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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 $ker(H)$?)

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  • $\begingroup$ You missed a character in your statement of the theorem. (You mean to say $H/(H\cap N)$). $\endgroup$ – Jonny Lomond Mar 23 '14 at 0:07
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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.

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    $\begingroup$ +1. Nice explanation. Does the third isomorphism theorem (in my post) also come from a similar, natural problem? $\endgroup$ – hengxin Apr 23 '14 at 11:28
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    $\begingroup$ @hengxin: Yes, actually. After asking following question you can reach the result. We know that image of subgroup is subgroup of image. What about normal subgroups? Is image of normal subgroup is normal in image ? Or Inverse image of normal is normal in G?. After that when you answer these questions, you should wonder is there any relation between the quatient in $G$ and the quatient in $G/K$. $\endgroup$ – mesel Apr 24 '14 at 18:08
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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 $HN/N$.

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  • $\begingroup$ It may help to describe $HN$ as the set of all elements of $G$ that are equivalent (modulo $N$) to an element of $H$. So switching from $H$ to $HN$ doesn't change anything from the perpsective of working modulo $N$, but it is a nicer object to work with because it contains $N$. $\endgroup$ – Hurkyl Mar 23 '14 at 10:19
  • $\begingroup$ @Hurkyl I find it more interesting to combine the explanation on $H/(H \cap N)$ of sea turtles with that on $HN/N$ of mesel. $\endgroup$ – hengxin Apr 23 '14 at 11:49

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