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.