Non-unique representation of permutations.

Question

Permutations are stated to be having non-unique representation as given below, in the text of Stahl, titled: Introductory Modern Algebra, on pg. #164; as shown here.

$\begin{pmatrix} 1 & 2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & 3 \end{pmatrix}= \begin{pmatrix} 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 3 \end{pmatrix} \begin{pmatrix} 1 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 3 & 4 \end{pmatrix} \begin{pmatrix} 2 & 4 \end{pmatrix} \begin{pmatrix} 3 & 4 \end{pmatrix}= \begin{pmatrix} 2 & 3\end{pmatrix} \begin{pmatrix} 1 & 3 \end{pmatrix} \begin{pmatrix} 2 & 3 \end{pmatrix} \begin{pmatrix} 3 & 4 \end{pmatrix} \begin{pmatrix} 2 & 4 \end{pmatrix} \begin{pmatrix} 3 & 4 \end{pmatrix}$

Am unable to understand it, and want to understand some part of it to get the reasoning behind it:
1. Say, can understand the starting part, i.e. for $\begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & 3 \end{pmatrix}$, have first (taking mappings occurring from right cycle to left) $\begin{pmatrix} 2 & 3 \end{pmatrix}$, then $\begin{pmatrix} 1 & 2 \end{pmatrix}$; i.e. first $2 \longrightarrow 3$, then $1 \longrightarrow 2$.
2. But, the next one is not comprehensible to me, as $\begin{pmatrix} 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2\end{pmatrix}$; which should mean that $1 \longrightarrow 2, (2\longrightarrow 1)$, then $1 \longrightarrow 3, (3\longrightarrow 1)$. This means $1$ maps to both $2,3$ in that ordering.
3. Similarly, the 3rd one is confusing, as it states $1 \longrightarrow 3, (3\longrightarrow 1)$, then $2 \longrightarrow 3, (3\longrightarrow 2)$.
This means $3$ is mapped to by both $1, 2$ in that ordering.
I hope that if the confusion in the 2nd, 3rd points above are removed; then would help me in understanding the rest too.

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In view of the answer by @G. Smith, have made an addendum. I want to make an algorithmic sort of evaluation for the same, & request modification of my method below:
Assumptions:- There is going to form relationships between elements in a cycle, with origin_end, & recipient_end, shown as left & right member of the arrow. so, $a\longrightarrow b$ has $a$ as the origin_end, $b$ as the recipient_end.

1- Proceed from right to left, taking one cycle at a time with right cycle given priority to decide
$\,\,\,\,$ relationship over left one for common origin_end, or recipient_end between $\ge$ two cycles.
2- $\,\,\,\,$ List the cycle members with two sets of relationships,
$\,\,\,\,\,\,\,\,\,$ (a) explicitly given, (b) implicitly given.
$\,\,\,\,\,\,\,\,\,$ E.g. $(a\,\, b)$ with explicit relation as $a \longrightarrow b$, & implicit relation as $b \longrightarrow a$.
$\,\,\,\,\,\,\,\,\,$ The implicit relation can be modified by earlier left, or further right cycles.
3- $\,\,\,\,$ The

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Update: In view of the answer by @Siong, have got a graph approach with vertices representing nodes, and edges representing the transpositions given. The edges are bi-directional, and the graph is limited by the transpositions (edges) given between the vertices.

The graph for the example 1.1 (2nd) link referred in comments below, is shown by a graph here. The transpositions are numbered from left-to-right, and the common ones are having the same edge; e.g. edge #1,4,10 stand for edge between nodes: 1, 2.

Answer 1

$\begin{pmatrix} x_0 & x_1 & \ldots & x_{n-1}\end{pmatrix}$ means $x_i \to x_{i+1 \pmod n} $ and if any term doesn't appear, identity is being applied to the element.

Hence $\begin{pmatrix} x_0 & x_1 \end{pmatrix}$ means $x_0 \to x_1$ and $x_1 \to x_0$.

Given an element $x$,$$\begin{pmatrix} x_0 & x_1\end{pmatrix}x=\begin{cases} x_1 & , x=x_0 \\ x_0 &, x=x_1 \\ x &, \text{Otherwise.}&\end{cases}$$

Now let's evaluate $\begin{pmatrix}2 & 3 \end{pmatrix}\begin{pmatrix}1 & 3 \end{pmatrix}$.

$$\begin{pmatrix}2 & 3 \end{pmatrix}\begin{pmatrix}1 & 3 \end{pmatrix}[1]=\begin{pmatrix}2 & 3 \end{pmatrix}[3]=2$$ $$\begin{pmatrix}2 & 3 \end{pmatrix}\begin{pmatrix}1 & 3 \end{pmatrix}[2]=\begin{pmatrix}2 & 3 \end{pmatrix}[2]=3$$ $$\begin{pmatrix}2 & 3 \end{pmatrix}\begin{pmatrix}1 & 3 \end{pmatrix}[3]=\begin{pmatrix}2 & 3 \end{pmatrix}[1]=1$$

We can see that it is equivalent to $\begin{pmatrix} 1 & 2 & 3\end{pmatrix}$.

Answer 2

Your explanation of the first one isn't clear, so I think that's why you don't understand the second one.

Let's look at the first one, $(12)(23)$. What happens to $1$, when first applying $(23)$ and then $(12)$? Well, $(23)$ doesn't change $1$ and then $(12)$ maps $1$ to $2$. So $1\to2$. What happens to $2$? $(23)$ maps it to $3$ and then $(12)$ doesn't change $3$. So $2\to3$. What happens to $3$? $(23)$ maps it to $2$, and then $(12)$ maps $2$ to $1$. So $3\to1$. Putting this together, we have $1\to2$, $2\to3$, $3\to1$, which is what $(123)$ means.

Now let's look at the second one, $(13)(12)$. What happens to $1$? $(12)$ maps it to $2$, and then $(13)$ leaves $2$ alone, so $1\to2$. What happens to $2$? $(12)$ maps it to $1$, and then $(13)$ maps $1$ to $3$, so $2\to3$. What happens to $3$? $(12)$ leaves it $3$, and then $(13)$ maps $3$ to $1$, so $3\to1$. So again we have $1\to2$, $2\to3$, $3\to1$ or $(123)$.

You should be able to follow the same logic for the others.