How would I go about constructing a homomorphism between a group and its subgroup. For example, the group multiplicative group $\mathbb{Z}/15\mathbb{Z}$ and the subgroup of the squares of its elements $\{1,4\}$.
Also, in general how would I construct a homomorphism between two groups without generators.
(I'm just at A-level, so I would like a simple explanation)
If your group $G$ is a finite abelian group then every subgroup of $G$ is a homomorphic image of $G$. This follows from the classification of these groups (the fundamental theorem for finite abelian groups). The multiplicative group of $\mathbb{Z}/15\mathbb{Z}$ is abelian so there exists a homomophism to the subgroup consisting of the squares of elements: it is simply the map $x\mapsto x^2$, as was pointed out in the comments.
However, in general if $H$ is a subgroup of $G$ there does not exist a surjective homomorphism $G\twoheadrightarrow H$. For example, consider the alternating group $G=A_5$. This is a subgroup of the permutation group $S_5$ (if you are going through a text book, these groups will almost certainly be covered!). Now $A_5$ is simple. This means that every normal subgroup is either trivial or the whole group; in terms of homomorphic images, this means that every homomorphism $\phi: A_5\rightarrow H$ has trivial image or its image is isomorphic to $A_5$ (by the first isomorphism theorem). In particular, if we have $\phi:A_5\rightarrow H$ where $H$ is a subgroup of $A_5$ then the image of $\phi$ is trivial. (Note that the group $A_5$ contains proper, non-trivial subgroups; abelian simple groups, such as the cyclic group of order $3$, do not contain any proper, non-trivial subgroups.)