代數學的證明 G=Z a*b

1(a) (a*b)*c = a - 2b - 2c, a*(b*c) = a - 2(b - 2c) = a - 2b + 4c

(a*b)*c ≠ a*(b*c)

G is not a group

(b) a * 0 = a and so the identity element is 0

Consider a * b = 0 => a + b + ab = 0 => b = -a/(a + 1)

As -a/(a + 1) may not be an integer and so G is not a group

(c) Similar reason as (b) and G is not a group

(d) G is closed under the operation *

a * 0 = 0, the identity element is 0 ∈ G

The inverse is a^(-1) = -a/(a + 1) ∈ G (a ≠ -1)

(a*b)*c = a + b + ab + c + c(a + b + ab) = a + b + c + ab + bc + ac + abc

a*(b*c) = a + (b + c + bc) + a(b + c + bc) = a + b + c + ab + bc + ac + abc

Associative law is satisfied.

So, G is a group