# Why is $\pi_1(F_g)^{ab} = \Bbb Z \langle a_1, b_1, \ldots , a_g, b_g \rangle$?

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 $\pi_1(F_g) / [\pi_1(F_g), \pi_1(F_g)] = \Bbb Z \langle a_1, b_1, \ldots , a_g, b_g \rangle$.

• 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 $a_i, b_i$. Dec 17, 2015 at 5:18

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

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

to

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

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

• 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 $[a_i, a_j] = [b_i, b_j] = [a_i, b_j] = e\ \forall\ i, j$. Dec 17, 2015 at 5:35
• Do you know what $\mathbb{Z}\langle a_1, b_1, \dots, a_g, b_g\rangle$ means? Dec 17, 2015 at 5:36
• So if $\Bbb Z = \langle x \rangle$, then $\Bbb Z \langle a_1, b_1, \ldots , a_g, b_g \rangle = \langle xa_1, xa_2, \ldots , xa_g, xb_g \rangle$? Dec 17, 2015 at 5:38
• No, that doesn't really make sense. I have added a few comments about the notation. Hopefully it makes it clear. 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 $$s_i, t_i\in{\mathbb Z}$$.