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.