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The braid group $B_n$ and the pure braid group $P_n$ sits in a short exact sequence $$1\to P_n\to B_n\to S_n\to 1.$$ The pure braid group $P_n$ has abelianization $\mathbb Z^{n\choose 2}$, with isomorphism (I believe) given by $P_n\to \mathbb Z^{n\choose2}$, where for any subset $\{i,j\}\subset\{1,\dots,n\}$, one obtains a map $P_n\to P_{\{i,j\}}$ which simply forgets all other strands, and pure braid group $P_{\{i,j\}}$ on two strands is isomorphic to $\mathbb Z$.

What is the quotient $B_n/[P_n,P_n]$? It sits in a short exact sequence $$1\to\mathbb Z^{n\choose 2}\to B_n/[P_n,P_n]\to S_n\to 1,$$which is classified by a class in $H^2(S_n,\mathbb Z^{n\choose2})$; what is it?

I believe it is not split (i.e., the cohomology class is nontrivial), since if the exact sequence split there would be a homomorphism $B_n\twoheadrightarrow \mathbb Z^{n\choose2}\rtimes S_n\to\mathbb Z$ which sends $(a_{ij})\in\mathbb Z^{n\choose2}$ to $\sum_{i<j}a_{ij}$. But this would be a homomorphism $B_n\to \mathbb Z$ which is surjective upon restricting to $P_n$, which cannot exist.

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Let $G_n \subseteq S_n \ltimes \frac{1}{2}\mathbb Z^{\oplus{ n \choose 2}}$ be the subgroup consisting of $(\sigma, (\frac{a_{ij}}{2})_{i<j})$ where $a_{ij} \in \mathbb Z$ satisfies the condition that $a_{ij}$ is odd if and only if $\sigma$ interchanges the order of $i,j$.

There is a homomorphism $B_n \to G_n$ which takes a braid $b$ to its associated permutation and its tuple of "half-winding numbers:" the standard generator $s_i$ maps to $((i, i+1), (c_{ij})_{i<j})$ where $c_{i,i+1} = 1/2$ and all other entries are zero. This map is surjective, and the kernel is contained in $PB_{n}$, is equal to the joint kernel of all of the pairwise winding number homomorphisms, hence is $[PB_n, PB_n]$.

You can directly see the non-split extension $\mathbb Z^{{n \choose 2}} \to G_n \to S_n$, given by projection.

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  • $\begingroup$ Thank you for your answer! I was looking for such an explicit description! $\endgroup$ Commented Jan 15, 2025 at 1:07
  • $\begingroup$ Actually I'm not convinced $G_n$ is a subgroup; when $n=3$ we have $((12),(\frac12,0,0))\cdot((23),(0,\frac12,0))=((132),(\frac12,\frac12,0))$, which is no longer in $G_3$ since $(132)$ preserves the order of $1$ and $2$. $\endgroup$ Commented Jan 15, 2025 at 1:44
  • $\begingroup$ Maybe you meant the subgroup of $(\sigma,(a_{ij}/2))$ such that $\mathrm{sgn}(\sigma)=(-1)^{\sum a_{ij}}$? $\endgroup$ Commented Jan 15, 2025 at 3:28
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    $\begingroup$ @KentaSuzuki: I’m not sure you’re multiplying correctly in the semi-direct product. It appears that you are parameterizing the second coordinate by triples (a,b,c) where a is the coordinate of the set 12, b is the coordinate of the set 23, and c is the coordinate of the set 13. Remember that the transposition (2,3) takes the set 12 to the set 13, so your product should be ((132),(0,1/2,1/2)). $\endgroup$ Commented Jan 15, 2025 at 3:46
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This extension and other related extensions are studied extensively in Quotients of the braid group that are extensions of the symmetric group by Day and Nakamura, which I believe should answer all your questions.

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  • $\begingroup$ Thank you! It seems the reference is Section 4 (particularly Theorem 4.10) of Nakamura's paper The Cohomology of the Mod 4 Braid Group. $\endgroup$ Commented Jan 15, 2025 at 0:52
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    $\begingroup$ @KentaSuzuki: Happy to help! Day is a long-time collaborator of mine and explained this work with his student Nakamura to me when he visited last year, so I was working off of memory rather than a detailed read of the paper itself. $\endgroup$ Commented Jan 15, 2025 at 3:20
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I convinced myself Phil's group $G_n$ really is a subgroup (thank you Andy for pointing out an error in my original computation), and this is a bit too long for a comment so I will prove this here.

For $i<j$ let $$\chi_{ij}(\sigma):=\begin{cases} 1&\text{if }\sigma(i)>\sigma(j)\\ 0&\text{if }\sigma(i)<\sigma(j).\end{cases}$$ There is an injection $$S_n\hookrightarrow \mathbb F_2^{n\choose2}\rtimes S_n$$defined by $\sigma\mapsto (\chi_{ij}(\sigma))_{i<j},\sigma)$. Indeed, to prove it is a group homomorphism, it suffices to check the cocycle condition $$\chi_{ij}(\sigma\tau)=\chi_{ij}(\tau)+\chi_{\tau i,\tau j}(\sigma)\in\mathbb F_2.$$ For this note that $\chi_{ij}(\sigma\tau)-\chi_{ij}(\tau)$ tracks whether $(\tau(i),\tau(j))$ and $(\sigma\tau(i),\sigma\tau(j))$ are in different orders. But this is exactly what $\chi_{\tau i,\tau j}(\sigma)$ does.

Now there is an obvious surjection $$\frac12\mathbb Z^{n\choose2}\rtimes S_n\to \mathbb F_2^{n\choose2}\rtimes S_n$$ and $G_n$ is the pullback of the subgroup $S_n$.

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