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Let $K$ be the group generated by four elements $x_1,\cdots,x_4$ with relations that each generator commutes with all its conjugates. (An equivalent relation is, any simple commutator with repeated generator is trivial; for example, $[[x_2,[x_1,x_3]],x_3]=1$.) It can be proved that $K$ is finitely presented.

Let $A$ be the subgroup of $K$ generated by the following elements:

- $x_1x_2x_3x_4$;
- $[x_1,x_2][x_1,x_3][x_1,x_4][x_2,x_3][x_2,x_4][x_3,x_4]$;
- $[x_1,x_2,x_3][x_1,x_2,x_4][x_1,x_3,x_4][x_2,x_3,x_4]$;
- $[x_1,x_3,x_2][x_1,x_4,x_2][x_1,x_4,x_3][x_2,x_4,x_3]$;
- $[x_1,x_2,x_3,x_4]$;
- $[x_1,x_2,x_4,x_3]$;
- $[x_1,x_3,x_2,x_4]$;
- $[x_1,x_3,x_4,x_2]$;
- $[x_1,x_4,x_2,x_3]$;
- $[x_1,x_4,x_3,x_2]$.

Now I would like to compute the abelianization of $A$; namely, the group $A/[A,A]$, where $[A,A]$ is the commutator subgroup of $A$. My question is:

- Group Operations/ Group Actions
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**Is there any way that I can use GAP to find the abelianization of $A$?**

As I never used GAP before, I am not familiar with how to input my question into GAP. A note of my notation: in my question $[a,b]=aba^{-1}b^{-1}$ and $[a,b,c,d]$ means $[[[a,b],c],d]$. If it is hopeless to apply GAP to my problem, is there any way that I can attack my problem by hand?

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