Chou asked in this paper whether The Higman group $H$ has a maximal normal subgroup $N$ such that $H/N$ has no (non-abelian) free subgroups (or is amenable). Is it known now if such subgroups exists or not?
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asked Sep 26, 2013 at 22:35
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5$\begingroup$ This is certainly an open question. Indeed, before this (arxiv.org/abs/1204.2132) recent paper by Juschenko and Monod, no example was known of a nontrivial amenable group with no nontrivial finite quotient. So if the answer to your question was positive it would yield new examples. If the answer was negative, it would be a surprisingly nice result; such results are usually only known for Kazhdan Property T groups or using related methods. $\endgroup$– YCorCommented Sep 27, 2013 at 3:04
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2$\begingroup$ @YvesCornulier, it seems to me that your comment would make a very nice answer. $\endgroup$– HJRWCommented Sep 27, 2013 at 4:34
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1 Answer
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For amenability, this is certainly an open question. Indeed, before this recent paper by Juschenko and Monod, no example was known of a nontrivial finitely generated amenable group with no nontrivial finite quotient. So if the answer to your question was positive it would yield new examples. If the answer was negative, it would be a surprisingly nice result; such results are usually only known for Kazhdan Property T groups or using related methods, e.g. for irreducible lattices in products.
Edit: for no free subgroups, maybe it's more doable. Indeed, Higman's group is known to be SQ-universal (Paul E. Small cancellation theory over free products with amalgamation. Math. Ann. 193 (1971), 255–264), and this is based on a suitable decomposition as an amalgam. This means that there is a lot of flexibility to construct quotients; maybe, if I'm not too optimistic, this can be used to construct a nontrivial quotient of the Higman group with no free subgroup, e.g. by forcing enough elements to be torsion in the image.
(I don't claim this would work in any SQ-universal group! for instance if we take Thompson's group $T$, then $T\ast T$ is SQ-universal but all its nontrivial quotients contain a copy of $T$ and hence contain a free subgroup.)
