This is a parity argument in the case where p = 2, where it simplifies to: In a group of even order, elements that are not their own inverse come in inverse pairs. That leaves an even number of elements that are there own inverses, which include the identity and an odd (and therefore nonzero) number of elements of order 2. Andy On Wed, Jan 29, 2020 at 6:34 PM Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
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
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