Consider the group $G$ having a presentation $G=\langle x,y,z~|~x^2y^2z^2 \rangle$. I am trying to find all subgroups of $G$ of index 6 using covering space theory. It is well-known that the connected sum $X=3\Bbb RP^2$ of three projective planes has fundamental group isomorphic to $G$. Also for each subgroup of $G=\pi_1(X)$, there is a covering space $p:\tilde{X}\to X$ such that $p_*(\pi_1(\tilde{X}))=H$, and if the index $[G:H]$ is $n$, then $p$ is $n$-sheeted. Thus the question reduces to find all $6$-sheeted covering spaces of $X$, but I can't see a way because I've never seen a covering spaces of connected sums. Any hints?

  • 3
    $\begingroup$ One way would be to consider the presentation complex of $\langle x, y, z\mid x^2y^2z^2\rangle$, then determine all the ways this group can surject onto a group of order $\leq6$. These surjections correspond to covers of the complex, and then presentations for the corresponding subgroups can be found. I guess you can use the space $X$ instead, but I personally am less comfortable with that :-) $\endgroup$
    – user1729
    May 14 '20 at 14:25
  • $\begingroup$ @user1729 How can we determine the ways the group can surject onto a group of order $\leq 6$? $\endgroup$
    – user302934
    May 18 '20 at 5:23
  • 1
    $\begingroup$ By thinking very hard. At the moment, I can't see any method apart from brute force (take your finite group and try to map the generators of the infinite group to each element individually, whilst rejecting obvious failures etc.), but I'm sure some thinking will reveal a better way. However, you shouldn't expect the number of maps/subgroups to be a small number; there are already $7$ surjections to the cyclic group of order two (corresponding to killing zero, one or two of the generators), and approximately twice this for the cyclic group of order four. $\endgroup$
    – user1729
    May 18 '20 at 6:44
  • 1
    $\begingroup$ (Also, forgot to say that this only gives you the normal subgroups; for non-normal subgroups you gave to mess around with cosets. Look up Reidemeister-Schreier for more details.) $\endgroup$
    – user1729
    May 18 '20 at 6:51
  • $\begingroup$ I suspect, in the question (which looks like a homework or a question from an old exam) you are asked to find all such subgroups up to an isomorphism, which is indeed a topological question. $\endgroup$ May 21 '20 at 21:07

The following is too long for a comment (and is, hence, a CW post).

There is at most $2,362$ subgroups of index at most $6$.

gap> F:=FreeGroup(3);
<free group on the generators [ f1, f2, f3 ]>
gap> rels:=[(F.1)^2*(F.2)^2*(F.3)^2];
[ f1^2*f2^2*f3^2 ]
gap> G:=F/rels;
<fp group of size infinity on the generators [ f1, f2, f3 ]>
gap> Size(LowIndexSubgroupsFpGroup(G,6));

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.