TITLE: Visual Group Theory's Intuitive Proof - Cayley's Theorem - Nathan Carter pp. 85, Theorem 5.1
QUESTION [5 upvotes]: Theorem 5.1. Cayley's Theorem: Every group is isomorphic to a collection of permutations.

Figure 5.31. A multiplication table for the group $V_4$, with nodes numbered 1 through 4 to facilitate analyzing how the arrows permute the elements. The permutation for each arrow color is shown on the right. Figure 5.32. A multiplication table made up of the permutations created in Figure 5.31. Each cell highlights the destination to which the permutation sends 1, using the corresponding color from Figure 5.31, emphasizing what the colors of the arrows already showed: The two tables contain the same pattern.
This proof can be summarized as two steps: Create a permutation for each column in
a group’s multiplication table, and then inspect how those permutations treat the group’s
identity element. Figure 5.32 illustrates these two steps. It shows a multiplication table
comprised of the permutations from Figure 5.31, and each cell of the table highlights the
result of applying the permutation to the identity element 1. The correspondence between
Figures 5.31 and 5.32 is clear: If I remove all but the highlighted elements from Figure
5.32, all that remains is the multiplication table from Figure 5.31.
This proof is more casual but I want to understand this before Fraleigh. The aim is to prove a multiplication table made out of the permutations of any group behaves the same as the mutliplication table of any group. viz. $p_i \cdot p_j = p_k \iff i \cdot j = k.$ For some reason, I understand figures 5.31 and 5.32 but I'm still unconvinced.
(1.) I think I understand the first three paragraphs. I'm confounded by the last two. Why does the proof only "consider how the permutations treat the identity element from the original group"? What about the other elements mapped by the permutations?
(2.) The proof says applying to $p_k$ to $1$ means multiplying $1$ by $k$. Is this because the permutation for column $k$ maps $1$ to $k$? I can see this is true for $V_4$ in Figure 5.32. $V_4$ was defined to have this property. However, how's it true in general? We don't know what the permutation $p_k$ is? 
(3.) Same question as (2.) for $p_i \cdot p_j$.
(4.) I don't fully know why I'm unsettled hence what other things am I missing?

REPLY [3 votes]: I wondered the exact same thing when I read through that part of Visual Group Theory as well. Here's what I came up with after reading aPaulT's answer.

First let's just look at the multiplication table from the book and extend it with the indexes from the proof. We don't really need e.g. that $2\ast3=4$, we only need the column and row that is multiplied with the identity element, so the following argument works for any multiplication table.

$
 \begin{array}{c|cccccccccccc}
\ast & 1 & 2 & 3 & 4 & \dots & j & \dots & k & \dots & i & \dots & n\\
\hline 1 & 1 & 2 & 3 & 4 &  & j &  & k &  & i &  & n\\
2 & 2 & 1 & 4 & 3\\
3 & 3 & 4 & 1 & 2\\
4 & 4 & 3 & 2 & 1\\
\vdots &  &  &  &  & \ddots\\
j & j &  &  &  &  & 1\\
\vdots &  &  &  &  &  &  & \ddots\\
k & k &  &  &  &  &  &  & 1\\
\vdots &  &  &  &  &  &  &  &  & \ddots\\
i & i &  &  &  &  & i\ast j &  &  &  & 1\\
\vdots &  &  &  &  &  &  &  &  &  &  & \ddots\\
n & n &  &  &  &  &  &  &  &  &  &  & 1
\end{array}$

If we create permutations $p_1,p_2,\dots,p_n$ for the above multiplication table, we can rewrite the table as:

$\begin{array}{c|cccccccccccc}
 & p_{1} & p_{2} & p_{3} & p_{4} & \dots & p_{j} & \dots & p_{k} & \dots & p_{i} & \dots & p_{n}\\
\\
\ast & 1 & 2 & 3 & 4 & \dots & j & \dots & k & \dots & i & \dots & n\\
\hline 1 & p_{1}(1) & p_{2}(1) & p_{3}(1) & p_{4}(1) & \dots & p_{j}(1) &  & p_{k}(1) &  & p_{i}(1) &  & p_{n}(1)\\
2 & p_{1}(2) & p_{2}(2) & p_{3}(2) & p_{4}(2) & \dots & p_{j}(2) &  & p_{k}(2) &  & p_{i}(2) &  & p_{n}(2)\\
3 & p_{1}(3) & p_{2}(3) & p_{3}(3) & p_{4}(3) & \dots\\
4 & p_{1}(4) & p_{2}(4) & p_{3}(4) & p_{4}(4) & \dots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots\\
j & p_{1}(j) & p_{2}(j) &  &  &  & p_{j}(j)\\
\vdots &  &  &  &  &  &  & \ddots\\
k & p_{1}(k) & p_{2}(k) &  &  &  &  &  & p_{k}(k)\\
\vdots &  &  &  &  &  &  &  &  & \ddots\\
i & p_{1}(i) & p_{2}(i) &  &  &  & p_{j}(i) &  &  &  & p_{i}(i)\\
\vdots &  &  &  &  &  &  &  &  &  &  & \ddots\\
n & p_{1}(n) & p_{2}(n) &  &  &  &  &  &  &  &  &  & p_{n}(n)
\end{array}$

Notice that $p_j(i)$ is in the same spot as $i\ast j$; $p_j(i)$ has to map to $i\ast j$ because we are making the permutations be this way. Notice also that $p_k(1)$ is in the same place as $k$, it is indeed we who are defining $p_k(1)$ to be equal to $k$.
Let's look at a table for the permutations, the operation is now composition:

$\begin{array}{c|cccccccccccc}
 & p_{1} & p_{2} & p_{3} & p_{4} & \dots & p_{j} & \dots & p_{k} & \dots & p_{i} & \dots & p_{n}\\
\hline p_{1} & p_{1} & p_{2} & p_{3} & p_{4} & \dots & p_{j} &  & p_{k} &  & p_{i} &  & p_{n}\\
p_{2} & p_{2} & p_{1} & p_{2}p_{3} & p_{2}p_{4} & \dots & p_{2}p_{j} &  & p_{2}p_{k} &  & p_{2}p_{i} &  & p_{2}p_{n}\\
p_{3} & p_{3} & p_{3}p_{2} & p_{1} & p_{3}p_{4} & \dots\\
p_{4} & p_{4} & p_{4}p_{2} & p_{4}p_{3} & p_{1} & \dots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots\\
p_{j} & p_{j} & p_{j}p_{2}\\
\vdots &  &  &  &  &  &  & \ddots\\
p_{k} & p_{k} & p_{k}p_{2}\\
\vdots &  &  &  &  &  &  &  &  & \ddots\\
p_{i} & p_{i} & p_{i}p_{2} &  &  &  & p_{i}p_{j}\\
\vdots &  &  &  &  &  &  &  &  &  &  & \ddots\\
p_{n} & p_{n} & p_{n}p_{2} &  &  &  &  &  &  &  &  &  & p_{1}
\end{array}
 $

Now we just assume that $p_ip_j=p_k$. If it is true that $p_ip_j=p_k$, it must be so that $i\ast j=k$, because $p_ip_j(1)=p_k(1)\Longleftrightarrow i\ast j=k$.
Because we assumed that $p_ip_j=p_k$, this works for any element, i.e. $p_ip_j(m)=p_k(m)$. One element that $m$ can be is identity element 1, and then we have that $p_ip_j(1)=p_k(1)$, which gives us $i\ast j=k$
If $p_ip_j(1)=p_k(1)\Rightarrow i\ast j=k$, and then somehow at the same time for $r\neq 1$, $p_ip_j(r)=p_k(r)\Rightarrow i\ast j\neq k$, then what element lies in the box in the $i$th row and the $j$th column of the multiplication table?