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


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




, and


Please help :)

  • $\begingroup$ What do the "$\langle\,\rangle$" brackets mean in this situation? $\endgroup$ – Paul Sinclair Nov 19 '17 at 21:13
  • $\begingroup$ They denote group presentations, @PaulSinclair. $\endgroup$ – Shaun Nov 19 '17 at 21:27
  • 1
    $\begingroup$ Your original group presentation does not appear to be isomorphic to the presentation labelled (1), because they have different abelianizations. It would be easier to help if you provided these preesentations in GAP format so that they could be easily copied and pasted. $\endgroup$ – Derek Holt Nov 19 '17 at 22:22
  • $\begingroup$ Thank you, @DerekHolt. That's odd. I probably copied it out wrong. I'll sort it out tomorrow. $\endgroup$ – Shaun Nov 19 '17 at 22:30
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    $\begingroup$ In the first example, if the group is finite it must be of order at least 163063385327123630078804811957339555008348160000 Thus I'm putting my money onto infinite (but don't have a proof). $\endgroup$ – ahulpke Nov 20 '17 at 22:10

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 ];

and submitting results in the output


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