Re: FLTMA: A little group theory

Chip Eastham wrote:

The Dougster wrote:

Orders: 2^6 == -1 mod 13; 7^6 == -1 mod 13; 9^2 == 1 mod 10; 2^6 == 1
mod 7

The exponents shown are the minimal ones possible
(note 2*7 = 1 mod 13), except that the order of 2 in
Z/7Z* is only 3.

Thank you. Repaired two errors and added exponents:

X = 3; Y = 4; Z = 5
Phi(3) = 2; Phi(4) = 2; Phi(5) = 4

3 * 2 = 6 == 1 mod 5; 4 * 4 = 16 == 1 mod 5
3 * 3 = 9 == 1 mod 4; 4 * 1 = 4 == 1 mod 3

4 * 2 = 8 == 3 mod 5; 3 * 3 = 9 == 4 mod 5
3 * 3 = 9 == 5 mod 4; 4 * 2 = 8 == 5 mod 3

2^2 = 4 == -1 mod 5; 3^2 = 9 == -1 mod 5
2^4 = 16 == 1 mod 5; 3^4 = 81 == 1 mod 5
3^2 = 9 == 1 mod 4; 2^2 = 4 == 1 mod 3

What remains is to determine solution n.

Doug





Doug

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