5
$\begingroup$

We know $\pi_1(\mathbb {T^2} \sharp \mathbb T^2)=\langle\alpha_1,\beta_1,\alpha_2,\beta_2|\alpha_1 \beta_1 \alpha_1^{-1} \beta_1^{-1}\alpha_2\beta_2\alpha_2^{-1}\beta_2^{-1}=1\rangle$.

My question is:How to find a subgroup of it with index $2$?

I think we need to find a subgroup H of $\pi_1(\mathbb {T^2} \sharp \mathbb T^2)$ which has two generator.And $aH \cup bH=\pi_1(\mathbb {T^2} \sharp \mathbb T^2)$ for $a,b\in \pi_1(\mathbb {T^2} \sharp \mathbb T^2)$.

But how to deal with the equivalent relation?Thank you!

  • 3
    Have you tried thinking about this geometrically, i.e., what surface covers this space with index 2?2012-06-13
  • 0
    @SteveD Yes,yes!A double cover of $\mathbb T^2 \sharp \mathbb T^2$ corresponds to a subgroup of index 2.But I'm sorry it's a little hard for me to find a double cover of it :(2012-06-13
  • 0
    And I'm still curious how to find it through groups2012-06-13
  • 4
    Dear Jiangnan, A subgroup of index two is normal, so why not try to find a homomorphism from $\pi_1$ to a group of order two, and then determine the kernel. Finding a homomorphism *from* a group described by generators and relations is normally easier than directly desribing subgroups inside, because the generators and relations tell you precisely what kind of homomorphic images are allowed! Regards,2012-06-13
  • 0
    To add to Matt E's comment, every subgroup of index 2 arises this way, and once you've figured out your homomorphism, a Reidemeister-Schreier rewriting will give you a presentation for that subgroup.2012-06-13
  • 0
    @MattE Thank you very much for your hints!Now I can figure it out:))2012-06-13
  • 0
    What $\sharp$ means?2018-01-22

1 Answers 1