Suppose φ : Z50 → Z15 is a homomorphism and φ(7) = 6.(a) What is φ(x) for any x ∈ Z50?(b) What is Image φ?(c) What is Ker φ?(d) Show that φ−1(3) is given by the coset 1 + Ker φ in Z50⧸Ker φ

(a) Since φ is a homomorphism, it preserves the operation of the group. In this case, the operation is addition. So, for any x ∈ Z50, we can write x as 7k + r for some integers k and r with 0 ≤ r < 7. Then φ(x) = φ(7k + r) = kφ(7) + φ(r) = 6k + φ(r). Since 0 ≤ r < 7 and φ is a homomorphism, φ(r) must be in Z15. Therefore, φ(x) is determined for all x ∈ Z50.

(b) The image of φ, denoted by Im(φ), is the set of all elements in Z15 that can be obtained by applying φ to some element in Z50. Since φ is a homomorphism and Z50 is a finite group, Im(φ) is a subgroup of Z15.

(c) The kernel of φ, denoted by Ker(φ), is the set of all elements in Z50 that are mapped to the identity element in Z15 by φ. In other words, Ker(φ) = {x ∈ Z50 | φ(x) = 0}.

(d) To show that φ−1(3) is given by the coset 1 + Ker φ in Z50⧸Ker φ, we need to show that every element in φ−1(3) is in the coset 1 + Ker φ and vice versa. Since φ is a homomorphism, φ−1(3) is a coset of Ker φ. Therefore, every element in φ−1(3) can be written as 1 + k for some k in Ker φ, which means φ−1(3) is a subset of the coset 1 + Ker φ. Conversely, every element in the coset 1 + Ker φ can be written as 1 + k for some k in Ker φ, which means the coset 1 + Ker φ is a subset of φ−1(3). Therefore, φ−1(3) is given by the coset 1 + Ker φ in Z50⧸Ker φ.