I think p was chosen to imply it had to be prime :-) On Fri, Jan 31, 2020 at 1:02 PM Tomas Rokicki <rokicki@gmail.com> wrote:
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.
This doesn't quite work here; there are also p-tuples that are periodic with period of a factor of p. For instance, mod 10, using 6-tuples: (1,2,2,1,2,2). There are only three unique cyclic permutations of this 6-tuple.
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