# Show the abelianization of braid broup $B_n$,$n\geq 2$ is isomorphic to $\mathbb{Z}$

I need to show that the abelianization of braid group $$B_n$$, for $$n\geq2$$ is isomorphic to $$\mathbb{Z}$$, and that the commutator subgroup $$[B_n,B_n]$$ is exactly the set of braids represented by words with total exponent sum zero in the generators $$\sigma_{i}$$. I was able to show all generators $$\sigma_i$$ are conjugate to each other.

Here is the question:

The fact that all generators are conjugate (which you have shown) implies that all generators have the same image in $$G/[G,G]$$ (where $$G=B_n$$), so $$G/[G,G]$$ is cyclic and generated by any one of the generators.
On the other hand, the map $$\phi$$ sending every generator $$\sigma_i$$ to $$1 \in {\mathbb Z}$$ induces a surjective homomorphism, so $$G/[G,G]$$ is infinite, and hence infinite cyclic.
Note that $$\ker \phi = [G,G]$$ is the set of elements with total exponent sum $$0$$ in the generators.
• Because $G/[G,G]$ is abelian, and two conjugate elements in an abelian group are equal. Commented Nov 11, 2018 at 19:22
• Also, Now that I think more about it, I see why $ker(\phi)$ consists of elements with total exponent sum 0 in the generators but I still don't understand why $ker(\phi) =[G,G]$. Commented Nov 14, 2018 at 11:02
• $G/K$ infinite cyclic implies $[G,G] \le K$, but $G/[G,G]$ is cyclic, and all proper quotients of cyclic groups are finite, so $[G,G]=K$. Commented Nov 14, 2018 at 17:09
Given a group by a presentation: $$G=\langle S\mid R\rangle,$$ its abelianization is $$G/[G,G]=\langle S\mid T,R\rangle,$$ where $$T$$ is the list of commutation relations between all generators: $$T=\{[u,v]\mid u,v\in S\}.$$ As a particular case, after some easy simplifications: $$B_n/[B_n,B_n]=\langle\sigma_i(1\le i $$=\langle\sigma\mid~\rangle=\Bbb Z.$$