If $N$ and $H$ are normal subgroups of $G$, then $N\cap H$ is also normal in $G$ and $G/(N\cap H)$ is isomorphic to a subgroup of $(G/N)\times(G/H)$? I've already shown that $N\cap H$ is a normal subgroup of $G$, so I just need to determine an isomorphism.
My problem is that I'm not entirely sure I understand what $(G/N)\times(G/H)$ looks like. I know that $G/N$ is the collection of left cosets of $N$ in $G$ and similarly $G/H$ is the collection of left cosets of $H$ in $G$. So what is $(G/N)\times(G/N)$ the collection of exactly?
In constructing an isomorphism I need to define bijective mapping $f:G/(N\cap H) \to (G/N)\times(G/H)$ which also preserves the operation of $G/(N\cap H)$. I feel I need to understand the structure of the $(G/N)\times(G/H)$ in order to do this.
 A: Well to see that $N\cap H$ is normal (it is obviously a subgroup), let $g\in G$ and $s\in N\cap H$. Since $s\in N$ and $N$ is normal, $gsg^{-1}\in N$. Since $s\in H$ and $H$ is normal, $gsg^{-1}\in H$. Thus $gsg^{-1}\in H\cap N$. This is precisely the definition of a normal subgroup.
Define $\phi:G\to G/H\times G/N$ by $g\mapsto (gH, gN)$. This map is a well-defined group homomorphism. It is not true that $\phi$ is surjective in general. The kernel of this map is the set of $s\in G$ such that $sH=H$ and $sN=N$, which is precisely the set $H\cap N$, i.e. $\ker(\phi)=H\cap N$ (since kernels of group homomorphisms are always normal subgroups, this is another proof that $H\cap N$ is normal).
Let's see why this map is not necessarily surjective. Simply take an abelian group, say $G=\mathbb{Z}$ so every subgroup is normal and then take $H=\{0\}$, the trivial subgroup and $N=2\mathbb{Z}$. Then $\phi$ is the homomorphism
$$\mathbb{Z}\to\mathbb{Z}\times\mathbb{Z}_2$$
defined by $n\mapsto(n,[n]_2)$. In particular, the element $(2,[1])$ of $\mathbb{Z}\times\mathbb{Z}_2$ does not lie in the range of $\phi$.
Let's also see why $\frac{G}{H\cap N}$ is not isomorphic to $\frac{G}{H}\times\frac{G}{N}$ in general. In the same example, if that were true we would have that $\mathbb{Z}\cong\mathbb{Z}\times\mathbb{Z}_2$. This is obviously false (why?).
