Q:

Permutation as a product of transposition

On my textbook, the following permutation for 8 particle systems
\begin{equation}
\mathrm{P}\boldsymbol{=}
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\
1 & 2 & 4 & 5 & 6 & 3 & 8 & 7
\end{pmatrix}
\tag{01}\label{01} 
\end{equation}
can be represented as P = T(3,4)T(4,5)T(5,6)T(6,3)T(7,8). 
Why does it not have T(8,7)? 

A:

$T(a,b)$ transposes (or permutes) $a$ and $b$.  Clearly permuting $a$ and $b$ is the same a permuting $b$ and $a$: thus $T(a,b)=T(b,a)$.  Indeed the index notation like $T(a,b,c,d)$ is meant to denote the cycle 
$a\to b\to c\to d\to a$ : $a$ goes to $b$, which goes to $c$, which goes to $d$, which goes back to $a$.  $(a,b)$ thus denotes the 2-cycle $a\to b\to a$, which is the same as $b\to a\to b$.
It is conventional to denote transposition by $T(a,b)$ where $a<b$.

