I have came across this before and just again now, in the same form of which I'm struggling to understand. Although I know it's link to parity, as a perm group pi:
$$ \pi = \begin{pmatrix} 0 & 1 & 2 & 3 & \dots & n \\ a_{0} & a_{1} & a_{2} & a_{3} & \dots &a_{n} \end{pmatrix}$$
$$ (-1)^{\pi} = \prod^{n}_{j}f(x_{j+1}) - f(x_{j})$$
There are repeatedly questions as such:
Example
Consider the composition (12)(23) of the two cycle (12) in the set $S_3$, and the 2-cycle (23) in $S_3$.
The number 1 is mapped to itself by (23) and then to 2 by (12). The number 2 is mapped to 3 by (23), which is mapped to itself by (12) The number 3 is mapped to 2 by (23), which is mapped to itself by (12). So overall, 1 maps to 2, which maps to 3, which maps to 1, or:
(12)(23) = (123)
How did he make those assumptions? I have cycled through the Levi-Civita tensor before however I can't see these exact mappings.
Are you asking why $(12)(23)=(123)$ ?
$(i,j)$ means a transposition which makes $\pi = \begin{pmatrix} 0 & 1 & 2 & 3 & \dots & i & \dots & j & \dots & n \\ a_{0} & a_{1} & a_{2} & a_{3} & \dots & a_{i} & \dots & a_{j} & \dots &a_{n} \end{pmatrix}$ into $\pi = \begin{pmatrix} 0 & 1 & 2 & 3 & \dots & i & \dots & j & \dots & n \\ a_{0} & a_{1} & a_{2} & a_{3} & \dots & a_{j} & \dots & a_{i} & \dots &a_{n} \end{pmatrix}$
i.e $(a_{i},a_{j})$ changes to $(a_{j},a_{i})$ here.
and $(i,j,k)$ is a three cycle i.e $(i->j->k->i)$ which basically maps $(a_{i}, a_{j},a_{k})$ into $(a_{j}, a_{k},a_{i})$ .
Now why $(i,j)(j,k)=(i,j,k)$ ?
For that apply $ \pi_{1} = (j,k)$ on $ \pi_{2} = (i,j)$ and check it by yourself that $\pi = (i,j,k)$ = $ \pi_{2}o\pi_{1}$ . (function composition). This would extend for general case for n transpositions or cycles. Also remember the following easy to handle formulations :
Hope I could clarify.