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I have been stuck on this proof and I keep getting more and more confused.
The theorem asserts

$H$ is a subgroup of $G$ and $N\triangleleft G$. $HN$ is a subgroup of G and $(H\cap N)\triangleleft H$. Then $$H/(H\cap N)\cong HN/N$$

I understand $HN$ is a subgroup of $G$ and $H\cap N\triangleleft H$. No worries there.

  1. The proof proceeds to define a map $\phi:H\to HN/N$ such that $\phi(h)=hN$.
  2. By first isomorphism Theorem, $HN/N=\phi(H)\cong H/\text{ker }\phi$.
  3. $\text{ker }\phi=H\cap N \therefore HN/N\cong H/H\cap N$

My queries
1. From Step 1,

$\phi:H\to HN/N$ such that $\phi(h)=hN$

If $\phi(h)= hN$, then $\phi$ should be mapped from $H$ to $H/N$. I know that for a quotient group, we need a normal group. For example, if $N\triangleleft H, H/N$ is a quotient group defined as above. But following the same logic, $N$ needs to be normal to $HN$. I concluded that $N\triangleleft HN$. Is that correct?
2. Following the same vein of thought, is $H/N=HN/N$? I don't think so but maybe I am missing something.
3. From Step 3,

$\text{ker }\phi=H\cap N \therefore HN/N\cong H/H\cap N$ I know that $\text{ker }\phi \triangleleft H$ and also any group normal in H belongs to $\text{ker }\phi$. Hence $\text{ker }\phi=H\cap N$ but is it true that for any homomorphism, the kernel will include the entirety of all normal subgroups? Is it possible, say, that there is another normal subgroup in $H$ that is not part of $\text{ker }\phi$?

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    $\begingroup$ Your statement of the homomoprhism theorem is incorrect. $H\cap N$ is not necessarily normal in $G$. The correct statement is that it is normal in $H$. $\endgroup$ Commented Dec 4, 2019 at 19:37
  • $\begingroup$ @ArturoMagidin Sorry. Corrected that. Please check again. $\endgroup$
    – PythonSage
    Commented Dec 4, 2019 at 19:41

1 Answer 1

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In general, if $N\leq K\leq G$, and $N\triangleleft G$, then $N\triangleleft K$: to see this, note that for every $g\in G$ we have $gNg^{-1}=N$, and therefore for every $k\in K$ we also have $kNk^{-1}=N$ (since $k\in G$ as well). So the fact that $N\subseteq HN\subseteq G$ and that $N\triangleleft G$ guarantees that we also have $N\triangleleft HN$.

On the other hand, because you do not know if $N\subseteq H$, then you cannot state that $N\triangleleft H$: in order to be a normal subgroup, you must be a subgroup; and in order to be a subgroup, you must be a subset. Since we do not have any information about whether $N$ is contained in $H$ or not, you cannot assert that $N\triangleleft H$; in particular, "$H/N$" may not even make sense.

(Note however, that if $N\subseteq H$, then you will have $HN=H$)

I do not understand what you mean when you say "... every normal subgroup of $H$ belongs to $\mathrm{ker}(\phi)$." What does it mean for a normal subgroup to "belong" to something? It is not true in general that every normal subgroup of $H$ is contained in $\mathrm{ker}(\phi)$; if you've somehow reached that conclusion, then your argument is incorrect.


My view is that the "right" way to think about the Second Isomorphism Theorem is as a counterpart to the Lattice (or Fourth) Isomorphism Theorem. The Lattice isomorphism theorem tells you that if $N\triangleleft G$, then there is a one-to-one, inclusion preserving correspondence between subgroups of $G$ that contains $N$, and subgroups of $G/N$; and moreover that this correspondence identifies normal subgroups with normal subgroups. And that this correspondence is induced by $\phi$; that is, it also tells you what $\phi$ does to subgroups of $G$ that contain $N$.

This should lead one to wonder: "Okay, that's what is going on with subgroups of $G$ that contain $N$. I understand what $\phi$ does to subgroups of $G$ that contain $N$. What about other subgroups of $G$? What does $\phi$ do to them?"

And the Second Isomorphism tells you: what happens to $H$ is the same thing as what happens to $HN$, which happens to be a subgroup of $G$ to contains $N$; namely, $H$ is mapped to $H/(H\cap N)$, and this is isomorphic to $HN/N$.

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