Michi’s blog » Modular representation theory – when Maschke breaks down

Modular representation theory – when Maschke breaks down

• April 21st, 2007
• 1:43 pm

This post is dedicated to Janine Kühn and her Proseminar-lecture.

We had, in my first representation theory post, a mention of Maschke’s theorem. This states that if the characteristic of our field doesn’t divide the group order, then simple and irreducible mean the same thing.

Now, obviously, the actual proof you normally see first deals with a construction that works for when the characteristic doesn’t divide the group order – which uses 1/|G| at one point. So, what happens when this is impossible to work with? When the conditions of Maschke simply do not hold?

The very simplest answer is that then we can get modules that are glued together by simple modules with some meshing. Such that they aren’t direct sums any more. The ways we can glue together modules are through extensions – i.e. we can glue together A,C by forming a short exact sequence
0 → C → B → A → 0
and we’ll have that B is a module such that B/C=A. Now, the typical such module is the direct sum of A and C – and if Maschke holds, this is indeed all there is.

If Maschke does not hold, then the set of all such B forms an abelian group, called Ext1(A,C), with addition formed in a slightly messy way called Bair summation. This is actually the same Ext as in all my cohomology posts. So the dimension 1 cohomology actually measures what kind of modules you can have at all.

Suppose the field k has characteristic p and the group order is pn for some n. Then an order argument, and a few properties of modules will yield that there is only one simple module: the trivial module k. So the only extensions really interesting will be in Ext1(k,k).

And at this point I can bring in my knowledge of group cohomology: if the characteristic is 2, and the group is C4, then the cohomology ring was mentioned in my post on the higher multiplication structure of the cohomology rings of groups of order 4. It is – so each degree has dimension 1, so there are precisely two extensions of the trivial module over the cyclic 4-group. One is the direct sum, and the other is an irreducible, but not simple module.

Precisely this example, however, I feel I won’t quite explore to the potential it is worth right now, so I’ll pick an example where I know that I won’t be telling anything wrong.

Consider the symmetric group S3. This has order 6, so the interesting cases are mod 2 and mod 3. For the 3-case, we have the trivial representation and the sign representation – since these always occur when they have a chance to. These are both 1-dimensional, simple, and irreducible.

At this point, we’ll start needing various theorems for the analysis. First off, we can count simple modules over the group algebra by counting conjugacy classes in the group where the order of the elements isn’t divided by the field characteristics. This way, we’ll know how many simples we have.

Furthermore, there is a theorem that states that regardless of Maschke, we certainly have a bijection between simples and irreducibles. I’m not going to go into how this is established, but if you’re interested, the relevant keyword is the radical of a module. Furthermore, if we decompose the algebra itself into a sum of irreducible modules, then each module occurs with a multiplicity corresponding to the dimension of the simple module it belongs to.

I’ve tried three times now to start writing out in painful detail precisely what’s going on here – but I can’t really, today, get a single example worked through from start to end. For S3, though, we know the conjugacy classes: the identity, the transpositions and the 3-cycles. For each of the modular cases one of these disappears, and so for the 2-case, we get the trivial module and one 2-dimensional simple; and for the 3-case, we get the trivial module and the sign representation. These correspond to irreducibles, that make up the group algebra – in both cases two kinds of irreducibles, and for the 2-case, one occuring once, and one occuring twice, and for the 3-case both occuring just once.

Writing down the irreducibles just doesn’t seem to work out for me today. I blame the flu.

Edit: Fixed a typo pointed out by Dylan Thurston in the comments. Again, I blame the flu.

2 People had this to say...

• Dylan Thurston
• April 21st, 2007
• 14:23

There’s a crucial typo in the second paragraph: you mean the characteristic does NOT divide the order of the group!

• Michi
• April 21st, 2007
• 14:27

Fixed. Any corrections to this post are very welcome – it’s evidently not a good idea to write in the blog while ill.