Showing $(NM)/M \cong N/(N\cap M)$ for $N,M \triangleleft G$ This is problem 2.7 #6 from the second edition of Herstein's Topics in Algebra.

If $N,M$ are normal subgroups of $G$, prove that $(NM)/M \cong N/(N\cap M)$.

Any hints in the right direction?
 A: Directly and without much mulling, define:
$$\phi: NM\to N/(N\cap M)\;\;\text{by}\;\;\phi(nm)=n(N\cap M)$$
(1) Show the map is well defined, i.e.: $$nm=n_1m_1\implies \phi(nm)=\phi(n_1m_1)$$
(2) What's the kernel of $\,\phi\,$?: $\,n(N\cap M)=N\cap M\iff n\in N\cap M\iff\ldots\,$
(3) Show that $\,\phi\,$ is onto
(4) Apply the first isomorphism theorem...
Note: if you will, you could define $\,\phi(nm):=m(N\cap M)\,$...it works just the same.
A: Here's an intuitive, straightforward way to see the isomorphism. Note that for $n \in N$, you have $n(N \cap M) = nN \cap nM = N \cap nM$. So in $N/N \cap M$ multiplication looks like this:
$(N \cap nM)(N \cap n'M) = N \cap nn'M$
Also, $N \cap nM = N \cap n'M$ if and only if $nM = n'M$.
Thus you can think of the group $N/N \cap M$ as the group of elements $nM$ (where $n \in N$) under multiplication. But this is the group $NM/M = \{nM : n \in N\}$. Finding an isomorphism $\phi: N / N \cap M \rightarrow NM/M$ should be easy now.
Of course, you could skip all this and just apply the first isomorphism theorem as in the other answer.
A: Hint 1:
The final theorem in section 2.7 says: (paraphrased)

If $\phi : G \to \bar{G}$ is a homomorphism, $\bar{N}$ is a normal subgroup of $\bar{G}$, and $N$ is the preimage of $\bar{N}$ under $\phi$, then $G/N \cong \bar{G}/\bar{N}$.

With this in mind, try to design a homomorphism $\phi : NM \to N$ where $M$ is the preimage of $N\cap M$ under $\phi$. Then, by the theorem, $(NM)/M \cong N/(N\cap M)$.
Note that you must justify that $NM$ is indeed a group and that $N\cap M$ is a normal subgroup of $N$ in order for the theorem to apply.
Hint 2:

 If you're stuck, try the homomorphism $\phi(nm) = n$. Prove that $M$ is indeed the preimage of $N\cap M$ under it.

A: Hint I: Consider the homomorphism: $\phi: G\to G/M$.
Hint II: If $N\le G$, then $\phi(N)=NM/M$.
Hint III: The kernel of $\phi$ is...what? And what is the intersection of the kernel with $N$?
Hint IV: We have $\phi:N\to NM/M$, so what does isomorphism theorem tell us?  

Hint V: The answer to III is $N\cap M$, and the isomorphism theorem tells us that $N/N\cap M\cong \phi(N)$.  

Barring mistakes. Thanks in advance.
