Final answer:
Homomorphisms from ℤ/12ℤ to ℤ/4ℤ include the trivial homomorphism and a mapping that doubles elements modulo 4. From ℤ/12ℤ to ℤ/11ℤ, the only homomorphism is the trivial one, mapping all elements to zero because the order 11 does not divide 12.
Step-by-step explanation:
To find all homomorphisms from ℤ/12ℤ to ℤ/4ℤ, and from ℤ/12ℤ to ℤ/11ℤ, we must understand that a group homomorphism is a function between two groups that preserves the group operation. The group ℤ/nℤ is a cyclic group consisting of n elements for the integers modulo n.
For the group ℤ/12ℤ to ℤ/4ℤ, a homomorphism must map the generator of ℤ/12ℤ (which can be taken as 1 + 12ℤ) to an element of ℤ/4ℤ such that the order of the image divides both 12 and 4. The possible orders of elements in ℤ/4ℤ are 1 and 2. As such, the homomorphisms are the mappings where 1 + 12ℤ is sent to 0 + 4ℤ (the trivial homomorphism), and where 1 + 12ℤ is sent to 2 + 4ℤ, doubling each element of ℤ/12ℤ modulo 4.
As for the homomorphisms from ℤ/12ℤ to ℤ/11ℤ, since 11 is prime, the only elements in ℤ/11ℤ are of order 11 or 1. Since 11 does not divide 12, the only homomorphism is the trivial one which maps every element of ℤ/12ℤ to 0 in ℤ/11ℤ.