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]; 
  • $\begingroup$ Could you perhaps find some means of making your definition of the group in GAP available, to avoid the necessity of having to type it all in? You could either just include it in the post, so that it could be cut and pasted, or provide a link to it. Have you tried computing subgroups up to a given index, using the GAP command LowIndexSubgroupsFpGroup? $\endgroup$
    – Derek Holt
    Apr 18 '18 at 11:00
  • $\begingroup$ @DerekHolt: I added in the body of the question the part of the Gap script defining the group . LowIndexSubgroupsFpGroup is very slow for this group, at least on my computer. $\endgroup$ Apr 18 '18 at 11:38
  • $\begingroup$ Have you tried the finite simple group of order 168? $\endgroup$
    – Will Sawin
    Apr 18 '18 at 13:20
  • $\begingroup$ I tried that. In fact there are no nonabelian simple quotients of order at most $10000$. $\endgroup$
    – Derek Holt
    Apr 18 '18 at 13:27
  • $\begingroup$ Both the anwers that I received are really useful and informative. They are somewhat complementary (Derek's is more computational, Will's more theoretical), and I definitely like both of them. Having to accept one, I choose Derek's just because he answered first. $\endgroup$ Apr 20 '18 at 9:07

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);
gap> Order(Image(phi, Br.5));

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 $3^5$.

  • $\begingroup$ Thank you for your nice answer, I will think about it. Is it straightforward to obtain from this an example of $G$ having even order? $\endgroup$ Apr 18 '18 at 12:36
  • 2
    $\begingroup$ You group $B_2(C_2)$ has a quotient $G/N$ of order $2$ not containing $\sigma$. So, if my quotient of order $3^5$ is $G/M$, then $G/(M \cap N)$ has order $2.3^5$. $\endgroup$
    – Derek Holt
    Apr 18 '18 at 12:48
  • 2
    $\begingroup$ Also, by running $\mathtt{pQuotient(Br,2,3)}$ and factoring out part of the centre you can get a quotient of order $2^{14}$ in which the image of $\sigma$ has order $4$. $\endgroup$
    – Derek Holt
    Apr 18 '18 at 12:52
  • $\begingroup$ ok, thank you again! Just a last question: are you conjecturing that $3^5$ is the minimum possible order for $G$? $\endgroup$ Apr 18 '18 at 12:54
  • $\begingroup$ Yes, I conjecture that $3^5$ is the order of the smallest quotient in which the order of the image of $\sigma$ is at least $3$. It might be difficult to prove this. I expect order $81$ could be ruled out using the same method as you if you waited long enough, but $128$ could be tricky. $\endgroup$
    – Derek Holt
    Apr 18 '18 at 13:16

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.

  • 1
    $\begingroup$ Thank you for your nice answer. Just to understand: in the image of $\pi_1(C_2 \times C_2-\Delta) \to \pi_1(X- \delta)$ there is no $\sigma$, but rather $\sigma^2$. Are you claming that your construction provide a group $G$ such that the image of $\sigma^2$ has order $p$? $\endgroup$ Apr 18 '18 at 12:39
  • 1
    $\begingroup$ @FrancescoPolizzi Yes, that's what I am claiming. $\endgroup$
    – Will Sawin
    Apr 18 '18 at 12:42
  • $\begingroup$ Your "associated extension" should be $$1 \to \mathbb{Z}/p \to G \to H_1(C_2 \times C_2, \, \mathbb{Z}/p) \to 1,$$ right? $\endgroup$ May 3 '18 at 13:59
  • 1
    $\begingroup$ @FrancescoPolizzi That is what I meant, sorry. $\endgroup$
    – Will Sawin
    May 3 '18 at 14:50
  • $\begingroup$ Working on the construction that you suggest, I have considered the cup-product map $$\wedge H^1(C_2 \times C_2, \, \mathbb{Z}/p) = \wedge H^1(C_2 \times C_2 - \Delta, \, \mathbb{Z}/p) \longrightarrow H^2(C_2 \times C_2-\Delta, \, \mathbb{Z}/p).$$ Since the left-hand side has dimension 28 and the right-hand side has dimension 17, there is a kernel $K$ that has dimension $\geq 11$. For every element of this kernel, your argument shows that the image of the obstruction class vanishes...[continue] $\endgroup$ Jun 6 '18 at 8:48

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