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Is there a conceptual reason why (non-abelian) Coxeter groups are never simple?

For example is there some obvious normal subgroup that can be defined? Or perhaps it is for some reason clear that the commutator subgroup of a Coxeter group is always proper?

I was looking through the list of Coxeter groups and noticed that a lot of them are pretty close to simple, but not quite. For example the Coxeter groups of type $ A_n $ are $ S_{n+1} $ which has the simple index $ 2 $ alternating subgroup $ Alt_{n+1} $ for $ n \geq 4 $.

Coxeter groups of type $ E_6, E_7, E_8, H_3 $ also have index 2 simple subgroups: $ PSp_4(3), Sp_6(2), GO_8^+(2),Alt_5 $ respectively. The Coxeter groups of type $ B_n=C_n $ and type $ D_n $ have a single non-abelian simple composition factor: $ Alt_n $, but these groups are not simple.

The Coxeter groups of type $ F_4 $ and $ I_2(n) $ are solvable. Finally, the Coxeter group of type $ H_4 $ has an index $ 4 $ subgroup $ Alt_5 \times Alt_5 $. So by exhaustion no Coxeter group is simple. I was wondering if there is a nice conceptual reason for that, especially since some of them get so close to being simple (like $ A_n, E_6,E_7,E_8, H_3 $), but aren't quite.

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    $\begingroup$ Could the person downvoting and voting to close explain why? This seems like a very reasonable MSE level question to me. Is there some extra detail you want? $\endgroup$
    – Ian Gershon Teixeira
    Commented Feb 17, 2023 at 17:10

2 Answers 2

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Yes there is an obvious normal subgroup of index $2$ consisting of elements defined by words of even length in the generators.

The same applies to any group defined by a presentation in which all defining relators have even length.

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  • $\begingroup$ Ah I see because relations are even length means that they preserve length mod 2 of any product of generators. So mod 2 parity of a group element is well defined. That's a very cool perspective I have not seen before on why sign of a permutation is well defined! Could a similar argument be used to show that a (non abelian) complex reflection group cannot be simple? $\endgroup$
    – Ian Gershon Teixeira
    Commented Feb 17, 2023 at 17:08
  • $\begingroup$ A nice rephrasing of your argument is given here math.stackexchange.com/a/4644407/758507 The normal index 2 subgroup is just the subgroup of determinant 1 matrices (kernel of the determinant map), and that subgroup is always proper because reflections have determinant $ -1 $ $\endgroup$
    – Ian Gershon Teixeira
    Commented Feb 22, 2023 at 16:42
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The parity of the miniumum length of a sequence of generators whose product expresses an element of the Coxeter group is a non-trivial group homomorphism. In the case of the symmetric group, this is the alternating subgroup. How obvious it is depends on the definition.

Starting from the presentation in terms of generators and relations, it is easy to see from the fact that the the braid relations are between expressions of the same length, and that the squaring relation is between an expression of length two and an expression of zero length.

Starting from the exchange or deletion conditions, it is easy to see that the difference in length between a reduced expression and a non-reduced expression for the same element is even. From this it follows that the length of the reduced word for a product of two reduced words is the same parity as the product of the parities.

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