Interpretation of Second isomorphism theorem

Solution 1:

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.

Solution 2:

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 $HN/N$.