$\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\PSp{PSp}$The braid groups $ B\_1=1 $ and $ B\_2\cong \mathbb{Z} $ have no perfect quotients.


$ B\_3 $ has quotients $ \SL(2,\mathbb{Z}) $ and $ \PSL(2,\mathbb{Z})\cong B\_3/Z(B\_3) $. As a result, $ B\_3 $ has infinitely many perfect quotients $ \SL(2,p) $ and $ \PSL(2,p) $ (of course $ p \neq 2,3 $ for it to be perfect). Indeed since $ B\_3/Z(B\_3) \cong \PSL(2,\mathbb{Z}) \cong C\_2 \* C\_3 $ then every $ (2,3) $ generated group is a quotient of $ B\_3 $. This includes all except for a few of the infinite families of finite simple groups see [Is every finite simple group a quotient of a braid group?](https://math.stackexchange.com/questions/4726854/is-every-finite-simple-group-a-quotient-of-a-braid-group) and
[On $(2,3)$-generation of finite simple classical groups](https://mathoverflow.net/questions/365374/on-2-3-generation-of-finite-simple-classical-groups?_gl=)


Does this hold for all $ n\geq 3 $? In other words, do all braid groups $ B\_n, n \geq 3 $ have infinitely many non-isomorphic (finite) perfect quotients?


Update: Here is an update based on the accepted answer from Ian Agol. Considered the reduced Burau representation <https://en.wikipedia.org/wiki/Burau_representation> evaluated at $ t=-1 $
$$
\rho: B\_n \to U(n-1)
$$
Note that this representation is not continuous i.e. the image is not a closed subgroup of $ U(n-1) $. The image of $ \rho $ turns out to be a finite index subgroup of $ \Sp(2m,\mathbb{Z}) $ by some topological/geometric argument I don't quite understand about preserving a symplectic form on the homology. Also the way that $ m $ is related to $ n $ is somehow topological. I guess $ m $ is the rank of the first homology $ H\_1 $ of some hyperelliptic curve. Again not exactly sure what's going on with the topology/geometry/algebraic curve stuff.


An example of this phenomenon where there is a quotient of $ B\_n $ which is a (finite index subgroup of) $ \Sp(2m,\mathbb{Z}) $ can already be seen for the first non-trivial case $ n=3 $. For $ B\_3 $, there is a central quotient of $ B\_3 $ isomorphic to $ \SL(2,\mathbb{Z}) \cong \Sp(2,\mathbb{Z}) $.


So for $ n=3, B\_3 $ we have $ m=1 $ and the index of $ \rho(B\_n) $ in $ \Sp(2,\mathbb{Z}) $ is just $ 1 $.


For higher values of $ n $ the values of $ m $ increase strictly monotonically with $ n $ (at least I think so, again there is some topology I don't quite understand here). And for higher values of $ n $ the index $ [\Sp(2m,\mathbb{Z}): \rho(B\_n)] $ also form a strictly increasing sequence of integers, an explicit form for the sequence is given in a corollary to theorem 1 in the paper "Tresses, monodromie et le groupe symplectique."


Now that you have a quotient of $ B\_n $ which is (finite index) subgroup of $ \Sp(2m,\mathbb{Z}) $ you can take congruence subgroups modulo $ p $ prime. Since $ p $ is prime then the image of $ \rho(B\_n) $ in the simple group $ \Sp(2m,p) $ must be either everything or trivial. Direct inspection of the matrices for the reduced Burau representation <https://en.wikipedia.org/wiki/Burau_representation> reveals that they are not congruent to the identity mod any $ p $ (there is a $ 1 $ off the diagonal). Thus the image is all of $ \Sp(2m,p) $.



Perfect quotients of braid groups
A’Campo showed that the finite symplectic group $Sp(2m,p)$, $p>2$ prime, is a quotient of the braid group $B\_n$ for some $m$ depending on $n$. Hence the finite groups $PSp(2m,p)$ are quotients of $B\_n$. [These groups are simple](https://groupprops.subwiki.org/wiki/Projective_symplectic_group_is_simple) non-abelian hence perfect for most $m, p$.


*A’Campo, Norbert*, [**Tresses, monodromie et le groupe symplectique**](https://doi.org/10.1007/BF02566275), Comment. Math. Helv. 54, 318-327 (1979). [ZBL0441.32004](https://zbmath.org/?q=an:0441.32004) [MR0535062](https://mathscinet.ams.org/mathscinet/article?mr=0535062)


