This question is somehow related to my previous MO question [Explicit description of a subgroup of the braid group $\mathsf{B}\_2(C\_2)$](https://mathoverflow.net/questions/254151/explicit-description-of-a-subgroup-of-the-braid-group-mathsfb-2c-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]; 

```


Epimorphisms from the genus $2$ surface braid group to finite groups
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 $3^5$.


