The Gilbert-Howie group H(9,7) has 37 elements

The Gilbert-Howie group $H(9,7)$ is defined by a presentation

$$H(9,7) = \langle x_0,\ldots,x_8 \mid x_ix_{i+7}=x_{i+1}\quad \ (0 \leq i < 9) \rangle$$

with addition understood modulo $9.$ A recent publication¹ notes it is unknown whether $H(9,7)$ is hyperbolic.

In this note we will demonstrate that it is isomorphic to the cyclic group $C_{37}.$

Define the group $G$ to be a semidirect product $H(9,7) \rtimes C_9$ with the presentation

$$\langle x_0,\ldots,x_8, b\mid b^9=1,\ x_ix_{i+7} = x_{i+1} = bx_{i}b^{-1}, \quad(0 \leq i < 9) \rangle.$$

Using Tietze moves, we can simplify this presentation substantially.

First, eliminate $x_3$ since it can be expressed as $x_2x_0.$ Similarly, $x_4 =x_3x_1=x_2x_0x_1$ and so forth. We continue this procedure until our remaining generators are $b,x_0,x_1,x_2$ and the relations are

$$\begin{align*} bx_2x_0x_1x_2x_2x_0x_2x_0x_1x_2x_0x_1x_2b^{-1}&=x_0\\ bx_0b^{-1}&=x_1\\ bx_1b^{-1}&=x_2\\ bx_2b^{-1}&=x_2x_0\\ bx_2x_0b^{-1}&=x_2x_0x_1\\ bx_2x_0x_1b^{-1}&=x_2x_0x_1x_2\\ bx_2x_0x_1x_2b^{-1}&=x_2x_0x_1x_2x_2x_0\\ bx_2x_0x_1x_2x_2x_0b^{-1}&=x_2x_0x_1x_2x_2x_0x_2x_0x_1\\ bx_2x_0x_1x_2x_2x_0x_2x_0x_1b^{-1}&=x_2x_0x_1x_2x_2x_0x_2x_0x_1x_2x_0x_1x_2\\ \end{align*}$$

and

$$\begin{align*} x_2x_0x_1x_2x_2x_0x_2x_0x_1x_2x_0x_1x_2x_2x_0x_1x_2x_2x_0&=x_0\\ x_0x_2x_0x_1x_2x_2x_0x_2x_0x_1&=x_1\\ x_1x_2x_0x_1x_2x_2x_0x_2x_0x_1x_2x_0x_1x_2&=x_2\\ \end{align*}$$

Now, define the free group automorphism $\varphi : F_3 \to F_3$ as $x_0,x_1,x_2 \mapsto x_1, x_2, x_2x_0.$ It is easy to check that all of the $b$-conjugation relations are consequences of this definition, thus making most of these relation redundant.

Also note that the relation

$$bx_2x_0x_1x_2x_2x_0x_2x_0x_1x_2x_0x_1x_2b^{-1}=x_0$$

is equivalent to $\varphi^9(x_0) = x_0.$ which is redundant, given that $b^9 = 1.$ Thus, we keep only the following $b$-conjugation relations

$$\begin{align*} bx_0b^{-1}&=x_1\\ bx_1b^{-1}&=x_2\\ bx_2b^{-1}&=x_2x_0 \end{align*}$$

Now, we can cancel the redundant generators in the three relations:

$$\begin{align*} x_2x_0x_1x_2x_2x_0x_2x_0x_1x_2x_0x_1x_2x_2x_0x_1x_2x_2 && (r_3)\\ x_0x_2x_0x_1x_2x_2x_0x_2x_0 && (r_1)\\ x_1x_2x_0x_1x_2x_2x_0x_2x_0x_1x_2x_0x_1 && (r_2)\\ \end{align*}$$

Observe that $\varphi(r_1) = r_2,$ hence this relator is redundant. Observe also that $\varphi(r_2) = x_2r_3x_2^{-1},$ so this relator is redundant as well.

Finally, note that

$$\varphi^9(x_0) = x_2x_0x_1x_2x_2x_0x_2x_0x_1x_2x_0x_1x_2x_2x_0x_1x_2x_2x_0$$

and hence

$$x_2x_0x_1x_2x_2x_0x_2x_0x_1x_2x_0x_1x_2x_2x_0x_1x_2x_2=1.$$

But

$$\varphi^2(r_1)=x_2x_2x_0x_1x_2x_2x_0x_2x_0x_1x_2x_0x_1x_2x_2x_0x_1x_2$$

and thus

$$x_2^{-1} \varphi^2(r_1) x_2 =\varphi^9(x_0)x_0^{-1} = 1.$$

This proves $r_1$ to be redundant.

Finally, $x_1$ and $x_2$ can be eliminated, leaving us with just $b x_0 b^{-3} = x_0 b^{-2} x_0.$

Let $a=x_0b^{-3}.$ Using this relation, we eliminate $x_0.$ The resulting relator is $ba=ababbb.$

This leaves us a short presentation: $\langle a,b \mid b^9, ba=ababbb\rangle.$

Observe that $ba=ababbb$ can be put into the form $(ba)^{-1} a^{-1} (ba) = b^3.$ Thus, $b^9=1$ is equivalent to $a^3 = 1.$ This leaves us with our final short presentation.

$$\langle a,b \mid a^3=1, ba=ababbb\rangle$$

Now, Tomas Rokicki demonstrated, in a MathOverflow answer ² , that the group with the presentation

$$\langle a,b \mid a^3=1, \quad ab^{-3}a^{-1}b^{-1}a^{-1}b=1 \rangle$$

is finite and has exactly $333$ elements. Clearly, this group is identical to $G.$

We observe that $333 / 9 = 37$ and so $|H(9,7)| = 37.$ Thus, $H(9,7) \cong C_{37}$ as claimed.

$\blacksquare$

Notes

More on the group

We can easily verify that the $333$ element group has the small groups ID $[333,3],$ generated by, for example,

a = (2,11,27)(3,21,16)(4,31,5)(6,14,20)(7,24,9)(8,34,35)(10,17,13)(12,37,28)(15,30,32)(18,23,36)(19,33,25)(22,26,29)

b = (1,34,23,2,9,19,28,25,26)(3,21,15,17,4,33,11,6,20)(5,8,7,32,36,10,31,24,14)(12,18,16,29,37,22,27,13,30)

From these, we can see that the generator shifting action of $C_9$ on $H(9,7)$ is faithful. GroupNames page provides a nice overview of this group.

Software

Rokicki’s proof uses group theory software (kbmag, MAF) to find confluent and terminating rewrite system for this group. I was able to replicate this computation. This presentation (taken as a monoid presentation) was also discovered by Slava Pestov³. The page dedicated to this monoid includes a complete rewriting system and a certificate deriving the rewriting rules directly from the monoid presentation. As such, it constitutes an independent proof that this group is indeed finite, being an enveloping group of a finite monoid.

Attribution

All of the difficult work was done by Stefan Kohl ⁴ who asked about finiteness of this presentation of $G,$ and Tomas Rokicki⁴ who proved finiteness of $G.$ I am also deeply grateful to Slava Pestov, whose monoid word problem project led me to the discovery of Gilbert-Howie groups embedding into semidirect product groups that can have exceptionally short presentations. Analyzing multiple exceptional cases encountered in Pestov’s work, I was able to identify the pattern and discover the hidden significance of these peresentations. I am also thankful for him bringing the MathOverflow answer⁴ to my attention, and his indepependent verification of the finiteness of the monoid cousin of $G.$

Footnotes

1. Chinyere, Ihechukwu and Williams, Gerald (2021) Hyperbolic groups of Fibonacci type and T(5) cyclically presented groups. Journal of Algebra, 580. pp. 104-126. DOI https://doi.org/10.1016/j.jalgebra.2021.04.003 ↩

2. Tomas Rokicki (https://mathoverflow.net/users/536277/tomas-rokicki), Catalogue of groups with short finite presentations, URL (version: 2024-08-28): https://mathoverflow.net/q/477760 ↩

3. Pestov, S. (2026). #22271 ⟨a, b | aaa=1, ababbb=ba⟩. Slava’s Monoid Zoo. https://monoids.net/2,2/22271.html ↩

4. Stefan Kohl (https://mathoverflow.net/users/28104/stefan-kohl), Catalogue of groups with short finite presentations, URL (version: 2022-06-30): https://mathoverflow.net/q/423541 ↩ ↩² ↩³