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Dec 29, 2016 at 17:47 answer added YCor timeline score: 5
Dec 29, 2016 at 5:31 answer added Izaak Meckler timeline score: 2
Aug 11, 2015 at 13:48 vote accept Andrew Penland
Aug 9, 2015 at 15:42 answer added Igor Rivin timeline score: 3
Aug 9, 2015 at 11:22 answer added Andreas Thom timeline score: 20
Aug 9, 2015 at 6:08 comment added HJRW 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 9, 2015 at 4:35 review First posts
Aug 9, 2015 at 5:47
Aug 9, 2015 at 4:32 history asked Andrew Penland CC BY-SA 3.0