In the Correspondence Theorem, does the modulo subgroup need to be the kernel? Let $f: G \twoheadrightarrow H$ be a surjective homomorphism with kernel $K$. If $S$ is a subgroup of $G$ which contains $K$, and $S^*=\{f(x) \in H \mid x \in S\}$, then $S/K \cong S^*$. This is how my book states the Correspondence Theorem.  However, no other sources say that $K$ must be the kernel of $f$. Other sources just say that $K \vartriangleleft G$ and $K \subset S$.
What is going on here? Does $K$ need to be the kernel, or just a normal subgroup contained in $S$?

Yes, $K$ must be the kernel of $f$.
But, to be clear, this isn't really a restriction.  When talking about the quotient group $G/K$, where $K$ is any normal subgroup, there is a natural map $f\colon G \to G/K$ defined by $x \mapsto xK$.  This map is a surjective homomorphism and it's kernel is exactly $K$!
So in one sense you could say that in the correspondence theorem you choose $f$ and then the subgroup $K$ must be the kernel of $f$.  But in another sense you could say that you choose $K$ and then the map $f$ is, up to isomorphism, just the unique surjective map with kernel $K$.  So $f$ and $K$ have to "match up" in the end, but this isn't a restriction.  You can choose either and then the other is determined uniquely (up to isomorphism).
