graph theory - Is there a purely set-theoretic expression of the Euler characteristic? idea – The Higman group retort

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Higman’s group isn’t unostentatious. Indeed, in case you contemplate at Higman’s paper, you’ll graze that his group is an amalgamated product of two teams $K_{1,2}=langle a_1, a_2, b_2mid a_1^{-1}a_2a_1=a_2^2, a_2^{-1}b_2a_2=b_2^2rangle$ and $K_{3,4}=langle a_3, a_4, b_4mid a_3^{-1}a_4a_3=a_4^2, a_4^{-1}b_4a_4=b_4^2rangle$ with free amalgamated subgroups $langle a_1, b_2rangle$ and $langle a_3, b_4rangle$ with $a_1=b_4, b_2=a_3$. Now maintain $K_{1,2}$ and add the relations $w(a_1,b_2)=1, w(b_2,a_1)=1$ for some “complicated” phrase $w$. You purchase a non-trifling group $Okay$. Similarly stately $w(a_3,b_4)=1=w(b_4,a_3)$ on $K_{3,4}$ you purchase a non-trifling group $Okay’$. Then there’s a homomorphism from the Higman group to the amalgamated product of $Okay$ and $Okay’$ with a non-trifling kernel. The proven fact that $Okay$ (and, equally, $Okay’$) isn’t trifling isn’t patent however it isn’t very troublesome to show.

A simplification. Instead of including two relations, in reality it is sufficient to add a relation $w(a_1,b_2)=w(b_2,a_1)$ and equally for $a_3,b_4$. For occasion, $(a_1b_2)^3=(b_2a_1)^3$.

Update 1. Probably the simplest path to breathe satisfied that $Okay$ ($=Okay’$) is non-trifling is to enter its presentation into GAP or Magma (you assassinate not necessity 3 within the simplified relation above, 2 is sufficient).

Update 2. An airplane simpler route is to ordain additional relations $a_2=b_2=1$ on $Okay$ (as within the simplification above). The factor-group is the innumerable cyclic group. Hence $Okay$ is innumerable (and $Okay’$ is innumerable too).

Update 3. I forgot to say that one too wants that after we ordain the relation as within the simplification, we should always make optimistic that the subgroup of $Okay$ generated by $a_1,b_2$ has an automorphism switching $a_1$ and $b_2$ (in any other case we can’t design the succesful amalgamated product of $Okay$ and $Okay’$). That is why we can’t supplant 3 within the Simplification above by $1$.

Update 4. Gilbert Baumslag instructed me that Higman didn’t know whether or not his group is unostentatious, however Paul Schupp proved it. And certainly, Schupp, Paul E. diminutive cancellation idea over free merchandise with amalgamation. Math. Ann. 193 (1971), 255–264 proved mighty extra: Higman’s group is SQ-universal, so each countable group embeds into certainly one of its quotients.

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