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This question is related to the question 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 $h$.

edited Feb 2 at 15:19

asked Feb 1 at 19:29

Lev Glebsky
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the abelianization of $G_{q,r}$ has the abelian group presentation $\langle a,b,w\mid a^{q-1}=ba^{-1}=w^r=1\rangle$. In particular if $q>1$ this is a finite group. So I don't understand your statement that $G_{2,4}$ is an HNN-extension. – Yves Cornulier Feb 2 at 11:05

PS by "abelian group presentation" I mean presentation in the category of abelian groups. To get a group presentation, just add as relators the commutators between generators. – Yves Cornulier Feb 2 at 11:06

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It's just a split extension of the Higman group by a cyclic group of order 4 - not an HNN extension. – Derek Holt Feb 2 at 12:27

@Yves and @Derek. Oops, you are right. I will make the corresponding corrections. Thank you. – Lev Glebsky Feb 2 at 15:15

Ashot Minasyan · Accepted Answer · 2013-02-03 00:03:44Z UTC

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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.

answered Feb 3 at 0:03

Ashot Minasyan
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	@Ashot. This answers my first question! Thank you. – Lev Glebsky Feb 3 at 5:52
	@Lev: one can apply a similar argument to the groups $G_{q,r}$. For $r \ge 4$ and $q \ge 2$, $G_{q,r}$ contains a Higman-like subgroup $H$ of index $r$ with $r$ generators $H=\langle w^i a w^{-i} \mid i=1,\dots,r\rangle$. Like before, $H$ will be an amalgam of two subgroups (say, one with 3 generators and the other with $r-1$ generators). I would guess that Schupp's argument still applies to show that $H$, and hence $G_{q,r}$ is SQ-universal. – Ashot Minasyan Feb 3 at 18:33
	The above also works if $q=1$ and $r \ge 4$, in which case $H$ will be the right angled Artin group corresponding to a cycle of length $r$. A lot of things are known for such $H$. – Ashot Minasyan Feb 3 at 20:35

The Higman group II

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