I often hear people saying things like:
- one only really understands groups if one looks at group homomorphisms between them
- one only really understands rings if one looks at ring homomorphisms between them
- …
Of course, these statements are just special cases the category theoretic slogan that what really counts is the morphisms not the objects. I can appreciate that it's quite cool that one can characterize constructions such as the free group or the direct product of groups just in terms of their relation to other groups (and in this sense, the morphisms from and to that construction help to understand the construction better). But besides, I'm struggling to appreciate the usefulness of homomorphisms. I understand that what one is interested in is groups up to isomorphism (one wants to classify groups), so the notion of isomorphism seems to me to be very fundamental, but the notion of a homomorphism seems to me in some sense just to be a precursor the fundamental notion of an isomorphism.
I guess it would help if some of you could point me to bits and pieces of group theory where homomorphisms (instead of isomorphisms) are essential. In which sense do group homomorphisms help us to understand groups itself better?
Of course, I could ask the same question about ring theory or some other subfield of mathematics. If you have answers why morphisms matter in these fields, then feel free to tell me! After all, what I'm interested in is examples of the usefulness of homomorphisms from down to earth concrete mathematics, so what I don't want is just category theoretic philosophy jabbering (this is not to say I don't like category theory, but for the purpose of this question I'm interested in why morphisms matter in specific subfields of mathematics such as group theory).
Best Answer
Even if all you wish to do is to classify groups up to isomorphism, then there is a very important collection of isomorphism invariants of a group $G$, as follows: given another group $H$, does there exist a surjective homomorphism $G \mapsto H$?
As a special case, I'm sure you would agree that being abelian is an important isomorphism invariant. One very good way to prove that a group $G$ is not abelian is to prove that it has a homomorphism onto a nonabelian group. Many knot groups are proved to be nonabelian in exactly this manner.
As another special case, the set of homomorphisms from $G$ to the group $\mathbb Z$ has the structure of an abelian group (addition of any two such homomrophisms gives another one; and any two such homomorphisms commute), this abelian group is called the first cohomology of $G$ with $\mathbb Z$ coefficients, and is denoted $H^1(G;\mathbb Z)$. If $G$ is finitely generated, then $H^1(G;\mathbb Z)$ is also finitely generated, and therefore you can apply the classification theorem of finitely generated abelian groups to $H^1(G;\mathbb Z)$. Any abelian group isomorphism invariants applied to $H^1(G;\mathbb Z)$ are (ordinary) group isomorphism invariants of $G$. For example, the rank of the abelian group $H^1(G;\mathbb Z)$, which is the largest $n$ such that $\mathbb Z^n$ is isomorphic to a subgroup of $H^1(G;\mathbb Z)$, is a group isomorphism invariant of $G$; this number $n$ can be described as the largest number of "linearly independent" surjective homomorphisms $G \mapsto \mathbb Z$.
I could go on and on, but here's the general point: Anything you can "do" with a group $G$ that uses only the group structure on $G$ can be turned into an isomorphism invariant of $G$. In particular, properties of homomorphisms from (or to) $G$, and of the ranges (or domains) of those homomorphisms, can be turned into isomorphism invariants of $G$. Very useful!