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1d
awarded  Good Answer
2d
comment Can we say that $A$ is a complement for a group $G$?
@GeoffRobinson, would you like to turn your comment into an answer?
2d
comment Can we say that $A$ is a complement for a group $G$?
@YemonChoi, if Geoff wants to turn his comment into an answer, that would be even better.
2d
comment Can we say that $A$ is a complement for a group $G$?
Since this question appears to have been answered in comments, I'm going to vote to close.
Nov
24
comment Periodic Growth behaviours of Cayley graphs
Have you considered non-residually finite examples? For instance, Higman's famous fp group with no non-trivial finite quotients?
Nov
14
comment Analogues of the dihedral group
@SteveD, that's a nice idea. You're right that $F_2/K$ can't be free, but it could still be virtually free, so I don't see how to finish the argument...
Nov
12
comment countably-infinite-index subgroup of a strongly complete profinite group
What does 'strongly complete' mean?
Nov
10
comment Explicitly showing that a free group is LERF
This question has been answered in comments.
Nov
10
comment Explicitly showing that a free group is LERF
In that case, I'm going to vote to close this question as 'answered in comments'.
Nov
9
comment Explicitly showing that a free group is LERF
To put it another way: the topological proof is really just language. Perhaps it would help if you would tell us which proofs you have looked at, and why you are unhappy with them.
Nov
9
comment Explicitly showing that a free group is LERF
@Pablo, if you read the standard 'topological' proof carefully, it's easy to construct the homomorphism you want (to the group of permutations of the vertices of the graph you build). Alternatively you could read Marshall Hall's original (1949) paper, in which he uses no topology.
Nov
6
comment New relator in hurwitz group
@Thomas, no , it would be a so-called almost simple group. These are rather easier to construct than infinite simple groups.
Nov
4
comment New relator in hurwitz group
And, now I think about it, the abelianization of G, hence H, is certainly trivial. Perhaps you got 'all subgroups of index at most 2'?
Nov
4
comment New relator in hurwitz group
Thomas - there's something wrong with what you said above. If H has a subgroup of index two then the abelianization of H is not trivial (since H maps onto Z/2).
Nov
4
comment New relator in hurwitz group
Well, if that's true then it certainly implies that your group is infinite (since no finite group is isomorphic to a proper subgroup of itself).
Nov
4
comment New relator in hurwitz group
Have you tried computing the abelianization of low-index subgroups? If you get lucky, you might show that it's infinite this way. (Note: it's a fact that this works for sufficiently large n, so this strategy is hopeful.)
Nov
4
awarded  Yearling
Nov
3
awarded  Excavator
Nov
3
revised What makes four dimensions special?
Corrected spelling of 'Freedman'.
Nov
2
comment Failure of Mostow rigidity in dim. 2
I believe the answer to Question 2 is precisely the 'quasi-symmetric' ones, but I'm not expert enough to tell you what this means.