One of the interesting problems in abstract polytope theory is to determine, for a given finite group, when that group is the automorphism group of a regular abstract polytope. This is equivalent to the following question: Given a finite group G, when is G generated by involutions $\rho_0, \ldots, \rho_n$ such that $(\rho_i \rho_j)^2 = 1$ if $|i - j| \geq 2$ and such that for all $I, J \subset \{0, \ldots, n\}$ we have $\langle \rho_i \mid i \in I \rangle \cap \langle \rho_i \mid i \in J \rangle = \langle \rho_i \mid i \in I \cap J \rangle$?

The last property can be difficult to check, so let's relax that requirement for now. If a finite group G is generated by n involutions such that non-adjacent generators commute, what can we say about the structure or size of G? Of particular interest: what if G is simple?

Here are a few simple observations:

  1. For each n, the smallest (abstract) n-polytope has an automorphism group that is isomorphic to the direct product of n copies of $C_2$, corresponding to the trivial Coxeter diagram on n nodes. So a finite group G cannot be the automorphism group of an abstract regular n-polytope for $n > \log_2(|G|)$.
  2. A (nontrivial) group generated by involutions has even order.
  3. The abelianization of a group generated by involutions such that nonadjacent generators commute is a quotient of the group in (1), the direct product of n copies of $C_2$.
  • $\begingroup$ I am having trouble following your simple observations. I thought the smallest $n$-polytope was the $n$-simplex, with symmetric automorphism group. Also, the trivial group has odd order but is generated by involutions (zero of them). $\endgroup$ – S. Carnahan May 29 '10 at 5:51
  • $\begingroup$ In 1), I meant the smallest abstract polytope, not the smallest convex polytope. In 2), I meant nontrivial. I'll update to clarify. $\endgroup$ – Gabe Cunningham May 31 '10 at 3:23
  • $\begingroup$ what is an abstract polytope? $\endgroup$ – Mariano Suárez-Álvarez May 31 '10 at 4:24
  • $\begingroup$ Okay, Wikipedia has enlightened me a bit. I guess the smallest abstract polytope has two cells of each dimension $0,\ldots,n-1$. I don't understand the "strongly connected" axiom well enough to see why this puts a lower bound on the size of the automorphism group. Aren't there abstract $n$-polytopes with trivial automorphism group for all $n>1$? $\endgroup$ – S. Carnahan May 31 '10 at 6:44
  • $\begingroup$ Oh, the regularity hypothesis implies the lower bound. Never mind. $\endgroup$ – S. Carnahan May 31 '10 at 6:47

Really quite a few finite simple groups are generated by three involutions, two of which commute (n=3).

For instance, this papers provides a (revised) proof that almost all sporadic groups have such a generating set:

Mazurov, V. D. "On the generation of sporadic simple groups by three involutions, two of which commute." Sibirsk. Mat. Zh. 44 (2003), no. 1, 193–198; translation in Siberian Math. J. 44 (2003), no. 1, 160–164 MR1967616 DOI:10.1023/A:1022028807652

Its references provide a large list of other simple groups with this property:

  • almost all groups of lie type in char 2: MR1131150
  • almost all alternating groups: MR1172472
  • almost all groups of lie type in odd char: MR1454692 (low rank exceptions) and MR1601503 (all large rank)

Since one of the reviews was inaccurate, I quickly checked through the sources I had access to, and the following is probably quite close to accurate: Every finite simple group other than:

  • A6, A7, A8, S4(3)=U4(2), M11, M22, M23, McL
  • L3(q), U3(q) for all prime powers q
  • L4(q) for even prime powers q

has a generating set consisting of three involutions, two of which commute. In particular the Monster group does have such a generating set (a short proof in the first paper, an earlier proof due to Simon Norton in a letter).

I (quickly) verified that A6, A7, A8, S4(3), M11, M22, M23 have no generating set of involutions where non-adjacent involutions commute (even for more than three generators). You can bound n by the 2-rank of the group: having lots of commuting involutions means you have a large elementary abelian subgroup, and so n ≤ 5 for these groups.

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  • $\begingroup$ Thank you for the detailed answer, Jack. I'll look into these to see if the satisfy the intersection property as well. $\endgroup$ – Gabe Cunningham May 31 '10 at 15:48
  • $\begingroup$ I believe it is now known that every non-Abelian finite simple group other than ${\rm PSU}(3,3)$ is generated by three involutions, but this does not impose the condition that two of these commute. $\endgroup$ – Geoff Robinson Aug 10 at 9:06

Have you read "Regular Polytopes" by H.S.M. Coxeter? or any other text book on reflection groups or Coxeter groups?

In a nutshell, the strategy is to write down the Gram matrix of an inner product that is preserved by the reflections and so by the group generated by the reflections. Then the group is finite if and only if this inner product is positive definite.

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  • $\begingroup$ The groups I'm dealing with are not just Coxeter groups but rather quotients of Coxeter groups. Also, the question here isn't so much "when are these groups finite" but "which finite groups arise this way". For example: is there a presentation for the Monster group satisfying the above criteria? $\endgroup$ – Gabe Cunningham May 28 '10 at 19:26
  • $\begingroup$ To be precise, without the last condition (involving subgroups generated by involutions indexed by $I$ and $J$), they are finite quotients of a single $\textit{right-angled}$ Coxeter group with $n$ generators with RACG graph $K_n\setminus C_n.$ I am mystified why they should classify the automorphism groups of regular abstract polytopes, though. $\endgroup$ – Victor Protsak May 28 '10 at 21:48
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    $\begingroup$ It is only with the last condition that they classify the automorphism groups of regular abstract polytopes. In fact, you can build a regular abstract polytope out of such a group out of the cosets of certain subgroups. The precise construction is given in chapter 2 of Schulte and McMullen's book Abstract Regular Polytopes. $\endgroup$ – Gabe Cunningham May 29 '10 at 1:44

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