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];
```

veryslow for this group, at least on my computer. $\endgroup$ – Francesco Polizzi Apr 18 '18 at 11:38