I am trying to see if the group


$G = \langle\ a,b,c,d\ \mid\ a^3 = b^3=c^3=d^3=$


$\qquad =(ab)^4= (ab^{-1})^4= (bc)^4= (bc^{-1})^4= (cd)^4= (cd^{-1})^4= (da)^4= (da^{-1})^4\ \rangle$


is trivial or not. I have tried all methods and tools I could but have not succeeded.


Knuth -- Bendix in Magma does not finish:



```
F <a,b,c,d> := FreeGroup(4);

L3  := [ w^3 : w in [ a, b, c, d ] ];
L4  := [ w^4 : w in [ a*b, a*b^-1, b*c, b*c^-1, c*d, c*d^-1, d*a, d*a^-1 ] ];

G := quo< F | L3 cat L4 >;

GR := RWSGroup(G : MaxRelations := 500000, TidyInt:=1000 );
// Warning: Knuth Bendix only partly succeeded

```

Determining the index of a large subgroup is inconclusive:



```
H := sub < G | a, b, c, (c*d)^2 >; 
Index(G,H:CosetLimit:=100000000,Hard:=true);
// 0  

```

GAP fails as well:



```
gap> F:=FreeGroup("a","b","c","d");;
gap> AssignGeneratorVariables(F);
#I  Assigned the global variables [ a, b, c, d ]
gap> 
gap> L3  := List ( [ a, b, c, d ], w -> w^3 );;
gap> L4  := List ( [ a*b, a*b^-1, b*c, b*c^-1, c*d, c*d^-1, d*a, d*a^-1 ], w -> w^4 );;
gap> 
gap> R:=Concatenation(L3,L4);;
gap> G:=F/R;
<fp group on the generators [ a, b, c, d ]>
gap> 
gap> gg := GeneratorsOfGroup(G);; 
gap> 
gap> LoadPackage("kbmag");
───────────────────────────────────────────────────────────────────
Loading  kbmag 1.5.4 (Knuth-Bendix on Monoids and Automatic Groups)
by Derek Holt (http://homepages.warwick.ac.uk/staff/D.F.Holt/).
Homepage: https://gap-packages.github.io/kbmag
───────────────────────────────────────────────────────────────────
gap> R := KBMAGRewritingSystem(G);;
gap> 
gap> H := Subgroup(G,gg{[1,2,3]});; # < a, b, c >
gap> HR := SubgroupOfKBMAGRewritingSystem(R,H);;
gap> A := AutomaticStructureOnCosets(R,HR); 
#Knuth-Bendix program failed or was inconclusive. Giving up.
false

```

Are there any other methods I could try?



Checking triviality of a finitely presented group
It's nontrivial, indeed infinite.



> 
> $G\neq 1$; moreover it is an amalgam (actually a double) over a subgroup of index $\ge 3$ (thus is infinite and contains nonabelian free subgroups).
> 
> 
> 


Indeed, write $$H=H(t,u)=\langle t,u\mid t^3,u^3,(tu)^4,(t^{-1}u)^4\rangle.$$


Conversely, let us first show that $H\neq 1$ implies $G\neq 1$. Note that $H\neq 1$ means that $t,u$ have order 3. Define $$L=L(t,u,v)=\langle t,u,v\mid t^3,u^3,v^3,(tu)^4,(t^{-1}u)^4,(uv)^4,(uv^{-1})^4\rangle.$$


Then $L$ is the amalgam of $H(t,u)$ and $H(u,v)$ over the cyclic subgroup of order 3 $\langle u\rangle$ (which by the way has index $\ge 3$, since $H$ is perfect and nontrivial and hence of cardinal $\ge 60$). 


Let $M$ be the subgroup of $L(t,u,v)$ generated by $t$ and $v$. Consider the double $D$ of $L=L(t,u,v)$ over $M$, that is, the amalgam $L\ast\_M L$. Then it has the presentation 


$$D=\langle t,u,v,t\_2,u\_2,v\_2\mid t^3,u^3,v^3,(tu)^4,(t^{-1}u)^4,(uv)^4,(uv^{-1})^4,t\_2^3,u\_2^3,v\_2^3,(t\_2u\_2)^4,(t\_2^{-1}u\_2)^4,(u\_2v\_2)^4,(u\_2v\_2^{-1})^4,t=t\_2,v=v\_2\rangle.$$
Eliminate $v\_2(=v)$ and $t\_2(=t)$ and write $w=u\_2$. This yields
$$D=\langle t,u,v,w\mid t^3,u^3,v^3,(tu)^4,(t^{-1}u)^4,(uv)^4,(uv^{-1})^4,w^3,(tw)^4,(t^{-1}w)^4,(wv)^4,(wv^{-1})^4\rangle,$$
where we precisely recognize $G$. Hence $G\neq 1$.


Now let us check that $G$ is a nontrivial amalgam- actually double- (over subgroups of index $\ge 3$). Indeed, let me discuss. If $M$ has index $\ge 3$, it's OK. $M$ can't have index 2 because $L$ is perfect. If $M=1$ (sounds unlikely but I haven't checked), then $L=M$ itself was described as an amalgam, actually a double.


Finally, to check that $H\neq 1$, just observe that in the symmetric group $\mathfrak{S}\_7$, $t=(123)(456)$ and $u=(234)(576)$ satisfies the given relators of $H$.


