(e)
An abelian group must include an identity element e and each element must have an inverse. The identity element may also be idempotent, that is, within the axioms of the group, e²=e. The closure requirement implies for a ℤ₃₀ group that e²=e+30m where m is a positive integer. This creates the quadratic equation:
e²-e-30m=0, and e=(1+√(120m+1))/2.
From this, e=1, 6, 10, 15, 16, 21, 25 (for m=0, 1, 3, 7, 8, 14, 20) for e∈ℤ₃₀.
Also, e must satisfy: e×g=30m+g ∀g∈G.
Divide through by g: e=30m/g+1.
When g=2, e=15m+1, and e=16 when m=1. 16 is idempotent because 16²=16 (mod 30).
e×g=16g=30m+g, 15g=30m, g=2m, which can be used to generate G1={2,4,8,14,16,22,26,28}, and produce the multiplication table below:
|
|
2
|
4
|
8
|
14
|
16
|
22
|
26
|
28
|
|
2
|
4
|
8
|
16
|
28
|
2
|
14
|
22
|
26
|
|
4
|
8
|
16
|
2
|
26
|
4
|
28
|
14
|
22
|
|
8
|
16
|
2
|
4
|
22
|
8
|
16
|
28
|
14
|
|
14
|
28
|
26
|
22
|
16
|
14
|
8
|
4
|
2
|
|
16
|
2
|
4
|
8
|
14
|
16
|
22
|
26
|
28
|
|
22
|
14
|
28
|
26
|
8
|
22
|
4
|
2
|
16
|
|
26
|
22
|
14
|
28
|
4
|
26
|
2
|
16
|
8
|
|
28
|
26
|
22
|
14
|
2
|
28
|
16
|
8
|
4
|
Note that the table shows inverses. The product of a and b is 16 when b is the inverse of a and vice versa. Note that in some cases an element is the inverse of itself (e.g., 4, 14 and 26).
Insufficient space to continue answer. More to follow in comment...