[Math] Interpretation of Second isomorphism theorem

abstract-algebragroup-theory

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 $ker(H)$?)

Best Answer

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.