most recent 30 from http://mathoverflow.net 2013-05-20T09:30:29Z http://mathoverflow.net/feeds/question/120535 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120535/the-higman-group-ii Lev Glebsky 2013-02-01T19:29:38Z 2013-02-03T00:03:44Z

<p>This question is related to the question <a href="http://mathoverflow.net/questions/87347/the-higman-group" rel="nofollow">The Higman group</a> (with a nice answer by M. Sapir). So for background, please, see the above cited question.</p> <p>The Higman group has an automorphism $h(a_j)=a_{j+1}$ ($j+1$ is mod 4). Does the Higman group have a nontrivial normal subgroup $N$, satisfying $h(N)=N$?</p> <p>Motivation. It seems to be an open question if the Higman group is hyperlinear. I seem to know how to construct a nontrivial almost representation of it in the sense of hyperlinearity. I don't know if the almost representation is exact. The negative answer on the above question would imply the exactness of my almost representation...</p> <p>More general groups. Consider $G_{q,r}=\langle a,b,w\;|\;a^q=b^{-1}ab,\;b=w^{-1}aw,\; w^r=1\rangle$. What is known about such a groups? For $q=2,\;r=4$ it is a semidirect product of a cyclic group of order 4 acting on the Higman group by $h$. </p>

http://mathoverflow.net/questions/120535/the-higman-group-ii/120633#120633 Ashot Minasyan 2013-02-03T00:03:44Z 2013-02-03T00:03:44Z

<p>I think that Higman's group H has plenty of such normal subgroups. Indeed, let G be the extension of H with the automorphism h. Then H has index 4 in G. By Schupp's theorem, H is SQ-universal, hence the same is true about G (that SQ-universality is stable under a passage to finite index sub/over groups was proved by Peter Neumann, I think.). Therefore G has (uncountably) many proper infinite normal subgroups M. Take one such M (of infinite index) and let N be its intersection with H. Clearly N has index at most 4 in M and is normal in G. Hence it possesses all the required properties.</p>