Finding a homomorphism between two groups is often difficult especially for a first course in group theory. There are many misconceptions and we address some of those here.
Between any two groups there is at least one homomorphism defined by . This is known as the trivial homomorphism. The set of homomorphisms between groups G and H is known as Hom(G,H), or in the case H=G, End(G). Thus, . We are often concerned with finding non trivial ones, if there are any.
For finite groups, we can make use of counting theorems. For example, the index of is a divisor of the order of G and is equal to the order of . So we could not construct an epimorphism from to since 6 does not divide 8. Nor could we construct a homomorphism from to such that the image is a cyclic subgroup of 3 elements, since has no normal subgroup of index 3. The only choice would be that the image is one of the three cyclic subgroups of order 2 in . Then, the kernel of such a homomorphism would be a normal subgroup of order 4 in , and there are 3 such subgroups. Another application shows that there are no non trivial homomorphisms from a cyclic group of order m to one of order n whenever m and n are coprime.
Keep in mind that the homomorphism is completely determined by the images of a generating set for the domain G. Also, domain elements can only map to elements in the codomain of order a divisor of the domain element. This again shows why no non trivial homomorphism from to can have elements of order three in . But why can't all the elements of order 2 and the identity of be in the image? It's because they do not form a subgroup of . As another application, there is no monomorphism from to . As another, if is a homomorphism then the image of is either or an element of order 3.
If G is cyclic note that is also cyclic. To see this, let . Then there is a g in G with . As , say, then for some integer k. Then, . Hence is generated by showing the image of G is cyclic.
Similarly, it is easily shown that if G is abelian then is abelian. However, note that these two results do not say that H is cyclic, or H is abelian. So for an example, if you wanted to construct a homomorphism from then it could not be an epimorphism.
One final misconception is that the homomorphism must be onto. No, unless it is stated that it is an epimorphism (or onto) the image need only be a subgroup of the codomain.