Usefulness of a theorem involving quotient groups In my class today, we proved the following theorem:

Suppose $N \trianglelefteq G, \ N \leq K \leq G$. Let $q: G \rightarrow G/N$ be the quotient map. Then $\phi: K/N \rightarrow q(K): kN \rightarrow kN$ is an isomorphism.

Isn't this theorem obvious though? If I'm not mistaken $q(K) = K/N$, so obviously these two will be isomorphic. What am I missing?
 A: K presumably is a subgroup of G.  In order for the map you've indicated to be an isomorphism its necessary (and sufficiient) for N to be normal in K.
A: The motivating question here is: what are the subgroups of $G/N$? We know that in general the image of a homomorphism is a subgroup of the codomain, so in particular, given $q\colon G\to G/N$, one subgroup of $G/N$ is $q(G)$ which is the entire group $G/N$.
But what are the other subgroups of $G/N$? This theorem is saying that, if we start with a subgroup $K$ of $G$ such that $K$ contains $N$, then the map $\phi\colon K/N\to q(K)$ with $kN\mapsto q(k)$ is (1) well defined, (2) a group homomorphism, and (3) an isomorphism. This implies that $q(K)$ is precisely the subgroup $K/N$. The subtle points here are (1) and (2), as $\phi$ wouldn't be a well defined group homomorphism had $K$ not contained $N$ as a subgroup.
So are there other subgroups of $G/N$, i.e. such that they don't come from a subgroup $K$ of $G$ containing $N$? No, there are no such subgroups. This fact you might have already seen in your class, or will see very soon.
With this fact in mind, one might argue that the non-obvious facts are in the background, i.e. that there is a 1-1 correspondence between the subgroups of $G/N$ and the subgroups of $G$ containing $N$, and the correspondence is given by $K \leftrightarrow K/N $).
