# Can a computer algebra system help me with this group theory problem?

I have a problem in group theory which gave me the impression to be tractable by computer algebra methods. I have never used computer algebra systems and started reading into GAP. Now after several days I am thoroughly confused by specific questions on what the GAP commands actually do.

Before investing more time in learning this, I would like to know from more experienced computer algebra users, whether a computer algebra system can in principle help me with this problem at all.

Probably many of the details of the problem are irrelevant, but I cannot judge which ones, so I will expose the whole problem.

Let m,a,f be natural numbers, m even, a > 1, f odd and congruent 1 modulo $2^{2a-1}$ (these restrictions are probably not so important). We have a group G given by generators $x_1, ... , x_{m+2}$ and the single relation $x_1^{2+2a}[x_1,x_2][x_3,x_4][x_5,x_6]...[x_{m+1},x_{m+2}]=1$. (the brackets denote commutators) Now we look at the normal subgroup N generated by $x_1, x_2, x_3^f, x_4, ..., x_{m+2}$. It admits a presentation as a group (not as a normal subgroup) with generators $x_3^f$ and the conjugates of $x_1, x_2, x_4, ..., x_{m+2}$ by $x_3^i$ $(i=0...f-1)$ and the $f$ relations given by conjugating the above single relation with $x_3^i$ $(i=0...f-1)$ (Since the commutator [$x_3$, $x_4$] occurs in there, and $x_3$ is not an element of the subgroup, it is not immediately clear that these are relations between elements of the subgroup, but it is true). We can further pass to the quotient $N / [[N,N],N]$ and get an induced presentation of this.

Now one can show that $N / [[N,N],N]$, also admits a presentation by generators $y_0, ..., y_{f-1}, z_0, ..., z_{f-1}, x_3^f, w, u_{i,j} (0 \leq i \leq f-1, 5 \leq j \leq m+2)$ subject to the single relation

$y_{f-1}^{2+2a} \cdot [y_{f-1}, z_{f-1}] \cdot [x_3^f, w] \cdot \Pi_{v=0}^{f-2}[y_v, z_{v-1}^{-1} z_v y_{v+1}^{(1+2^{a-1})(1+2^a)}] \cdot \Pi_{v=0}^{f-1}(\Pi_{r=3}^{m+2/2}[u_{v,2r-1}, u_{v,2r}]) = 1$

I need to find such $y_s,z_t,w,u_{i,j}$, expressed in terms of the original generators $x_1,...x_{m+2}$.

I have no idea how feasible it is to answer this with computer help. In Magma, to which I have no access, there seems to be a function searching for isomorphisms between given groups. I could maybe use this to look for an isomorphism between the groups given by the two presentations above. After having read enough of the GAP manual, I have the impression that GAP can not help me at all with this problem. Am I right? But maybe the problem is too complex for computer algebra at all? Or one needs the groups given in a specific way which I don't have? In my first steps in GAP I kept running into these kinds of problem and now would like to ask whether I can hope for a solution at all. Welcome answers would be: A "Yes" or "No", if "No", then a reason why. If "Yes" then an indication which computer algebra system could help me; if there is a freely available one, then this would be preferred. Of course I wouldn't mind an outline of how to go about, too :-)

Thanks a lot!

Summary of the smallest sensible case ($m=4, f=9, a=2$ in the above description):

I have a group $G$ with 38 generators $y_0,y_1,y_2,y_3,y_4,y_5,y_6,y_7,y_8,x,w,z_0,z_1,z_2,z_3,z_4,z_5,z_6,z_7,z_8, u_{0,5},u_{1,5},u_{2,5},u_{3,5},u_{4,5},u_{5,5},u_{6,5},u_{7,5},u_{8,5},u_{0,6},u_{1,6},u_{2,6},u_{3,6},u_{4,6},u_{5,6}, u_{6,6},u_{7,6},u_{8,6}$ subject to the single relation

$y_{8}^{6} \cdot [y_{8}, z_{8}] \cdot [x, w] \cdot \Pi_{v=0}^{7}[y_v, z_{v+1}^{-1} z_v y_{v+1}^{15}] \cdot \Pi_{v=0}^{8}[u_{v,5}, u_{v,6}] = 1$.

Now I set $N:=G/G^{2^6}[G^{2^5}[G,G],G]$. (I found functions in GAP and Magma doing this quotient operation)

I have a second group $H$ with 46 generators $x_{1,0},x_{1,1},x_{1,2},x_{1,3},x_{1,4},x_{1,5},x_{1,6},x_{1,7},x_{1,8}, x_{2,0},x_{2,1},x_{2,2},x_{2,3},x_{2,4},x_{2,5},x_{2,6},x_{2,7},x_{2,8}, x_{3}, x_{4,0},x_{4,1},x_{4,2},x_{4,3},x_{4,4},x_{4,5},x_{4,6},x_{4,7},x_{4,8}, x_{5,0},x_{5,1},x_{5,2},x_{5,3},x_{5,4},x_{5,5},x_{5,6},x_{5,7},x_{5,8}, x_{6,0},x_{6,1},x_{6,2},x_{6,3},x_{6,4},x_{6,5},x_{6,6},x_{6,7},x_{6,8}$ subject to the 9 relations $x_{1,0}^6*[x_{1,0},x_{2,0}]*x_{4,1}^{(-1)}*x_{4,0}*[x_{5,0},x_{6,0}],\\ x_{1,1}^6*[x_{1,1},x_{2,1}]*x_{4,2}^{(-1)}*x_{4,1}*[x_{5,1},x_{6,1}],\\ x_{1,2}^6*[x_{1,2},x_{2,2}]*x_{4,3}^{(-1)}*x_{4,2}*[x_{5,2},x_{6,2}],\\ x_{1,3}^6*[x_{1,3},x_{2,3}]*x_{4,4}^{(-1)}*x_{4,3}*[x_{5,3},x_{6,3}],\\ x_{1,4}^6*[x_{1,4},x_{2,4}]*x_{4,5}^{(-1)}*x_{4,4}*[x_{5,4},x_{6,4}],\\ x_{1,5}^6*[x_{1,5},x_{2,5}]*x_{4,6}^{(-1)}*x_{4,5}*[x_{5,5},x_{6,5}],\\ x_{1,6}^6*[x_{1,6},x_{2,6}]*x_{4,7}^{(-1)}*x_{4,6}*[x_{5,6},x_{6,6}],\\ x_{1,7}^6*[x_{1,7},x_{2,7}]*x_{4,8}^{(-1)}*x_{4,7}*[x_{5,7},x_{6,7}],\\ x_{1,8}^6*[x_{1,8},x_{2,8}]*x_{3}^{(-1)}*x_{4,0}^{(-1)}*x_{3}*x_{4,8}*[x_{5,8},x_{6,8}]$

Then I set $M:=H/H^{2^6}[H^{2^5}[H,H],H]$

I know that $N$ is isomorphic to $M$ and I would like to find the isomorphism. It seems Magma can do it in principle, thanks to the function SearchForIsomorphism, but the free online version runs out of memory.

Questions:

1. Is this problem too complex for a computer algebra system running on a standard home computer?

2. Is there a free system which would allow me to do this?

• Sage is free and it has some support from group theory, apparently including groups with an arbitrary finite presentation. – Jair Taylor Oct 1 '16 at 23:03
• You can also try magma at magma.maths.usyd.edu.au/calc – verret Oct 2 '16 at 4:53
• @JairTaylor: Yes, the same is true of GAP, but after having read a lot of the manual I concluded that it cannot help me with my particular problem. I spent days finding this out and would not want to do the same with Sage... – Embarassed Guy Oct 2 '16 at 10:05
• I think the group theory in Sage consists of an interface to GAP, so it would not offer any advantages over GAP. – Derek Holt Oct 2 '16 at 10:54
• I am finding it too much effort to read through this post. Could you isolate the smallest or simplest instance of your problem, and state is perhaps as a single specific problem? – Derek Holt Oct 2 '16 at 10:56