This question is related to the question [The Higman group](https://mathoverflow.net/questions/87347/the-higman-group) (with a nice answer by M. Sapir). So for background, please,
see the above cited question.


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 $h(N)=N$?


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...


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 $h$. 



The Higman group II
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.


