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 $S$?
On
Let's pick a concrete example. We'll use $G = D_8$, the dihedral group with $8$ elements, and let's use the homomorphism $f: G \to (\{\pm1\}, \times)$ which sends rotations to $1$, and reflections to $-1$.
So, let's just let all of $G$ be our subgroup containing $\ker f$,the set of rotations; we'll let $S = G$.
Thus, $S^* = \{f(x) \in H : x \in S\} = \{\pm 1\}$. Then we do indeed have $G/\ker f \cong S^*$.
Now, if instead of modding by $\ker f$, we choose the normal subgroup $\{1, r^2\}$, then it's clear that $G/\{1, r^2\}$ is too big to be isomorphic to $S^*$; the quotient group having $4$ elements but $S*$ only having two.
The correspondence theorem really does require us to look at $S/\ker f$ when we're $f$ to define using $S^* = \{f(x) \in H : x \in S\}$.
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).