An isomorphism is a map which is: 1) bijective, 2) a homomorphism.
Since a bijection between finite sets implies that the sets have the same number of elements, if is isomorphic to then has order 14 and as a consequence 14 divides So that divisibility condition is a necessary condition, and since you're dealing with cyclic groups, try to see why it is also a sufficient condition ( think of what are the subgroups of ).
a) and b) are done using the two hypotheses given: for instance, a): abelian and one-to-one, let be elements of then so, since and using injectivity again.
c) can be deduced from a) and b).
Compute . Compute ? Any idea to find a isomorphism?