On 29/01/2020 14:44, James Propp wrote:
What are people’s favorite examples of existence proofs that show that a set is not empty by showing that its cardinality is odd?
Not exactly parity, but very much in the same mental pigeonhole: a group whose order is a multiple of p has an element of order p. Proof: consider p-tuples of elements whose product is 1. There are |G|^(p-1) of these because we can pick the first p-1 things in the tuple freely and then the last one is the inverse of their product. This is a multiple of p. The number of such tuples whose elements _aren't_ all equal is a multiple of p because we can permute cyclically. So the number whose elements _are_ all equal -- i.e., the number of elements of order 1 or p -- is also a multiple of p. And it's not 0 because of (1,...,1), so the number of elements of order p is p-1 (mod p) and in particular is nonzero. -- g