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.
Best Answer
As stated, it is false that if $\phi\colon G\to G'$ is a group homomorphism, then there is a bijective correspondence between the subgroups of $G'$ and those subgroups of $G$ that contain $\mathrm{ker}(\phi)$. What you are forgetting is that for the correspondence to exist, you must have $\phi$ onto $G'$.
For instance, take $G$ to be any group, $G'$ to be a nontrivial group, and $\phi$ to be the zero morphism. The only subgroup of $G$ that contains $\mathrm{ker}(\phi)$ is $G$ itself, but $G'$ has at least two subgroups (the trivial one and $G'$) so there can be no bijection.
What is true, however, is that there is a bijection between the subgroups of $\phi(G)$ and the subgroups of $G$ that contain $\mathrm{ker}(\phi)$.
The reason this follows from the theorem given by Herstein is that the inverse image function respects set-inclusions, and so if $K,K'$ are subgroups of $\phi(G)$, then $K\subseteq K'$ if and only if $\{g\in G\mid \phi(g)\in K\}\subseteq \{g\in G\mid \phi(g)\in K'\}$. The theorem from Herstein guarantees that these inverse images are subgroups that contain $\mathrm{ker}(\phi)$. This shows that there is an injection from the collection of subgroups of $\phi(G)$ and the collection of subgroups of $G$ that contain $\mathrm{ker}(\phi)$, and that this injection is inclusion preserving. On the other hand, if $H$ is any subgroup of $G$ that contains $\mathrm{ker}(\phi)$, then $\phi^{-1}(\phi(H)) = H$, and if $H\neq H'$ and both contain the kernel of $\phi$, then $\phi(H)\neq \phi(H')$. So the direct image is an injection going the other way, which is also inclusion preserving, and this gives you the bijection.