The infinite group of Higman which has no finite quotient is given by the presentation (with 4 generators and 4 relations):

G=⟨ai,i∈Z/4Z∣aiai+1(mod 4)a−1i=a2i+1⟩.

Its main feature (proved by Higman) is that [it is infinite and] it has no finite quotients. Another important feature (proved by Schupp, see this post) is that it is SQ-universal (:=”every countable group can be embedded in one of its quotient groups”).My question is: is it known whether this group has property (T) or not?

My lame attempts to get an insight led me to a first side question: does SQ-universal exclude property (T)?

Then I was looking into an algorithmic way of trying property (T) on Higman’s group (Zuk’s criterion). Which brought to the other side question: has this group a solvable word problem ? (Superficially, it looks like it could, by analogy with the Baumslag-Solitar groups).

Lastly, a presentation is rarely enough to get any information on the representation theory of the group. Are there any known facts about the representations of Higman’s group (except it has no finite-dimensional non-trivial representations [using finite generation + Malcev theorem to the effect that finitely generated linear groups are residually finite])

**Answer**

**Attribution***Source : Link , Question Author : ARG , Answer Author : Community*