I know of this version of the correspondence theorem for Groups (From Herstein's Abstract Algebra):
''Let $\phi$ be a group homomorphism from G onto G' with kernel $K$. If $H' \leq G'$ and $H = \{ a \in G : \phi(a) \in H '\}$, then $H$ is a subgroup of $G$ that contains $K$ and $H/K \cong H'$.
To illustrate this, consider the following homomorphism $\phi$ from $S_4$ to $S_3$: Partition the set of 4 indices $\{1,2,3,4\}$ into pairs of subsets in the following way:
$\Pi_1 = \{1,2\} \cup \{3,4\}$
$\Pi_2 = \{1,3\} \cup \{2,4\}$
$\Pi_3 = \{1,4\} \cup \{2,3\}$.
Then take say the cycle $(1234)$ in $S_4$ and let it act on each of the $\Pi_i's$.
It should be immediate that $(1234)$ switches $\Pi_1$ and $\Pi_3$ while fixing $\Pi_2$. So we see can define $\phi$ by the action of $(1234)$ on the $\Pi_i's$ (I don't even know if this is correct terminology). In addition we see that $(1234)$ is mapped to $(13)(2)$ in $S_3$.
So now if I consider the subgroup $H' = \{e, (12)\}$ of $S_3$, we can see that the kernel of $\phi$ contains the elements $\{e, (12)(34), (13)(24), (14)(23) \}$. So if we form the quotient $H/K$, it will just contain the elements $\{ [e], [(1324)] \}$. It is then apparent that $H/K$ is isomorphic to $H'$.
However I also know that there is a one to one correspondence between the subgroups of $S_4$ containing the kernel and $S_3$. For example another such correspondence would be between the alternating group $A_4 \leq S_4$ and the subgroup in $S_3$ generated by the cycle $(123)$.
For any groups $G$, $G'$ and $H, H'$ defined as above, how is $H/K \cong H'$ equivalent to there being a bijective correspondence between the subgroups of $G'$ and those of $G$ containing the kernel?
Thanks.