My group theory book teaches a strategy how to write a cycle as a series of transpositions. Although the strategy is not hard to understand, I appreciate that they included a proof that this strategy actually works. However I wasn't able to follow it until the end, so I would appreciate if someone could help me out.
The strategy and the proof as follows:
Theorem: If $a_1, a_2,..., a_r$ are symbols (where $r \ge 2$), then the composite of transpositions:
$$ \begin{pmatrix}a_1 &a_r \end{pmatrix} \circ \begin{pmatrix}a_1 &a_{r-1} \end{pmatrix} \circ \text{...}\circ \begin{pmatrix}a_1 &a_3 \end{pmatrix} \circ \begin{pmatrix}a_1 &a_2 \end{pmatrix} $$
is equal to the cycle
$$ \begin{pmatrix}a_1 & a_2 & ... & a_r \end{pmatrix} $$
Proof:
First we consider the symbol $a_1$. The first transposition $\begin{pmatrix}a_1 &a_2 \end{pmatrix}$ maps $a_1$ to $a_2$ and the remaining transpositions map $a_2$ to itself, so the composite maps $a_1$ to $a_2$.
Now we consider any symbol $a_s$, where $2 \le s \le r-1$. We see that:
- Each of the transpositions $$ \begin{pmatrix}a_1 &a_2 \end{pmatrix}, \begin{pmatrix}a_1 &a_3 \end{pmatrix}, ..., \begin{pmatrix}a_1 &a_{s-1} \end{pmatrix} $$ maps $a_2$ to itself.
Here I'm already getting some doubts. If we let $s=2$ then the transposition $\begin{pmatrix}a_1 &a_2 \end{pmatrix}$ certainly doesn't map $a_s = a_2$ to itself, it maps it to $a_1$. Or am I misunderstanding what they are saying?
The proof continues:
- the next transposition $\begin{pmatrix}a_1 &a_s \end{pmatrix}$ maps $a_s$ to $a_1$.
- then the next transposition $\begin{pmatrix}a_1 &a_{s+1} \end{pmatrix}$ maps $a_1$ to $a_{s+1}$
- and each of the remaining transpositions $$ \begin{pmatrix}a_1 &a_{s+2} \end{pmatrix}, ..., \begin{pmatrix}a_1 &a_r \end{pmatrix} $$ maps $a_{s+1}$ to itself. Hence the composite maps $a_s$ to $a_{s+1}$. It remains to find the image of $a_r$. [...]
I left out the last part because that is actually clear to me. I'm not following the line of reasoning that tries to show the composite maps $a_s$ to $a_{s+1}$.
In answer to your first question, note that the list of transpositions only goes up to $(a_1 \, a_{\color{red}{s-1}})$. So, if we consider $s = 2$, then the list would not contain $(a_1 \, a_2)$, as this is already $(a_1\, a_s)$. In fact the list would be empty. If $s = 3$, then there would only be one transposition in the list, i.e. $(a_1 \, a_2)$.
The claim that this list of transpositions fails to affect $a_s$ is indeed correct. The only symbols that appear in this list of transpositions are $a_1, \ldots, a_{s - 1}$, none of which are $a_s$ (unless $s = 2$, in which case the list is empty!).
To illustrate the argument, we are attempting to compute $$\color{blue}{(a_1 \, a_r)(a_1 \, a_{r-1})\ldots(a_1 \, a_{s+2})}(a_1 \, a_{s+1})(a_1\, a_s)\color{red}{(a_1 \, a_{s-1})(a_1 \, a_{s-2})\ldots(a_1 \, a_{2})}a_s.$$ According to the first dot point, none of the red transpositions will affect $a_s$, swapping only other symbols. So, partially evaluating this yields $$\color{blue}{(a_1 \, a_r)(a_1 \, a_{r-1})\ldots(a_1 \, a_{s+2})}(a_1 \, a_{s+1})(a_1\, a_s)a_s.$$ As per the second dot point, $(a_1\, a_s)$ maps $a_s$ to $a_1$, so we get $$\color{blue}{(a_1 \, a_r)(a_1 \, a_{r-1})\ldots(a_1 \, a_{s+2})}(a_1 \, a_{s+1})a_1.$$ Then, $(a_1 \, a_{s+1})$ maps $a_1$ to $a_{s+1}$ (dot point 3), so we get $$\color{blue}{(a_1 \, a_r)(a_1 \, a_{r-1})\ldots(a_1 \, a_{s+2})}a_{s+1}.$$ Finally, dot point 4 states that none of the blue-coloured transpositions affect $a_{s+1}$ either. Note that the symbols involved here are $$a_1, a_{s+2}, a_{s+3}, \ldots, a_r,$$ (unless $s = r - 1$, in which case the list is empty!), none of which are $a_{s+1}$. So, since all of these transpositions fix $a_{s+1}$, we simply get $a_{s+1}$ at the end. Thus $a_s \mapsto a_{s+1}$ as claimed.