Epimorphisms from the genus $2$ surface braid group to finite groups

Question

This question is somehow related to my previous MO question Explicit description of a subgroup of the braid group $\mathsf{B}_2(C_2)$; for the reader convenience, let me write down again the relevant set-up.

Let $C_2$ be a smooth curve of genus $2$ and $X:=\mathrm{Sym}^2(C_2)$ its second symmetric product. If $\delta \subset X$ is the diagonal, then the topological fundamental group $\pi_1(X-\delta)$ is isomorphic to the braid group $\mathsf{B}_2(C_2)$ on two strands on $C_2$.

Such a group is generated by five elements $a_1, \, a_2, \, b_1, \, b_2, \, \sigma$ subject to the following set of relations:

\begin{equation*} \begin{split} (R2) \quad & \sigma^{-1} a_1 \sigma^{-1} a_1= a_1 \sigma^{-1} a_1 \sigma^{-1} \\ & \sigma^{-1} a_2 \sigma^{-1} a_2= a_2 \sigma^{-1} a_2 \sigma^{-1} \\ & \sigma^{-1} b_1 \sigma^{-1} b_1 = b_1 \sigma^{-1} b_1 \sigma^{-1} \\ & \sigma^{-1} b_2 \sigma^{-1} b_2 = b_2 \sigma^{-1} b_2 \sigma^{-1}\\ & \\ (R3) \quad & \sigma^{-1} a_1 \sigma a_2 = a_2 \sigma^{-1} a_1 \sigma \\ & \sigma^{-1} b_1 \sigma b_2 = b_2 \sigma^{-1} b_1 \sigma \\ & \sigma^{-1} a_1 \sigma b_2 = b_2 \sigma^{-1} a_1 \sigma \\ & \sigma^{-1} b_1 \sigma a_2 = a_2 \sigma^{-1} b_1 \sigma \\ & \\ (R4) \quad & \sigma^{-1} a_1 \sigma^{-1} b_1 = b_1 \sigma^{-1} a_1 \sigma \\ & \sigma^{-1} a_2 \sigma^{-1} b_2 = b_2 \sigma^{-1} a_2 \sigma \\ & \\ (TR) \quad & [a_1, \, b_1^{-1}] [a_2, \, b_2^{-1}]= \sigma^2. \end{split} \end{equation*} The geometric interpretation for the above generators of $\mathsf {B}_2(C_2)$ is the following. The $a_i$ and the $b_i$ are the braids coming from the representation of the topological surface associated with $C_2$ as a polygon of $8$ sides with the standard identification of the edges, whereas $\sigma$ is the classical braid generator on the disk. In terms of the isomorphism with $\pi_1(X-\delta)$, the element $\sigma$ corresponds to the homotopy class in $\textrm{Sym}^2(C_2)-\delta$ of a topological loop that "winds once around $\delta$". For more details see P. Bellingeri's paper

On presentations of surface braid groups, Journal of Algebra 274 (2004), 543-563.

For some research problems related to algebraic surfaces, I would like to construct some group epimorphism $$\varphi \colon \mathsf {B}_2(C_2) \longrightarrow G, \quad (\ast)$$ where $G$ is a finite group. I also want that the element $s :=\varphi(\sigma)$ is not the identity of $G$.

Making some experiments with GAP4, I discovered that, up to order $|G|=64$, if $\varphi$ exists than the order of $s$ is at most $2$. This was a surprise, since I do not see anything in the presentation of $\mathsf {B}_2(C_2)$ forcing this behaviour. So I wonder if this happens just because $64$ is too small or if, instead, I am missing some conceptual point here, maybe some basic result of combinatorial group theory I am not aware of.

Of course, the time requested for the machine computations grows rapidly with the order of $G$, hence it is not possible to systematically check all cases when such a order becomes too large (already, $64$ takes a lot of time). So, let me ask the following

Q. Is it possible to construct a group epimorphism of type $(*)$, such that the order of $s:=\varphi(\sigma)$ is at least $3$? If yes, what is the minimum order of $G$ such that this happens? If not, why?

Note. Here is the Gap4 script I used to define $\mathsf {B}_2(C_2)$:

F:=FreeGroup("a1", "b1", "a2", "b2", "s");
a1:=F.1;  b1:=F.2;  a2:=F.3;  b2:=F.4; s:=F.5;
 R1 := s^(-1)*a1*s^(-1)*a1*(a1*s^(-1)*a1*s^(-1))^(-1);
 R2 := s^(-1)*a2*s^(-1)*a2*(a2*s^(-1)*a2*s^(-1))^(-1);
 R3 := s^(-1)*b1*s^(-1)*b1*(b1*s^(-1)*b1*s^(-1))^(-1);
 R4 := s^(-1)*b2*s^(-1)*b2*(b2*s^(-1)*b2*s^(-1))^(-1);
 R5 := s^(-1)*a1*s*a2*(a2*s^(-1)*a1*s)^(-1);
 R6 := s^(-1)*b1*s*b2*(b2*s^(-1)*b1*s)^(-1);
 R7 := s^(-1)*a1*s*b2*(b2*s^(-1)*a1*s)^(-1);
 R8 := s^(-1)*b1*s*a2*(a2*s^(-1)*b1*s)^(-1);
 R9 := s^(-1)*a1*s^(-1)*b1*(b1*s^(-1)*a1*s)^(-1);
 R10 := s^(-1)*a2*s^(-1)*b2*(b2*s^(-1)*a2*s)^(-1);
 R11 := a1*b1^(-1)*a1^(-1)*b1*a2*b2^(-1)*a2^(-1)*b2*s^(-2);
Br:=F/[R1, R2, R3, R4, R5, R6, R7, R8, R9, R10, R11]; 

Answer 1

I have found a finite image of order $3^9$ in which $\sigma$ has nontrivial image, by using the $\mathtt{pQuotient}$ function which, for a given prime $p$ and class $n$, computes the largest quotient of the group which is a $p$-group of exponent $p$-class at most $n$.

gap> pq := PQuotient( Br, 3,2);
  <3-quotient system of 3-class 2 with 9 generators>
gap> phi := EpimorphismQuotientSystem(pq);
  [ a1, b1, a2, b2, s ] -> [ a1, a2, a3, a4, a5 ]
gap> Image(phi, Br.5);
  a5
gap> Order(Image(phi, Br.5));
  3

In fact this quotient has elementary abelian centre of order $3^5$, so there are certainly solutions to your problem of size $3^5$, and perhaps smaller. But factoring out a normal subgroup of order $3^4$ that lies in the centre and does not contain the image of $\sigma$ gives an extraspecial group of order $3^5$.

Answer 2

Consider $C_2 \times C_2$ minus the diagonal, this is a double cover of your space. Given a symplectic form on $H_1( C_2 \times C_2 - \Delta, \mathbb Z/p)$, we can form an associated extension $1 \to G \to H_1( C_2 \times C_2, \mathbb Z/p)$ as a Heisenberg group, which will be a quotient of $\pi_1$ if and only if the image of the obstruction class in $H^2 ( \pi_1(C_2 \times C_2 - \Delta), \mathbb Z/p) = H^2 ( C_2 \times C_2 - \Delta, \mathbb Z/p)$ vanishes. The obstruction class should be the class induced by viewing symplectic forms on on $H_1$ as elements of $\wedge^2 H^1$ and applying the cup product.

To ensure that $\sigma$ has order $p$ on $G$, it is sufficient to show that the obstruction class does not vanish in $H^2( C_2 \times C_2 - \Delta, \mathbb Z/p)$. In other words, we want the obstruction class to be a multiple of the class of $\Delta$. This is possible as every class in $H^2$ is a cup product.

Then because $C_2 \times C_2 - \Delta$ is a double cover of your $X - \delta$, it's fundamental group will map to the wreath product of $\mathbb Z/2$ with $G$, and probably something simpler like a semidirect product if we can choose the form to be invariant.

The minimal group size for this method to work is something like $2 \cdot 3^9$, which explains why it wasn't seen by your brute-force search.