Sporadic and Related groups Lecture 14 – Presentations

Revision – Fabulous groups. Conway originally defined a “fabulous” group (Free Abelian Under Little!) inductively to be a presentation of a finite group where you are only allowed add a relator on a subset of the generators whose normal closure (in the group generated by those generators) is a free Abelian group. We will be varying this a little later today...

Various definitions. One thing we will want to vary is to allow the factoring out of normal abelian subgroups of finite order, as well as (in Conway's original definition) free Abelian subgroups. Another thing we wish to relax is that some subsets of the generators may (using only the relators involving those generators) be infinite. To start with though, we stick with Conway.

Revision – to the root systems 1 generator – must specify the order This gives us all the finite cyclic groups 2 generators–involutions and given product order This gives the real reflection groups. Next stage we get the extended Dynkin diagrams A n B n C n D n E 6 E 7 E 8 F 4 G 2

This is how we get the root systems. Let's look at the case where we have a set of involutions with relations giving the order of the product of any pair. We can model this as a (real) reflection group by taking vectors of norm 2 for the involutions, and giving the inner product as 2.cos(2π/n) if the product has order n.

Classification of root systems. If this is to be a finite group (respectively Free ABelian Under Little) this matrix must be positive definite (respectively positive semi-definite). We can classify such matrices combinatorially to get the usual classification of root systems, starting at dimension 1 and going up one dimension at a time. The determinant at every stage may not be negative.

Complex reflection groups A Complex reflection group is a finite group generated by complex matrices with only one eigenvalue different from 1.(that eigenvalue need not be -1). These were classified by Shephard and Todd in To summarize the result... Root systems as on previous slide X n.S n and its determinant 1 subgroup A few small exceptional groups, the largest being 6.U 4 (3).2 in six dimensions.

Hermitian forms. A Hermitian form is a bit like a quadratic form, except that (b,a) is the complex conjugate of (a,b) and (a,λb) is (a,b) times the complex conjugate of λ. Although the inner products are not, in general, real, the norms are, and it still makes sense to talk about positive definite Hermitian forms where all the norms are positive.

Hermitian (or Unitary) reflections. A vector V of norm 2 then defines a unitary reflection by x → x – (x,V).V. There is a little bit of “play” here, though, in that if we multiply V by a scalar of norm 1, it defines the same reflection. Similarly the order of the product is determined by the absolute value of the inner product, so the whole set-up is a bit less rigid.

This allows us an explicit model Nevertheless we can take a presentation and try to find Unitary reflections that satisfy it. If we can, we can then look at the positive-definite nature of the form to decide whether the group is finite, or has a free Abelian normal subgroup of finite index.

Classification of Complex reflection groups. Essentially we can do the same as we did with real ones. The reflection vectors themselves are a bit arbitrary (you can multiply by any scalar of norm 1) and you only know the absolute value of the inner products, but with a bit of care you can do the combinatorics and get the Shephard-Todd classification.

This gives us many of the next stage groups. For any extended Dynkin diagram, we can give complex reflections that satisfy the diagram presentation and any given order for the extending root. If this is to be a finite group, it must be on Shephard and Todd's list. As we see in the Y555 diagram for the monster, things can get quite interesting if we quotient out sublattices that are not spanned by the images of the extending root.

Now I want to relax the definitions Take any set of generators, and if any subset of them generate a group with a normal abelian subgroup of finite index, you may add the relator to factor it out.

Notation! Well... to be honest I haven't got one. What we would need is a little symbol to indicate all the different Abelian subgroups we may wish to factor out. There is essentially no bound to the complexity of this notation, so I will try not to rely on it too much. Please interrupt and ask if I have forgotten to explain one (or if you have forgotten the explanation).

First page of notation. O represents a generator of order two. O O (not joined at all) means they commute. O O means their product has order 3 O === O means product order 4 O /\/\/\/\ O means product order 5 O ≡≡≡≡ O means product order 6

Can “name” the central involution in a dihedral group O O ===== O This is 2 x D 8 as it stands, and this has a normal abelian subgroup of order two consisting of the product of the left-hand node and the square of the product of the other two. We can factor this out to get a D 8 diagram with three nodes.

Notation for D 8 O O ===== O ↑ ---- \_________|

2 x S 4 O O ==== O This is a presentation of 2 x S 4 and this group has, of course, a normal abelian subgroup of finite index, namely the central 2, and we may wish to factor it out. This is not a free abelian subgroup, so does not come under Conway's original definition, but it is interesting nevertheless. We indicate this relation as follows – a diagram for S 4 O O ===== O ----

2 x A 5 O O /\/\/\/\/\ O Similarly this is a presentation of 2 x A 5 with a normal abelian subgroup of finite index, again the central 2. O O /\/\/\/\/\ O ----

Simlarly 2 x A 5 with four nodes O O /\/\/\/\/\ O O ---- ↑ \______________/

L 2 (11) Here, then, is a pretty presentation for L 2 (11) O O /\/\/\/\/\ O O

J1J1 This leads us to consider the following diagram, which exhibits 2 x A 5 and L 2 (11) – both subgroups of J 1. Indeed it is a presentation of J1. O /\/\/\/\/\ O O /\/\/\/\/\ O O ↑ \_____________________/

Here is M 11 O / \\ / \\______________ / \\ \ / \\ ↓ O O /\/\/\/\/\ O O

ASCII art has its limitations. 3.M 22 = L 2 (11) plus E order 4 from A and 3 from C. BCE line (A5) / ABE line (S 4 ) / AECD is 2 3.S 4 not 2 4.S 4 Also (ABCE) 8 for simple group. Then F 4-B, 3-A (BF) 2 = D gives Higman-Sims

J2J2 2.(S 3 x J 2 ) Line A 5 B 3 C 8 D 3 E A=(CD) 4

Several others. Suffice it to say that the ATLAS gives many presentations of this form, including M 23, M 24, G 2 (4), McL, He, Suz, O'Nan, Co 3, Co 2, Co 1 as well as those coming from the Fischer groups and the Monster's Y555 diagram.

So fabulous presentations are good. Indeed one is led to the impression that if you start with involutions and repeatedly quotient out normal Abelian subgroups of finite index, you would very quickly find all the interesting groups. I have no idea why this might be.

Symmetrical presentations Rob Curtis has recently been experimenting with presentations where you have a fairly large number of generators (e.g. 24), and a permutation group acting on them (e.g. M 24 ), and require that all your relators imply their images under the group. These again lead to very concise presentations of large groups.

Free products with amalgamation If you take the group generated by two groups, and just add the relations identifying a subgroup (which is not all of either of them) the result is a group a bit like a free group. It is called a free product with amalgamation, and it is easy to work in.

Tree of groups This can be extended without difficulty to more than two groups, provided that the amalgamations are done in a “tree” - no loops.

For example You can take three groups A, B and C, and take any subgroups of A and B that are isomorphic and identify (amalgamate) them, and as well take any subgroups of B and C that are isomorphic and identify them also.

Graph of groups But as soon as you allow loops in your amalgamation (to get what is called a graph of groups) the results can be trivial, finite or infinite and the problem gets hard. These “amalgam” problems often arise in the classification.

The amagam for J 1 A B C D E O /\/\/\/\/\ O O /\/\/\/\/\ O O ↑ \_____________________/ Omit A and you have L2(11). Omit B and you have 2 x A5. Omit C and you have D10 x S3. Omit D and you have 2 x A5 (you can't omit E!). And every relator omits at least one of these, so the amalgam MUST actually be J 1.

How do amalgams arise. Perhaps you have two involution centralizers and a centralizer of an element of order 3, and know what all these groups must be and how they intersect. In this case it is worth looking to see if the group presented is finite, since if it is, you are basically finished. In particular, if it is trivial, there is no such group.