I suppose I'd like to focus on the theorems for groups and rings, first of all. In particular, I'd rather not see anything about modules, simply because I don't feel I know enough about them.
Anyway, in doing a lot of exercises in group theory and ring theory, I find that I use the first and third isomorphism theorems a lot more than the second and fourth. What I want to see are some really useful-looking applications of the second and fourth theorems that I may not have been exposed to before. Thanks!
EDIT: for clarity, the theorems I identify as the second and fourth are as follows for rings (cf. Dummit & Foote, 3ed p. 246)
The second:
Let $A$ be a subring of $R$ and $B$ be an ideal of $R$. Then $A+B = \{a+b|a \in a, b \in B\}$ is a subring of $R$, $A \cap B$ is an ideal of $A$, and $(A+B)/B \cong A/(A \cap B)$.
The fourth:
Let $I$ be an ideal of $R$. The correspondence $A \leftrightarrow A/I$ is an inclusion preserving bijection between the set of subrings $A$ of $R$ that contain $I$ and the set of subrings of $R/I$. Furthermore, $A$ (a subring containing $I$) is an ideal of $R$ if and only if $A/I$ is an ideal of $R/I$.
Best Answer
This answer includes an application of the 4th isomorphism theorem for rings, and one (well, kinda two) of the 2nd isomorphism theorem for groups. I also link to a really cool application of the 4th isomorphism theorem in groups (but is kinda the same as the application to rings).
Rings, 4th isomorphism theorem
An neat application of the 4th isomorphism theorem for rings is the following:
Proof: Let $R$ be a ring and let $I$ be a maximal ideal of $R$. Then consider the quotient $R/I$, and apply the correspondence theorem.
Indeed, if $R$ is commutative with identity then $R/I$ is a field. See here for a proof.
Groups, 4th isomorphism theorem
There is a rather cool trick in the theory of infinite groups, which was used by Higman to construct an infinite simple group. The idea is to appeal to Zorn's lemma and obtain a maximal normal subgroup, and then quotient this out to get a simple group. See here for a neat application of it. This is very similar to the above application to rings. (You quotient out a maximal subgroup/ideal to get a simple group/prime ring.)
Groups, 2nd isomorphism theorem
A nice application of the 2nd isomorphism theorem for groups is the following. It deals with soluble groups, which are an important class of groups (and are often taught in a second course on groups). A group $G$ is soluble (or solvable) if it possesses an abelian series, that is, a series $$1=G_0\lhd G_1\lhd\cdots\lhd G_n=G$$ where each factor $G_{i+1}/G_i$ is abelian.
Proof: Suppose $G$ is soluble with abelian series $1=G_0\lhd G_1\lhd\cdots\lhd G_n=G$.
Subgroups: If $H$ is a subgroup of $G$ then we can apply the 2nd isomorphism theorem to get the following. $$\frac{H\cap G_{i+1}}{H\cap G_i}\cong\frac{(H\cap G_{i+1})G_i}{G_i}\leq\frac{G_{i+1}}{G_i}$$ Hence, the groups $\frac{H\cap G_{i+1}}{H\cap G_i}$ are abelian so the set $\{H\cap G_i; i=0, 1, \ldots, n\}$ forms an abelian series for the group $H$, as required.
Homomorphic images: If $N$ is a normal subgroup of $G$ then the can apply the 2nd isomorphism theorem to get the following. $$\frac{G_{i+1}N}{G_iN}\cong\frac{G_{i+1}}{G_{i+1}\cap (G_iN)}$$ The subgroup $G_i$ is a subgroup of $G_{i+1}$ and of $G_iN$, and so $G_i\lhd G_{i+1}\cap (G_iN)$. Therefore, $\frac{G_{i+1}}{G_{i+1}\cap (G_iN)}$ is a homomorphic image of $\frac{G_{i+1}}{G_i}$, and hence is abelian. By the 3rd isomorphism theorem we have the following. $$\frac{G_{i+1}N/N}{G_iN/N}\cong \frac{G_{i+1}N}{G_iN}$$ Hence, the set $\{G_iN/N; i=0, 1, \ldots, n\}$ forms an abelian series for $G/N$, as required.