SLIDE 31 (Zp, +) is a cyclic group
For every prime p every element g ∈ Zp \ {0} is a generator: Zp = g = {g · i mod p | 0 ≤ i ≤ p − 1}
Note that this follows from the last fact on slide 59: Zp is of order p, which is prime.
Example in Z7: (1 · 0, 1 · 1, 1 · 2, 1 · 2, 1 · 4, 1 · 5, 1 · 6) = (0, 1, 2, 3, 4, 5, 6) (2 · 0, 2 · 1, 2 · 2, 2 · 2, 2 · 4, 2 · 5, 2 · 6) = (0, 2, 4, 6, 1, 3, 5) (3 · 0, 3 · 1, 3 · 2, 3 · 2, 3 · 4, 3 · 5, 3 · 6) = (0, 3, 6, 2, 5, 1, 4) (4 · 0, 4 · 1, 4 · 2, 4 · 2, 4 · 4, 4 · 5, 4 · 6) = (0, 4, 1, 5, 2, 6, 3) (5 · 0, 5 · 1, 5 · 2, 5 · 2, 5 · 4, 5 · 5, 5 · 6) = (0, 5, 3, 1, 6, 4, 2) (6 · 0, 6 · 1, 6 · 2, 6 · 2, 6 · 4, 6 · 5, 6 · 6) = (0, 6, 5, 4, 3, 2, 1) All the non-zero elements of Z7 are generators
- rd(0) = 1, ord(1) = ord(2) = ord(3) = ord(4) = ord(5) = ord(6) = 7
61
(Z∗
p, ·) is a cyclic group For every prime p there exists a generator g ∈ Z∗
p such that
Z∗
p = {g i mod p | 0 ≤ i ≤ p − 2} Note that this does not follow from the last fact on slide 59: Z∗
p is of order p − 1, which is usually
even, not prime.
Example in Z∗
7:
(10, 11, 12, 13, 14, 15) = (1, 1, 1, 1, 1, 1) (20, 21, 22, 23, 24, 25) = (1, 2, 4, 1, 2, 4) (30, 31, 32, 33, 34, 35) = (1, 3, 2, 6, 4, 5) (40, 41, 42, 43, 44, 45) = (1, 4, 2, 1, 4, 2) (50, 51, 52, 53, 54, 55) = (1, 5, 4, 6, 2, 3) (60, 61, 62, 63, 64, 65) = (1, 6, 1, 6, 1, 6) 3 and 5 are generators of Z∗
7
Fast generator test (p. 60), using |Z∗
7 | = 6 = 2 · 3:
36/2 = 6, 36/3 = 2, 56/2 = 6, 56/3 = 4, all = 1.
1, 2, 4, 6 generate subgroups of Z∗
7: {1}, {1, 2, 4}, {1, 2, 4}, {1, 6}
- rd(1) = 1, ord(2) = 3,
- rd(3) = 6, ord(4) = 3,
- rd(5) = 6, ord(6) = 2
The order of g in Z∗
p is the size of the subgroup g.
Lagrange’s theorem: ordZ∗
p (g) | p − 1 for all g ∈ Z∗ p 62