I want to calculate the fundamental group of of a $7$-gon with labelling scheme $abaaab^{-1}a^{-1}$. I will call the quotient space $X$ with reference point $x_0$. This is what I tried:

If you have the book topology of Munkres, Theorem 74.2 says that this fundamental group should be isomorphic to the quotient of the free group on the generators $a$ and $b$ by the least normal subgroup containing the element $aba^3b^{-1}a^{-1}$. So if $G_1=\langle a\rangle$ and $G_2=\langle b\rangle$ and $N$ is this least normal subgroup containing $aba^3b^{-1}a^{-1}$, then $\pi_1(X,x_0)=(G_1 *G_2)/N$. Now theorem 68.7 of the same book says that if you have two groups $H_1$ and $H_1$ and two normal subgroups $M_1\subset H_1$ and $M_2\subset H_2$ and suppose $M$ is the least normal subgroup of $H_1*H_2$ containing both $M_1$ and $M_2$ then we have the isomorphism $(H_1*H_2)/M\cong H_1/M_1*H_2/M_2$.

In my case, I have a normal subgroup $N$ of $G_1*G_2$ which contains the element $aba^3b^{-1}a^{-1}$. I see that this element is a conjugate: $(ab)a^3(ab)^{-1}$. So $a^3\in N$ and we must have that $N_1$ is the least normal subgroup of $G_1$ containing the element $a^3$, that is $N_1=\langle a^3\rangle$. I cannot find a normal subgroup $N_2$ of $G_2$ which is contained in $N$ except for the trivial group. So I think that $\pi_1(X,x_0)\cong \mathbb{Z}/3\mathbb{Z}*\mathbb{Z}$.

I am not sure whether this is correct. In particular, I am not sure whether $N_2$ is the trivial normal subgroup of $G_2$. Can you help me at this point? Thanks.

  • 1
    $\begingroup$ Have you tried drawing the n-gon? you should see that two of the $a$s cancel and both of the $b$s cancel (that is you can 'zip' them together). So you should be left with just a triangle with labelling scheme $aaa$. This is also easy to see algebraically because labelling schemes are invariant under cyclic permutation. $\endgroup$ – Dan Rust Aug 9 '14 at 11:44
  • $\begingroup$ @DanielRust So the fundamental group should be $\mathbb{Z}/3\mathbb{Z}$? What went wrong in my calculation then? $\endgroup$ – Badshah Aug 9 '14 at 11:54
  • 1
    $\begingroup$ @DanielRust is spot on, you've forgotten this condition from the statement of the theorem: "If $\pi$ maps all the vertices of $P$ to a single point $x_0$ of $X$ [...]". $\endgroup$ – Najib Idrissi Aug 9 '14 at 13:07
  • 1
    $\begingroup$ Ha, I think I have to take back the above comments. I was too quick to assume that you can just forget about the $a$ and $a^{-1}$ which appear next to each other, but this isn't the case because they are also identified with other edges in the $n$-gon so they can't just 'zip up' and disappear. The condition @NajibIdrissi mentioned is satisfied by your labeling scheme and so you can apply the theorem, in which case your calculation is correct! Apologies for leading you astray! $\endgroup$ – Dan Rust Aug 9 '14 at 14:04
  • 1
    $\begingroup$ @DanielRust Thanks, I now understand where I went wrong, indeed, you can't just forget about the $a$ and $a^{-1}$ because there are other edges with that label. Thank you very much for your time :) $\endgroup$ – Badshah Aug 9 '14 at 14:10

Your Answer

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

Browse other questions tagged or ask your own question.