# Quotients of the Higman Group

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?

• 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.
– YCor
Sep 27 '13 at 3:04
• @YvesCornulier, it seems to me that your comment would make a very nice answer.
– HJRW
Sep 27 '13 at 4:34

(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.)