$\DeclareMathOperator\PSL{PSL}$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:



$h_1h_2h_4h_3=h_3h_4h_1h_2$, 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 $(2, 3, 7; 9)$.

  • 1
    $\begingroup$ Looks like a problem for KBMAG: homepages.warwick.ac.uk/~mareg/download/kbmag2/kbmag.README $\endgroup$
    – Sam Nead
    Aug 26, 2022 at 13:02
  • $\begingroup$ I doubt the question “what is this group” has a better answer than the description you have already given. Can you focus your question? Are there some properties you would like to know about this group? (By the way, the “cyclic permutations” of your relations are redundant, since relations always imply their conjugated.) $\endgroup$
    – HJRW
    Aug 26, 2022 at 13:40
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    $\begingroup$ @HJRW I think OP means cyclically permitting the generators, not the words themselves. $\endgroup$ Aug 26, 2022 at 13:44
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    $\begingroup$ @HJRW They state that $H$ is normal in $G$ with $G_2\leq G$, so normality in $G_2$ is surely immediate? (Possibly they actually constructed a normal chain $H\lhd G_2\lhd G_1\lhd G$, as Reidemeister-Schreier is easier to handle when we have normality.) $\endgroup$
    – ADL
    Aug 26, 2022 at 15:32
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    $\begingroup$ It wasn't a normal chain, but it turns out you can use the coset diagram for $G_1 \leq G$ to find the appropriate coset diagram for $G_2 \leq G_1$, and then for $H$. The calculations were somewhat tricky, but not impossibly hard with a little patience. $H$ is indeed normal in $G$, which aided the computations significantly, and accounted for the cyclic structure of the relators. $\endgroup$
    – Thomas
    Aug 26, 2022 at 20:21

1 Answer 1


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);
> AQInvariants(K);
[ 0, 0, 0, 0, 0, 0, 0 ]
> NilpotencyClass(pQuotient(K,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) }
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    $\begingroup$ Actually, I did these computations by hand. There were some tricky terms in the middle, but fortunately the calculations didn't completely blow up, and since H is normal, the calculations simplified at the end (and created the cyclic permutation structure). I have access to the Magma online calculator, so if you have code in that to verify it, that would help. $\endgroup$
    – Thomas
    Aug 26, 2022 at 20:12
  • $\begingroup$ If the grammar question was in earnest, the age-old "try the sentence without it" trick works here: "$H$ is virtually abelian", so "$H$ and hence also $G$ is virtually abelian", because "and hence also $G$" is a parenthetical clause even though not formally set off as one (I would use commas, but I use more commas than most). $\endgroup$
    – LSpice
    Aug 26, 2022 at 22:09
  • $\begingroup$ Thank you! So we can say that $(2,3,7;9)$ has structure: $PSL(2,8).{\mathbb Z}^7.{\mathbb Z}/2{\mathbb Z}$ right? Does this mean there is a way to faithfully represent $(2,3,7;9)$ with $7$x$7$ matrices over some field? I'd be very keen to see a construction of such a representation, with explicit matrices for $a$ and $b$. $\endgroup$
    – Thomas
    Aug 27, 2022 at 1:26
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    $\begingroup$ @LSpice Thanks! I thought "is" sounded better, but I was momentarily unsure. $\endgroup$
    – Derek Holt
    Aug 27, 2022 at 8:21
  • $\begingroup$ @Thomas Yes, that is the structure, and there must be faithful representations, but I doubt whether it can be done in degree $7$. The quotient ${\rm PSL}(2,8).{\mathbb Z}^7$ certainly lies in ${\rm SL}(8,{\mathbb Z})$, but I would have to think about how to get that $2$ factor underneath. $\endgroup$
    – Derek Holt
    Aug 27, 2022 at 8:24

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