A. show that {a : 15a ≡ 18 (mod 24)} is not a subgroup of the group z24 (with operation +)
b. let g = zm (the operation is addition) and let b be some number with 1 ≤ b < m. show that the cyclic subgroup h generated by b is all of g if and only if the least common multiple of b and m is bm, if and only if (b, m) = 1.
c. show that the subgroup of z consisting of all solutions x in z of ax ≡ 0 (mod m) is the cyclic subgroup rz where r = m/(a, m).