group has a finite quotient (equivalently, has trivial profinite completion)