Technically, x^m is a member of the group. The set of all powers of x, {x^m}, is a cyclic group. And, since you are given that G is cyclic, it follows that {x^m} is a subgroup of G having index(x) members. Now, think about factors of ord(G).

I also know a monomorphism is an injective homomorphism, which is one to one as a function which relates maybe to f:G-->G.

When it asks to prove it is a group morphism does this mean I need to find the kernal and the image.

I know what the question means but just don't know the proof of it.

Any help would be great thanks