A cycle as a product of transpositions Can someone please explain how a cycle $(1234)$ can be written as a product of transpositions: $(14)(13)(12)$? And how they can be multiplied to (1234)? Thanks in advance.
 A: $$(14)=\begin{pmatrix}1&2&3&4 \\ 4&2&3&1\end{pmatrix}$$
$$(13)=\begin{pmatrix}1&2&3&4 \\ 3&2&1&4\end{pmatrix}$$
$$(12)=\begin{pmatrix}1&2&3&4 \\ 2&1&3&4\end{pmatrix}$$
$$(14)(13)=(14)\circ(13)=\begin{pmatrix}1&2&3&4 \\ 4&2&3&1\end{pmatrix}\circ\begin{pmatrix}1&2&3&4 \\ 3&2&1&4\end{pmatrix}=\begin{pmatrix}1&2&3&4 \\ 4&2&1&3\end{pmatrix}$$
$$(14)(13)(12)=((14)\circ(13))\circ(12)$$
$$(14)(13)(12)=\begin{pmatrix}1&2&3&4 \\ 4&2&1&3\end{pmatrix}\circ\begin{pmatrix}1&2&3&4 \\ 2&1&3&4\end{pmatrix}=\begin{pmatrix}1&2&3&4 \\ 4&1&2&3\end{pmatrix}=(1432)$$
A: This is not "getting transposed", instead, this is representing $(1234)$ as the product of transpositions.  I.e., $(1234)=(14)(13)(12)$.
Multiplication here is from right to left.  So, we have:


*

*$1 \mapsto 2$ from the cycle $(12)$, [and $2$ is fixed by $(13)$ and $(14)$]

*$2 \mapsto 1 \mapsto 3$ from the cycle $(12)$ then the cycle $(13)$, [and $3$ is fixed by the cycle $(14)$]

*$3 \mapsto 1 \mapsto 4$ from the cycle $(13)$ then the cycle $(14)$, [since $3$ is fixed by the cycle $(12)$]

*$4 \mapsto 1$ from the cycle $(14)$  [since $4$ is fixed by the cycles $(12)$ and $(13)$].

