Show an isomorphism If $N,M$ are normal subgroups of $G$ then $\frac{NM}{M} \simeq \frac{N}{N\cap M}$
I have been trying to build a function from $N$ to $\frac{NM}{M}$ and looking at is kernel, but i'm.strugling at what $\frac{NM}{M}$ looks like
 A: The idea is very good.
You can first observe that $NM=\{xy:x\in N, y\in M\}$. Clearly, the right-hand side is a subset of $NM$ and contains both $N$ and $M$. On the other hand, a product of the form $(x_1y_1)(x_2y_2)$, with $x_1,x_2\in N$ and $y_1,y_2\in M$, can be rewritten as
$$
\underbrace{x_1\underbrace{(y_1x_2y_1^{-1})}_{\in N}}_{\in N}
\underbrace{(y_1y_2)}_{\in M}
$$
from which it easily follows that the right-hand side is closed under products. It's also closed under inverses, because, for $x\in N$ and $y\in M$,
$$
(xy)^{-1}=y^{-1}x^{-1}=
\underbrace{y^{-1}xy}_{\in N}
y
$$
Hence the right-hand side is a subgroup of $G$ containing both $N$ and $M$, so it coincides with $NM$.
In particular an element of $NM/M$ is of the form $xyM=xM$, for $x\in N$ and $y\in M$.
Hence the surjective homomorphism $N\to NM/M$ you are looking for is
$$
x\mapsto xM
$$
whose kernel is
$$
\{x\in N:x\in M\}=N\cap M
$$
The homomorphism theorem now shows $N/(N\cap M)\cong NM/M$.
Note that it is not needed to assume $M$ is normal in $G$, but just a subgroup; only normality of $N$ is required.
A: In particular, $NM=\{nm \, | \, n \in N \text{ and } m \in M\}$, therefore the quotient set (in this case quotient group as things are given to be normal) $$NM/M=\{n\color{red}{mM} \, | \, n \in N \text{ and } m \in M\}=\{nM \, | \, n \in N\}$$
Now start with $N \longrightarrow NM/M$ by using the map $\phi(n)=nM$ (so you are sending an element of $N$ to the coset it belongs to). It is easy to see that this is a surjective homomorphism with kernel $N \cap M$ (check it yourself!!). Now apply the first isomorphism theorem.    
A: Let $\pi: G \to G/M$ be the canonical projection with kernel exactly $M$. Since $\pi$ is surjective, $\pi\restriction_N: N \to \pi(N)$ is surjective. By the first isomorphism theorem, 
$$
\pi(N) \simeq N/\ker{\pi\restriction_N} = N/(\ker \pi \cap N) = N/(M \cap N) 
$$ 
On the other hand, we can restrict the codomain, giving another surjective morphism,
$$
\pi \restriction^{\pi(N)} : \pi^{-1}(\pi(N)) \to \pi(N)
$$ 
and since $M = \ker \pi \subseteq \pi^{-1}(\pi(N))$, then $\ker \pi\restriction^{\pi(N)} = M$. Moreover, we have that $\pi^{-1}(\pi(N)) = NM$. In effect, from
$$
\pi(NM) = \pi(N)\cdot \{1\} = \pi(N)
$$
we see that $NM \subseteq \pi^{-1}(\pi(N))$ and the other inclusion is given by the fact that, if $\pi(x) \in \pi(N)$, there exists $n \in N$ with $\pi(x) = \pi(n)$, and so $xn^{-1} \in \ker\pi = M$ which implies $x \in M\cdot n \subseteq MN$. Thus once again, by the first isomorphism theorem,
$$
NM/ M \simeq \pi(N)
$$
and so
$$
NM/M \simeq\pi(N) \simeq N/(M\cap N)
$$
as desired.
