Question

the number of distinct homomorphism from Z/5Z to Z/7Z​

Answer 1

Answer:

Any homomorphism Z5→ϕZ7 yields a subgroup of Z7 given by ϕ(Z5).

Answer 2

Explanation:

Any homomorphism Z5→ϕZ7Z5→ϕZ7 yields a subgroup of Z7Z7 given by ϕ(Z5)ϕ(Z5). What are the subgroups of Z7Z7? Given the subgroups and their orders, what possibilites are there for ϕ(Z5)ϕ(Z5)? (You might want to use Lagrange's theorem.)

Both groups are cyclic and generated by 11. Therefore, a homomorphism ϕ:Z5→Z7ϕ:Z5→Z7 is determined by where 11 is sent (since any element k∈Z5k∈Z5 is equal to k⋅1=1+…+1k timesk⋅1=1+…+1⏟k times). So ϕ(1)ϕ(1) generates a subgroup of Z7Z7. Think about the possibilities for the size of the subgroup given by ⟨ϕ(1)⟩⊆Z7⟨ϕ(1)⟩⊆Z7 based on the orders of Z5Z5 and Z7Z7 (and hence the possibilities for ϕ(1)ϕ(1))

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