A curious group presentation $\DeclareMathOperator\PSL{PSL}$I'm studying the Hurwitz group $(2, 3, 7; 9)$, with presentation: $G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^9 \rangle$. This group has $\PSL_2(8)$ as a quotient, so I decided to try the Reidemeister-Schreier process to find the presentation of the normal subgroup $H$ with $G/H=\PSL_2(8)$. I did this in three phases, using the permutation representation of $\PSL_2(8)$ to first find an index 9 subgroup $G_1$ of $G$, which led to a clear choice of an index 8 subgroup $G_2$ in $G_1$. Finally $H$ is an index 7 subgroup of $G_2$.
What I ended up with was a very strange presentation that I need help understanding:
$H = \langle h_1, h_2, h_3, h_4, h_5, h_6, h_7\rangle$, with relations:
$h_1h_2h_6h_7h_4h_5h_2h_3h_7h_1h_5h_6h_3h_4=1$
$h_1h_3h_4h_6=h_3h_4h_6h_1$
$h_1h_2h_4h_3=h_3h_4h_1h_2$, and
$h_3h_4h_6h_7 = h_1h_3$
Along with their cyclic permutations ($h_1\to h_2\to\dots\to h_7\to h_1$).
What is this group? The generators seem to very nearly commute, but I'm not sure they quite do. I'd like to understand exactly what this group is, if there are simpler defining relations for it, and exactly how $\PSL_2(8)$ acts on it in $(2, 3, 7; 9)$.
 A: Yes you are right, the subgroup $H$ is very nearly abelian but not quite!
In fact it is nilpotent of class $2$ with centre of order $2$, and $H/Z(H) \cong {\mathbb Z}^7$. So in fact $H$ and hence also $G$ is (are?) virtually abelian.
Are you using GAP or Magma to do your computations? If you let me know, then I can add some code to verify  the above claims.
Here is Magma code to verify the claims. Note that the Magma notation for the commutator of $g$ and $h$ is $(g,h)$, not $[g,h]$.
> G := Group< a,b | a^2, b^3, (a*b)^7, (a,b)^9 >;
> h := Homomorphisms(G, PSL(2,8));
> K := Kernel(h[1]);
> K := Rewrite(G,K);
> Ngens(K);
7
> AQInvariants(K);
[ 0, 0, 0, 0, 0, 0, 0 ]
> NilpotencyClass(pQuotient(K,2,2));
2

So $K$ is nonabelian and $K/[K,K] \cong {\mathbb Z}^7$.
We now run the Knuth-Bendix algorithm on the presentation of $K$. This doesn't complete, but all identities among words that it derives are guaranteed to be correct. We verify (a bit inefficiently) that $K$ is nilpotent of class (at most) $2$ and that that $[K,K]$ is cyclic of order (at most) $2$.
> RK := RWSGroup(K);
Warning: Knuth Bendix only partly succeeded
> { (RK.i, RK.j) : i in [1..7], j in [1..7] };
{ RK.1 * RK.3 * RK.1^-1 * RK.3^-1, Id(RK) }
> { (RK.i, RK.j)^2 : i in [1..7], j in [1..7] };
{ Id(RK) }
> { ((RK.i, RK.j),RK.k) : i in [1..7], j in       [1..7], k in [1..7] };
{ Id(RK) }

