Interpretation of Second isomorphism theorem I have a question about the Second Isomorphism Theorem.(Actually my book called it the first), namely, let $G$ be a group, $N$ is a normal subgroup of $G$, and let $H$ be any subgroup of $G$, then $ (HN)/N \cong (H/ (H \cap N))$. So what's the main argument the theorem want to tell?
I understand that the first homomorphism theorem (For a homo from group $G_1$ to $G_2$, $G_1/ker(\phi) \cong \phi(G_1)$) basically try to describes the image of $G_1$ by using the partitions by $ker(\phi)$. So what about the Second Isomorphism Theorem? Is it only a "formula" like theorem ? Is $N$ being normal the key in this theorem? (Else $H \cap N $ is not the $ker(H)$?)
 A: The intuition, to my mind, is it describes two different ways of thinking about "$H$ mod $N$." Note the theorem is true even $N$ isn't normal, we just have to interptet $\cong$ differently. Quantitatively, the coset spaces are in canonical bijection, and qualitatively it is an isomorphism of so-called "$H$-sets" which means sets equipped with an action of the group $H$ (here, by left multiplication on cosets). 
One way to interpret the phrase "$H$ mod $N$" is to mod out $H$ by the relation that two elements are congruent if they are "related" by an element of $N$ (equivalently, define the same coset of $N$). It's not possible to get from one element of $H$ to another through an element outside of $H$, so all of the elements of $N$ outside of $H$ are irrelevant, so deleting superfluous data, we're really thinking of "$H$ mod $H\cap N$" which is already described as a coset space.
The second way is to project $H$ onto the space of cosets of $N$. These are collected simply in $HN/N$.
A: I guess it comes from very natural problem. 
Let $\phi$ be cononical homomorphism from $G$ to $G/N$. Let $H$ be any subgroup of $G$.
Question is that what is the image of $H$? If $N\leq H$ then answer is simple $\phi(H)=H/N$.
What if $H$ does not contain $N$? We can find answer in two different way and it gives us an equality. 
$1) $ image of $HN$ and $H$ are same since $\phi(hn)=\phi(h)\phi(n)=\phi(h)$ and since $HN$ includes $N$, $\phi(H)=\phi(HN)=(HN)/N$
$2)$ Let restriction of $\phi$ on $H$ is $f$ then $f$ is an homomorphism from $H$ to $G/N$.   What is the kernel of $f$? $Ker(f)=H\cap N$. Then by first ismomorphim theorem $f(H)\cong H/(H\cap N)$.
From $1$ and $2$, we have desired result.
