Let $N$ and $M$ be two normal subgroups of a group $G$. Then we can show that the set $NM \colon= \{\ nm \ | \ n \in N, \ m \in M \ \}$ is a subgroup of $G$, that $M$ is normal in $NM$, and that $N \cap M$ is normal in $N$, etc.
Now can we determine if the quotient groups $NM / M$ and $N / (N\cap M)$ are isomorphic or not? In which cases are these isomorphic or non-isomorphic?
This is just the second isomorphism theorem. You can find the proof all over the place, for example here:
http://www.proofwiki.org/wiki/Second_Isomorphism_Theorem/Groups
We want to use the first isomorphism to prove this, so first we want to define a surjective homomorphism (and hope it has kernel $N\cap M$. So we proceed as follows:
We define the map $f:N\rightarrow NM/M$ such that $f(n)=nM$.
Now $f(nm)=nmM=nMmM=f(n)f(m)$ so this is indeed a homomorphism.
Moreover to check it is surjective we pick an element in $NM/M$ say $nmM$ with $n\in N$ and $m\in M$ and need to find a pre-image for it. However $nmM=nM$ so just choose $n\in N$ and we have $f(n)=nM=nmM$ and so the map is surjective.
We now apply the first isomorphism theorem to get $N/Ker(f)\cong NM/M$
Now lets think about $ker(f)$. So we have:
$Ker(f)=\{n\in N|f(n)=M\}=\{n\in N| nM=M\}=\{n\in N|n\in M\}=N\cap M$. So finally we have:
$N/M\cap N\cong NM/M$ as desired