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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.

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2 Answers 2

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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.

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  • $\begingroup$ Sorry, why all generators are conjugate to each other implies all generators have the same image in G/[G,G]? Can you elaborate? $\endgroup$
    – user614287
    Commented Nov 11, 2018 at 17:34
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    $\begingroup$ Because $G/[G,G]$ is abelian, and two conjugate elements in an abelian group are equal. $\endgroup$
    – Derek Holt
    Commented Nov 11, 2018 at 19:22
  • $\begingroup$ 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 $ker(\phi) =[G,G]$. $\endgroup$
    – user614287
    Commented Nov 14, 2018 at 11:02
  • $\begingroup$ $G/K$ infinite cyclic implies $[G,G] \le K$, but $G/[G,G]$ is cyclic, and all proper quotients of cyclic groups are finite, so $[G,G]=K$. $\endgroup$
    – Derek Holt
    Commented Nov 14, 2018 at 17:09
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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: $$B_n/[B_n,B_n]=\langle\sigma_i(1\le i<n)\mid[\sigma_i,\sigma_j](1\le i<j<n), \sigma_i=\sigma_{i+1} (1\le i<n-1)\rangle$$ $$=\langle\sigma\mid~\rangle=\Bbb Z.$$

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