I am trying to determine the fundamental group of the n-holed torus

I know that the fundamental group of the torus is $\mathbb{Z} \times \mathbb{Z}$

The n-holed torus deformation retracts onto n copies of the circle linked together (i.e. an extended version of the figure 9 loop - with n loops)

What would its fundamental group be?

I am guessing something of the form $\mathbb{Z} \times \mathbb{Z} \times .... \mathbb{Z} $

  • 1
    $\begingroup$ I believe it is not even abelian, you can see this intuitively by noticing one can embed the n-fold figure eight(that is, a space with the same fundamental group as $\mathbb{R}^2-{x_1,..,x_n}$) in the n-fold torus. I believe (but am not sure) this induces an injection of fundamental groups, so the fundamental group of the n-fold torus has a non-abelian subgroup, hence it is not abelian. $\endgroup$
    – M. Van
    May 17, 2016 at 20:01

1 Answer 1


The $n$-holed torus has as fundamental group the group presented as

$$\langle a_1, b_1, \ldots, a_n, b_n \mid [a_1,b_1]\cdots[a_n,b_n] = 0\rangle$$

where $[a, b] = aba^{-1}b^{-1}$.

As an example, consider this octagon:


Identify all corners, then identify the edges as labeled, and you get a 2-holed torus. The sequence $a_1b_1a_1^{-1}b_1^{-1}a_2b_2a_2^{-1}b_2^{-1}$ of edge labelings immediately gives you the generating relation for the fundamental group.

  • $\begingroup$ For $n=2$, is that supposed to be $[a_1,b_1][a_2,b_2] = 0$ or $[a_1,b_1] = [a_2,b_2] = 0$? $\endgroup$
    – user14972
    May 17, 2016 at 20:10
  • 1
    $\begingroup$ It's supposed to be $[a_1,b_1][a_2,b_2] = 0.$ $\endgroup$
    – Anon
    May 17, 2016 at 20:12
  • $\begingroup$ Can you explain how you got this please? $\endgroup$
    – thinker
    May 17, 2016 at 20:18
  • $\begingroup$ Are $a_i$ and $b_i$ loops? Did you use the triangulation on simplicial complex to get this? $\endgroup$
    – thinker
    May 17, 2016 at 20:19
  • $\begingroup$ Edited the answer, added an example. $\endgroup$
    – Anon
    May 17, 2016 at 20:26

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