-

@ John Carlos Baez
2025-02-22 21:16:49
HARDCORE MATH POST
I knew a guy named John McKay who had a crazy good ability to perceive patterns in math. He came up with amazing results, but also some even more amazing conjectures which turned out to be true. One was recently proved by Britta Späth and Marc Cabanes. They proved it after 20 years of hard work, and in the process they fell in love and started a family together. The proof is massive and requires studying lots of special cases. I hope it's correct! It's way too hard for me to understand.
But I understand the statement.
If you have a finite group G whose cardinality |G| is divisible by p, there's a largest power of p that divides |G|, say pᵏ, and G is guaranteed to have a subgroup of cardinality pᵏ. This is called a 'Sylow p-subgroup' of G. There may be more than one, but they're all conjugate - so they all look alike in a very strong sense.
McKay made a conjecture about this. Understanding it requires knowing a bit more stuff, which I'll happily explain if you ask.
Here's what it says:
For any finite group K, let n(K) be the number of irreducible complex representations of K whose dimension is not divisible by p. Let G be any finite group, and let N be the normalizer of a Sylow p-subgroup of G. Then n(G) = n(N).
(Here of course I count isomorphic representations as the same.)
Now the sensible way to prove this would - seemingly! - be to set up a recipe to turn any irreducible complex representation of G whose dimension isn't divisible by p into one for N, and vice versa. But nobody knows how to do that, in general. So the only known proof requires separately studying every finite simple group... which is really scary.
https://www.quantamagazine.org/after-20-years-math-couple-solves-major-group-theory-problem-20250219/