Try using the fact that a homomorphism takes subgroups to subgroups. What are the subgroups of the cyclic group of order 7?
Let be a cyclic group of order 5. Let be a cyclic group of order 7. Let be a homomorphism. Why is the trivial homomorphism?
In my group, the trivial homomorphism means the map that takes every element of the domain and sends it to the element of the codomain.
I know what and look like.
But how is it that must be the trivial homomorphism?
I found this theorem in my book. Let be a group-homomorphism. The image of is a subgroup of .
So by this theorem, is a subgroup of .
Then I found this online: Let a cyclic group have order n. If r is a divisor of n then there is some number s such that rs = n. Let a be an element of the group with order n. (it must have one if it is cyclic) Then ar is of order s and therefore generates a cyclic subgroup of order s while as generates a subgroup of order r.
So if that is true, since 7 is a prime the only subgroups has is of order 1 or 7. This would imply that is the trivial homomorphism.
Is this correct?