Proving an isomorphism Let $N$ be a normal subgroup of $G$ and $G$ a group. Let $H$ be a subgroup of $G$ with $|(N)||(H)|=|(G)|$ and $N\cap H=\{1\}$.
Why is $G=HN$ and why are two subgroups of $G$ isomorphic?
Thanks a lot!
 A: We have $|G| = |N||H|$ and $N \cap H$ is trivial. To see that $G=HN$, consider that the number of possible products of elements in $H$ and elements of $N$ is $|G|$. Suppose that $hn$ and $h'n'$ are two distinct products, and that $hn = h'n'$. Use the normality of $N$ to prove that $h=h'$ and $n=n'$. Then $NH$ has $|G|$ many elements, and hence must be the entire group.
A: Hint: Consider $\phi: N \times H \to G$ given by $(n,h) \mapsto nh$. Prove that $\phi$ is injective. Conclude that it is surjective. Note that $\phi$ is not necessarily a homomorphism.
A: There is a result in every basic group theory book (for example, Isaacs Algebra book 4.17) that $|HN|=\frac{|H||N|}{|H\cap N|}$ (at least, for finite groups). 
In this case, that boils down to "$|HN|=|H||N|=|G|$". With $HN\subseteq G$, $|HN|=|G|$ implies $HN=G$.
The last part of your question is unclear, but there is something obvious that we can say which you might be looking for.
By an isomorphism theorem, $G/N=HN/N\cong H/(N\cap H)\cong H$.
