So in reading about category theory I'm starting to see this picture that it is just a higher level of abstraction where we consider similarities between mathematical structures by way of morphisms between them. In reading the beginning of Maclane's book, he mentions one of the useful results we get from category theory is the idea of universal constructions/properties.
My question is, what's an instance where a universal property has shortened a proof significantly?
On
This was the exercise that really convinced me that universal properties are magical:
Prove that $(\Bbb Z/(2\Bbb Z)) \otimes_{\Bbb Z} 2\Bbb Z$ is not the trivial module over $\Bbb Z$.
Using the usual definition of tensor products, i.e., via the equalities $$ na \otimes b = a \otimes nb\\ a \otimes (b + c) = a \otimes b + a \otimes c\\ (a + b) \otimes c = a \otimes c + b \otimes c $$ it's difficult to show that $1 \otimes 2 \neq 0 \otimes 0$.
However, with the universal property, it's easy to show that the two can't be equal.
(In case you haven't seen it, $\otimes$ is the tensor product of modules.)
Here is a simple example from group theory. Suppose $n \neq m$ are positive integers. A common exercise is to show that the free groups $F_n$ and $F_m$ are not isomorphic. There are lots of ways of doing this, but to my mind the cleanest way is to use their universal properties: $F_n$ is the free group on $n$ generators, or said another way, it represents the functor $G \mapsto G^n$ in the sense that
$$\text{Hom}(F_n, G) \cong G^n.$$
So if $G$ is a finite group of order $|G| \neq 1$ then there are $|G|^n$ homomorphisms $F_n \to G$ and $|G|^m$ homomorphisms $F_m \to G$ and these numbers are different.