I am told that for a surface with genus $g$, call it $F_g$, the abelianization of $\pi_1(F_g) = \langle a_1, b_1, \ldots , a_g, b_g \mid [a_1, b_1] \cdots [a_g,b_g] = e \rangle$ is $\pi_1(F_g)^{ab} = \Bbb Z \langle a_1, b_1, \ldots , a_g, b_g \rangle$.

The definition of abelianization of a group $G$ (from Wikipedia) is $G / [G, G]$. So the abelianization of $\pi_1(F_g)$ is $\pi_1(F_g) / [\pi_1(F_g), \pi_1(F_g)]$.

I don't see why $\pi_1(F_g) / [\pi_1(F_g), \pi_1(F_g)] = \Bbb Z \langle a_1, b_1, \ldots , a_g, b_g \rangle$.

  • 3
    $\begingroup$ There are more direct ways of showing it, but since we're talking about algebraic topology anyway: $\pi_1(F_g)^{ab} = H_1(F_g)= H_1((T^2)^{\# g}) = H_1(T^2)^{\oplus g} = \mathbb{Z}^{2g}$, and unraveling the isomorphisms above gives the explicit presentation of $\pi_1(F_g)^{ab}$ in terms of the $a_i, b_i$. $\endgroup$
    – anomaly
    Dec 17, 2015 at 5:18

2 Answers 2


Note that $[G, G]$ consists of all products of commutators of elements of $G$. In particular, $[a_1, b_1]\dots[a_g, b_g] \in [\pi_1(F_g), \pi_1(F_g)]$, so the relation $[a_1, b_1]\dots[a_g, b_g] = e$ becomes $e = e$ in the quotient group. This may lead you to believe that $\pi_1(F_g)^{ab}$ has presentation $\langle a_1, b_1, \dots, a_g, b_g\rangle$ and hence must be the free group on $2g$ generators, but that is false. For each $i$ and $j$, all of the elements $[a_i, a_j], [b_i, b_j], [a_i, b_j]$ also belong to $[\pi_1(F_g), \pi_1(F_g)]$, and hence become trivial in the quotient. Therefore

$$\pi_1(F_g)^{ab} = \langle a_1, b_1, \dots, a_g, b_g \mid [a_i, a_j] = [b_i, b_j] = [a_i, b_j] = e\ \forall\ i, j\rangle = \mathbb{Z}\langle a_1, b_1, \dots, a_g, b_g\rangle.$$

That is, $\pi_1(F_g)^{ab}$ is the free abelian group on $2g$ generators, i.e. $\pi_1(F_g)^{ab} \cong \mathbb{Z}^{2g}$.

Said another way, given a group $G$ with presentation

$$\langle r_1, \dots, r_m \mid s_1 = \dots = s_n = e\rangle,$$ then its abelianisation, $G^{ab}$, has a corresponding presentation

$$\langle r_1, \dots, r_m \mid s_1 = \dots = s_n = e, [r_i, r_j] = e\ \forall\ i, j\rangle.$$

In your case, the original group presentation had only one relation, $[a_1, b_1]\dots[a_g, b_g] = e$. When the commutator relations between the generators are added to the presentation for the abelianisation, the original relation becomes redundant (i.e. $[a_i, b_i] = e$ for all $i$ implies $[a_1, b_1]\dots[a_g, b_g] = e$). Therefore, you obtain the presentation for $\pi_1(F_g)^{ab}$ I wrote above, and hence conclude that $\pi_1(F_g)^{ab} \cong \mathbb{Z}^{2g}$.

Added Later: Judging from your comments, you seem to be misunderstanding the notation; for the sake of simplicity, I will only consider a finite set of generators $R = \{r_1, \dots, r_m\}$.

One can form the free group on $R$ which can be denoted simply by $F_R$ or $\langle r_1, \dots, r_m\rangle$. The elements of this group are finite strings in the generators and their inverses which are reduced (i.e. they do not contain products of the form $r_ir_i^{-1}$ or $r_i^{-1}r_i$). The group operation on such strings is concatenation (write one string after the other, then reduce).

Alternatively, one can form the free abelian group on $R$ which can be written simply as $\mathbb{Z}^{(R)}$, $\langle r_1, \dots, r_m \mid\ [r_i, r_j] = e\ \forall\ i, j\rangle$, or $\mathbb{Z}\langle r_1, \dots, r_m\rangle$. As before, the elements are reduced strings, but now, two strings which are rearrangements of each other are considered the same, e.g. $r_4r_1r_2^{-1}r_4$ is the same string as $r_1r_2^{-1}r_4^2$. As we can always reorder our strings, we can write every element in a unique way as $r_1^{k_1}r_2^{k_2}\dots r_m^{k_m}$ where $k_1, \dots, k_m \in \mathbb{Z}$. Then $\mathbb{Z}\langle r_1, \dots, r_m\rangle \cong \mathbb{Z}^m$ where the isomorphism is given by $r_1^{k_1}r_2^{k_2}\dots r_m^{k_m} \mapsto (k_1, k_2, \dots k_m)$. When $m = 1$, this is the map that shows that any infinite cyclic group is isomorphic to $\mathbb{Z}$.

In the comments you were unsure how I went from

$$\langle a_1, b_1, \dots, a_g, b_g \mid [a_i, a_j] = [b_i, b_j] = [a_i, b_j] = e\ \forall\ i, j\rangle$$


$$\mathbb{Z}\langle a_1, b_1, \dots, a_g, b_g\rangle.$$

As I've outlined above, they are merely two different ways of describing the same group: the free abelian group on the generators $\{a_1, b_1, \dots, a_g, b_g\}$.

  • $\begingroup$ Even with your explanation, I'm still having trouble seeing why $\pi_1(F_g)^{ab} = \langle a_1, b_1, \dots, a_g, b_g \mid [a_i, a_j] = [b_i, b_j] = [a_i, b_j] = e\ \forall\ i, j\rangle = \mathbb{Z}\langle a_1, b_1, \dots, a_g, b_g\rangle.$ I get that $\pi_1(F_g)^{ab} = \langle a_1, b_1, \dots, a_g, b_g \mid [a_i, a_j] = [b_i, b_j] = [a_i, b_j] = e\ \forall\ i, j\rangle$, but I don't see where the $\Bbb Z$ comes from or why you removed the relation $[a_i, a_j] = [b_i, b_j] = [a_i, b_j] = e\ \forall\ i, j$. $\endgroup$ Dec 17, 2015 at 5:35
  • 1
    $\begingroup$ Do you know what $\mathbb{Z}\langle a_1, b_1, \dots, a_g, b_g\rangle$ means? $\endgroup$ Dec 17, 2015 at 5:36
  • $\begingroup$ So if $\Bbb Z = \langle x \rangle$, then $\Bbb Z \langle a_1, b_1, \ldots , a_g, b_g \rangle = \langle xa_1, xa_2, \ldots , xa_g, xb_g \rangle$? $\endgroup$ Dec 17, 2015 at 5:38
  • $\begingroup$ No, that doesn't really make sense. I have added a few comments about the notation. Hopefully it makes it clear. $\endgroup$ Dec 17, 2015 at 6:04

The group presentation $G = \langle x_1, \ldots, x_n\mid r_1, \ldots, r_m\rangle$ is the complicated one. It says that $G=\langle x_1, \ldots, x_n\rangle/H$, where $H$ is the smallest normal subgroup of $\langle x_1, \ldots, x_n\rangle$ which contains $\{r_1, \ldots, r_m\}$. That's quite a mouthful, especially the normal bit. But the similar-looking presentation for an abelian group is much easier. The abelian group presentation $A={\mathbb Z}[ x_1, \ldots, x_n \mid r_1, \ldots, r_m]$ says that $A = {\mathbb Z}[x_1, \ldots, x_n]/{\mathbb Z}[r_1, \ldots, r_m]$, where $F = {\mathbb Z}[x_1, \ldots, x_n] = \{s_1 x_1 + \ldots + s_n x_n\}$ for any $s_i \in {\mathbb Z}$ and $B = {\mathbb Z}[r_1, \ldots, r_m]$ is the subgroup of $F$ consisting of the elements $\{t_1 r_1 + \ldots + t_n r_n\}$ for any $t_j\in{\mathbb Z}$. Note that I've changed the name of the operation in the group from $x_1 x_2$ (multiplication) to $x_1 + x_2$ (addition). Also note a subtlety in the definitions of $A$ and $B$. In the definition of $A$, the $x_i$ are unrelated indeterminates. In the definition of $B$, the $r_i$ are already elements of $F$ and you might or might not need all of them to generate $B$. But the point is that it's linear algebra now, and that's way easier.

Now back to your problem. Because you're calculating $G^{\rm ab}$, which is abelian, you can just pretend everything in sight commutes. So the answer is going to be ${\mathbb Z}[a_1, b_1, \ldots, a_g, b_g|0,...,0]$. Why $0$? Because that's what $[a_i, b_i]$ becomes, if everything commutes. The point is that the $a_i$ and $b_i$ commute as well, so $G^{\rm ab} = {\mathbb Z}[a_1, b_1, \ldots, a_g, b_g]$.

But you were asked to show that $G^{\rm ab} = {\mathbb Z}\langle a_1, b_1, \ldots, a_g, b_g\rangle$. In my opinion, this is a poor notation, and your comment about $\mathbb Z = \langle x\rangle$ so surely ${\mathbb Z}\langle a_1, b_1, \ldots, a_g, b_g\rangle = \langle x, a_1, b_1, \ldots, a_g, b_g\rangle$ demonstrates why this is a poor notation. In fact, it looks like a completely different kind of object to me (a group ring). In any case, what it means is ${\mathbb Z}[a_1, b_1, \ldots, a_g, b_g]$, but with the operation being written as $x_1 x_2$. In other words $\{a_1^{s_1} b_1^{t_1} \cdots a_n^{s_n} b_n^{t_n}\}$ for $s_i, t_i\in{\mathbb Z}$.


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