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Consider the hyperbolic (extended) triangle group $\Delta(2,3,7)=\langle a,b,c\mid a^2,b^2,c^2,(ab)^2,(bc)^3,(ca)^7\rangle$. I construct it in GAP as a finitely presented group, using the standard method:

FG:=FreeGroup("a","b","c");;
D:=FG/[FG.1^2,FG.2^2,FG.3^2,(FG.1*FG.2)^2,(FG.2*FG.3)^3,(FG.3*FG.1)^7];;

I next construct the set $W=\{\gamma_1,\ldots,\gamma_7\}\subset\Delta(2,3,7)$ as a particular set of seven (complicated) words in the generators $a,b,c$. I know from mathematical considerations that $W$ should generate an index-336 normal subgroup of $\Delta(2,3,7)$, i.e., $\Gamma=\langle W\rangle\triangleleft \Delta(2,3,7)$:

GAMMA:=Subgroup(D,gam);;

where gam is a list containing the seven words $\gamma_1,\ldots,\gamma_7$ above.

Now here is my problem. GAP correctly recognizes that $\Gamma$ is a subgroup of $\Delta(2,3,7)$ of index 336:

gap> IsSubgroup(D,GAMMA);
true
gap> Index(D,GAMMA);
336

but it doesn't seem to recognize that it is normal:

gap> IsNormal(D,GAMMA);
false

What am I doing wrong?

Shaun
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MathPhysGeek
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  • Just out of curiosity, because the number '336' is so unique: Klein quartic-related? And more relevant to the question: what happens if you ask for the normalizer of $\Gamma$ in $D$? – Steven Stadnicki Feb 23 '22 at 22:04
  • @StevenStadnicki Yes, $\Gamma$ should be isomorphic to the fundamental group of a genus-3 surface. 336 is the number of Schwarz triangles in Klein's "Hauptfigur" (hyperbolic 14-gon). If I compute the index of $\Gamma$ in Normalizer(D,GAMMA), I get 14. I am even more confused now... – MathPhysGeek Feb 23 '22 at 22:29
  • Can you get a presentation of Normalizer(D,GAMMA)? It sounds like you may have a small bug in your list of $\gamma_i$ somewhere and that's the most straightforward path I can think of to try and track it down. – Steven Stadnicki Feb 23 '22 at 22:39
  • I define (on purpose) the $\gamma_i$ as opposite-side pairings as opposed to pairings between sides $2k+1$ and $2k+6$ mod $7$, $k=0,\ldots,6$ in the actual Klein quartic. So maybe I shouldn't expect $\Gamma$ to be normal in $\Delta$? – MathPhysGeek Feb 23 '22 at 22:55
  • Ok, I think my $\Gamma$ as defined is just not normal in $\Delta$. If I use the actual side-pairing transformations of the Klein quartic, I do find that the corresponding $\Gamma$ is normal. Both are index-336 in $\Delta$ but the normality structure is different. Thanks @StevenStadnicki for the help. – MathPhysGeek Feb 23 '22 at 23:54

1 Answers1

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Without seeing the explicit list gam it is hard to give a certain reason, but a common error is to take elements of the free group (in your example FG) instead of the factor group (in your example G).

If you want to test, your quotient of order 336 is most likely $PGL(2,7)$:

gap> GQuotients(D,PGL(2,7));
[ [ a, b, c ] -> [ (2,5)(4,6)(7,8), (1,3)(2,7)(5,8), (3,6)(4,7)(5,8) ]]

Evaluate your words $\gamma_i$ in these permutations to see whether they are in the kernel.

ahulpke
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  • Thanks! I made sure to use elements of D and not FG. gam[i] in Kernel(GQuotients(D,PGL(2,7))[1]); for my originally defined gam returns false, but if I define new gamK[i] which are the standard side pairings of the Klein quartic, then gamK[i] in Kernel(GQuotients(D,PGL(2,7))[1]); returns true. So my original $\Gamma=\langle\gamma_1,\ldots,\gamma_7\rangle$ was not normal in $\Delta$, it turns out. – MathPhysGeek Feb 24 '22 at 23:16
  • Thank you for the confirmation. So the world is sane again. – ahulpke Feb 24 '22 at 23:35