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?


  • $\begingroup$ I don't understand why your penultimate paragraph begins with "However..."; how does the correspondence somehow "clash" with anything you've said up to that point? Or for that matter, I don't understand how everything you say about the map from $S_4$ to $S_3$ is related to your final question about why the lattice isomorphism theorem is equivalent to the statement given. $\endgroup$ Sep 7, 2011 at 4:37
  • $\begingroup$ @ArturoMagidin That map was just an example of a homomorphism from S4 to S3 that was onto. It was just an example used in Artin's Algebra to illustrate the correspondence theorem. By the way you're right the map has to be onto. I forgot to add that in the hypothesis, I'm sorry. $\endgroup$
    – user38268
    Sep 7, 2011 at 5:20

1 Answer 1


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.

  • $\begingroup$ I think I'm missing something, I get the bit on how the inverse images are subgroups that contain the kernel. Where does the bit on $H/K \cong H'$ come into all of this? $\endgroup$
    – user38268
    Sep 7, 2011 at 5:25
  • $\begingroup$ @D B Lim: The important part is that the isomorphism is induced by $H'$. Since we are assuming $\phi$ is onto, we deduce that $H\mapsto \phi(H)\mapsto \phi^{-1}(\phi(H))$ is the identity on subgroups that contain $K$; the fact that $\phi$ induces the isomorphism $H/K=H'$ is used to show that $H'\mapsto \phi^{-1}(H')\mapsto \phi(\phi^{-1}(H'))$ is also the identity on subgroups of the image. $\endgroup$ Sep 7, 2011 at 15:46
  • $\begingroup$ Did you mean to say that $\phi$ induces the isomorphism $H/K \cong H'$? How can I use $H/K \cong H'$ to show that $H'\mapsto \phi^{-1}(H') \mapsto \phi(\phi^{-1}(H'))$? $\endgroup$
    – user38268
    Sep 7, 2011 at 22:16
  • $\begingroup$ @DBLim: (i) Yes, I did. (ii) What you quote does not need "showing": it describes the composition of the correspondence that goes from $H'$ to $H=\{g\in G\mid \phi(g)\in H\}$, followed by the one that goes from $H$ to $\phi(H)$. But $\phi(H)$ is precisely $H'$ by the fact that $\phi$ is what induces the isomorphism $H/K\cong H'$. So that means that the composition of these two correspondences (between subgroups of $\phi(G)$ to subgroups of $G$ that contain $\mathrm{ker}(\phi)$, followed by the one from the latter to subgroups of $\phi(G)$) is the identity. $\endgroup$ Sep 8, 2011 at 2:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy