p. 146: We should acquire an intuitive feeling for this theorem in terms of $\color{red}{collapsing}$ one of the factors to the identity element.
p. 147 15.8 Theorem: $\hat{H} = \{(h, e) : h \in H\} \unlhd H \times K$, the direct product of groups H and K . Also $\frac{H \times K}{\hat{H}} $ is isomorphic to K in a natural way. Similarly, $\frac{H \times K}{\hat{K}} \simeq H$ is in a natural way.
Proof: Consider the homomorphism $\pi_2 : H x K \to K$, where $\pi_2(h, k) =k$.
(See Example 13.8). By means of p. 132 Corollary 13.20, $\ker$ is a normal subgroup hence $\ker(\pi_2) = H \; \unlhd H x K$. Because $\pi_2$ is onto K, Theorem 14.11 tells us that $\frac{H \times K}{K} \simeq H$. ♥
(1.) Can someone please flesh out the intuition? I understand the proof hence not asking about it.
(2.) Can someone please unfold $\color{red}{collapsing}$ ? What does it mean here?
(3.) What's 'natural way' here? Before finding a homomorphism and proving, how do you envisage and envision $\frac{H \times K}{\hat{H}}, \frac{H \times K}{\hat{K}}$ are isomorphic to $K, H$ respectively? Does 'natural way' mean this?
(4.) Why fret about $\hat{H}, \frac{H \times K}{\hat{H}}, \frac{H \times K}{\hat{K}}$? What do they mean intuitively?
Update Fev. 21 2014: I just want to clarify Sangeeta's exquisite informal answer.
(1) What this means is that you started out with a group of many small dots, but you were able to reorganize the dots in such a way that the new set of blobs you have is a group. And the normal subgroup of the original group is the identity of this group.
(1.1.) What are the 'many small dots'? Elements of the group? Normal subgroup?
(3) Now you could "factor out" $H \times K$. How? Collect all the pairs where $k=k_1$ say. In this collection your first co-ordinate will vary over all of $H$, but we don't care. What we will do is collect all the pairs of $H \times K$ in this way and send each blob to that element of $K$ which is in the 2nd place. This is the natural way of doing it, and this is what they mean when they say $\pi_2 (h,k)=k$.
(3.1) What does "factor out" mean here intuitively?
(3.2) What's $k = k_1$?
(3.3) What is the 'blob' here?
The intuition: factoring out a group by a normal subgroup is like treating a big blob as identity. "Factoring out" a group is an informal way of referring to the process of forming the quotient group. When I say "blob", I am simply referring to a collection of group elements, it is nothing formal and does not have a mathematical meaning in this context. Rather think of the english meaning of the word "blob". Copies of that blob (normal subgroup) are other big blobs (cosets of the normal subgroup). The things in the blob move together - no matter how you multiply them, integrity of the blobs is not violated i.e. blobs do not break up. What this means is that you started out with a group of many small dots, but you were able to reorganize the dots in such a way that the new set of blobs you have is a group. And the normal subgroup of the original group is the identity of this group.
That is what is collapsing. The normal subgroup may have a million dots in it, but in your new re-organized group it has been collapsed into identity element of the group. Similarly all its cosets have collapsed into single group elements.
Think about $H \times K$. This is the set of all ordered pairs of the form $(h,k)$ where $h \in H, k \in K$. In this set of ordered pairs collect all the things that have the identity element of $K$ in the 2nd place. You would think that this blob should be like $H$ and it is. Outside this blob are all pairs like $(h,k)$ where $k \neq e$. Now you could "factor out" $H \times K$. How? Collect all the pairs where $k=k_1$, where $k_1$ is any other element of the group $K$. In this collection your first co-ordinate will vary over all of $H$, but we don't care. What we will do is collect all the pairs of $H \times K$ in this way and send each blob to that element of $K$ which is in the 2nd place. This is the natural way of doing it, and this is what they mean when they say $\pi_2 (h,k)=k$.
Read ahead and you will understand the significance of direct products.