Confusion with showing that $HN/N \cong H/(H \cap N)$ Suppose $G$ is a group, $H \le G, $ and $N \triangleleft G$.
Show that
$$HN/N \cong H/(H \cap N)$$
Well, I know there was a similar question on math.stackexchange and I know there is a solution by first isomorphism theorem.
But I want to solve it using another method (by definitions).
I'll assume that we know that $HN$ is a subgroup in $G$ and that $N$ is a normal subgroup in $HN$ (since it is normal in $G$).
Firstly, using the definition of quotient:
$$HN/N=\{hnN\mid n\in N, h \in H\}$$
Since $nN=N$ we can say that
$$
HN/N=\{hN\mid h \in H\}
$$
But then it comes that $HN/N = H/N$ and not $HN/N=H/(N \cap H)$. Well, actually $H/N$ may not be even a group. But it seems all steps were right. Where is a bug here?
 A: First, you should really not write $H/N$, because this notation assumes that $N\subseteq H$, and that is not the case here. You can certainly look at cosets of $N$ represented by elements of $H$.
But now let's look more carefully at what you have. Just because you have $h\neq h'$, it does not follow that $hN\neq h'N$, right? So what cosets do you really have?
Well, $hN = h'N$ if and only if $h'^{-1}h\in N$. But certainly $h'^{-1}h\in H$, so you will in fact have $hN=h'N$ if and only if $h'^{-1}h\in H\cap N$.
That means that in fact you don't have as many cosets as there are elements of $H$: you really just have as many cosets as you have cosets of $H\cap N$ in $H$. Hmm.... where did we see a connection between those cosets again?
Second... you say "Well, $H/N$ may not even be a group"... actually, remember that "$H/N$" is a misnomer. It's really $HN/N$, and that is definitely a group, because it is a subgroup of $G/N$. If you define the operation on your collection of representatives, all of the form $hN$ with $h\in H$, by
$$(h_1N)(h_2N) = (h_1h_2)N$$
you will find that this is in fact a well-defined operation that gives a group. Just remember that you have multiple names for each element!
In fact, we just have one coset of $N$ represented by an element of $H$ for each coset of $H\cap N$ in $H$. It would be interesting to see whether, if we choose representatives for the cosets of $H\cap N$ in $H$ and use them and them alone, we get the same definition of the operation...
Note that the assertion is that you have an isomorphism, not an equality. So the fact that the set $\{hN\mid h\in H\}$ is not literally equal to the set $\{h(N\cap H)\mid h\in H\}$ is not an obstacle for there to be an isomorphism between them. You need a bijection between the two sets that respects the operations defined on them, of course, but you don't need an actual equality of underlying sets.
