Q: How many elements of the cyclic group GF(81)* are generators?
A: So here is what I got I could be wrong, and I am open to suggestions and/or corrections.
The corresponding multiplicative group of GF(81)* must be a cyclic group of order 80 (i.e. U(80)), which follows from pages 383-385 (if you are following along in the book). Moreover the generators of U(80) we know from chapter 4 are numbers that are smaller than 80 and relatively prime to 80. Since 80 = 2^4 *5 the values we are looking for must be odd, less than 80, and are not multiples of 80. Since there are 40 odd numbers less than 80 and of these 40, 8 of them are divisible by 5 (namely 5, 15, 25, 35, 45, 55, 65, 75) we have that there are 32 generators of GF(81)*
--Bakey 09:06, 30 November 2012 (UTC)
