Proving $B_n/[B_n,B_n]$ is infinite (cyclic) group.

Question

Let $B_n$ be the braid group. I have shown that all generators of $B_n$ are conjugate, thereby proving that the surjective map $f:B_n \rightarrow B_n/[B_n,B_n]$ takes all generators of $B_n$ to a single generator, say $xH$ where $H$ is the commutator subgroup $[B_n,B_n]$ and $x$ is in $B_n$, which shows that $B_n/H$ is cyclic. Now, how do I prove that it is infinite? I think we'll have to use surjectivity of $f$ now, but how?

Answer 1

Remember the function $\ell$ from last time? It is defined by taking the number of positive exponents and subtracting the number of negative exponents. To show that the abelianization is infinite, it suffices to prove that $\sigma_1^n$ is not a product of commutators whenever $n > 1$. Note that every product of commutators $\beta \in [B_n, B_n]$ satisfies $\ell(\beta) = 0$. But $\ell(\sigma_1^n) = n$, which is non-zero. So $\sigma_1^n$ is not a product of commutators, whence the quotient has infinitely many elements.

Edit: As noted in the comments it is important to check that $\ell$ is well-defined. This can be done by observing that applying the braid relations or commutativity relations does not change the value of $\ell$. From there it is also easy to see that the abelianization is infinite: Just check that $\ell$ is a homomorphism into an infinite abelian group.