Identifying $\langle \{x_h\mid h\in H\}\mid \{x_{h'}x_{h'a}x_{h'b}^{-1}\mid h'\in H\}\rangle$ if $H=\langle a,b\mid a^2, b^4, ab=ba\rangle$.

Question

Let $H$ be $C_2\times C_4=\langle a,b\mid a^2, b^4, ab=ba\rangle$ with identity $e$. What group is $$G_H(a, b):=\langle \{x_h\mid h\in H\}\mid \{x_{h'}x_{h'a}x_{h'b}^{-1}\mid h'\in H\}\rangle?$$

Thoughts . . .

For illustration, the group is

$$\langle E, A, R, S,T, X,Y,Z\mid EA=R, AE=X, RX=S, SY=T, TZ=E, XR=Y, YS=Z, ZT=A\rangle$$

with GAP code

F:=FreeGroup("e","a","r","s","t","x","y","z");
rels:=ParseRelators(F,"ea=r,ae=x,rx=s,sy=t,tz=e,xr=y,ys=z,zt=a");
G:=F/rels;`

The group is isomorphic to $$\langle s,t\mid tst^{-1}s(st)^2s^2t^{-1}s, st(s^2t^{-1})^2s^2tst^{-1}\rangle\tag{1}$$ by GAP's IsomorphismSimplifiedFpGroup. Here's the code for the relators:

[ t*a*t^-1*a*(a*t)^2*a^2*t^-1*a, a*t*(a^2*t^-1)^2*a^2*t*a*t^-1 ]

I don't know whether or not the group is finite.

---

Using inversions & cyclic permutations on $x_ex_ax_b^{-1}$, multiplication in the subscripts by $a$, and the substitutions $y_g:=x_g^{-1}$ & $z_g:=x_{g^{-1}}^{-1}$ altogether, one gets that $G_H(a,b)$ is isomorphic to all of $G_H(a,ab), G_H(a, ab^3),$ and $G_H(a, b^3)$, which have presentations

$$\langle u,v\mid uv(vu)^2uv^{-1}u^{-1}(u^{-1}v^{-1})^2v^{-1}, uv^2u(vu^2v)^2uv^2\rangle,\tag{2}$$

$$\langle m,n\mid mnm^{-1}n(nm)^2m^2n^{-1}m, nm(n^2m^{-1})^2n^2mnm^{-1}\rangle,\tag{3}$$

and $$\langle c,d\mid cdc^2d^{-1}cdcd^{-1}c^2d, dc^{-1}d^{-1}c^{-1}(c^{-1}d)^2cd^{-1}c^2d\rangle,\tag{4}$$

respectively, according to GAP's IsomorphismSimplifiedFpGroup; their code, respectively, is

F:=FreeGroup("e","a","r","s","t","x","y","z");
rels:=ParseRelators(F,"ea=x,ae=r,rx=y,sy=z,tz=a,xr=s,ys=t,zt=e");
G:=F/rels;`

,

F:=FreeGroup("e","a","r","s","t","x","y","z");
rels:=ParseRelators(F,"ea=z,ae=t,rx=a,sy=x,tz=y,xr=e,ys=r,zt=s");
G:=F/rels;

, and

F:=FreeGroup("e","a","r","s","t","x","y","z");
rels:=ParseRelators(F,"ea=t,ae=z,rx=e,sy=r,tz=s,xr=a,ys=x,zt=y");
G:=F/rels;

Please help :)

Answer 1

After experimenting with this group on the computer for a while, it became clear that it was infinite.

The GAP command $\mathtt{PQuotient}$ can be used to compute finite $p$-quotients of a finitely presented group for a fixed prime $p$. The largest abelian quotient of $G$ has order $15$, so $|G:[G,G]| = 15$.

Applying $\mathtt{PQuotient}$ with the prime $2$ to $[G,G]$ indicates that its largest $2$-quotient of class $k$ has order $2^{9k-3}$ for all $k \ge 1$, and if that is true then $G$ is clearly infinite. Unfortunately, we can only do this calculation for specific $k$, so we cannot conclusively prove that $G$ is infinite by this method. (There are methods that can be used to prove that groups that behave in this fashion are infinite, but they require some expertise, and specialized knowledge.)

I succeeded in proving it infinite using Magma, which has a command $\mathtt{L2Quotients}$ (written by Sebastian Jambor) that can be used to simultaneously compute all quotients of $G$ of the form ${\rm PSL}(2,q)$ or ${\rm PGL}(2,q)$. I don't know whether this has been implemented in GAP. It shows that $G$ has quotients of the form ${\rm PSL}(2,2^k)$ for arbitrarily large $k$, which of course proves that it must be infinite.

Magma is unfortunately not open source, but there is a Magma calculator that can be used to do this calculation. Typing in

F<a,t> := FreeGroup(2);
rels := [ t*a*t^-1*a*(a*t)^2*a^2*t^-1*a, a*t*(a^2*t^-1)^2*a^2*t*a*t^-1 ];
G:=quo<F|rels>;
L2Quotients(G);

and submitting results in the output

[
    L_2(2^infty)
]