Question

Let Z48 = ⟨x⟩, and let Z36 = ⟨y⟩. Prove that there exists a

(well defined) homomorphism φ : Z48 → Z36 satisfying φ(x) = y^a if
and only if 3 | a.

Answer 1

To determine the existence of a homomorphism from Z48 to Z36 such that φ(x) = y^a, it is necessary and sufficient that 3 divides a. This condition ensures the mapping φ preserves the group operation and results in a well-defined function.

Step-by-step explanation:

The question pertains to group theory and homomorphisms between two cyclic groups, Z48 and Z36. In order to prove the existence of a homomorphism φ : Z48 → Z36 such that φ(x) = y^a, we need to show that this mapping is well-defined and respects the group operation. For Z48 and Z36, which are cyclic groups generated by x and y respectively, a map φ is a homomorphism if it preserves the group operation, which, in this case, is addition.

A necessary and sufficient condition for φ to be a homomorphism is that 3 | a (3 divides a). To see why consider that any element in Z48 can be expressed as x^k for some integer k. The mapping under φ would send x^k to (y^a)^k = y^{ak} in Z36. For φ to be well-defined, we need to ensure that the result always lands in Z36. Since 48 is a multiple of 3, any k multiple of 48 gets mapped to zero in Z36 if 3 divides a, satisfying y^{48k} = (y^{36})^{k} = e, where e is the identity in Z36.