# Questions about 2nd isomorphism theorem

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 $$\text{ker }\phi$$?

• 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 $H$. Commented Dec 4, 2019 at 19:37
• @ArturoMagidin Sorry. Corrected that. Please check again. Commented Dec 4, 2019 at 19:41

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 $$HN/N$$.