Representing a 5-cycle as a product of transpositions

Question

Dr. Pinter's "A Book of Abstract Algebra" shows that:

$$(12345)$$

can be written as the following product of transpositions:

$$(54)(53)(52)(51)$$

How can the first representation, $(12345)$, be represented as $(54)(53)(52)(51)$?

Answer 1

Here's one way to see what's going on:

$$\begin{matrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \end{matrix} \quad \xrightarrow{(51)} \quad \begin{matrix} 5 \\ 2 \\ 3 \\ 4 \\ 1 \end{matrix} \quad \xrightarrow{(52)} \quad \begin{matrix} 5 \\ 1 \\ 3 \\ 4 \\ 2\end{matrix} \quad \xrightarrow{(53)} \quad \begin{matrix} 5 \\ 1 \\ 2 \\ 4 \\ 3 \end{matrix} \quad \xrightarrow{(54)} \quad \begin{matrix} 5 \\ 1 \\ 2 \\ 3 \\ 4 \end{matrix} $$

and $$\begin{matrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \end{matrix} \quad \xrightarrow{(12345)} \quad \begin{matrix} 5 \\ 1 \\ 2 \\ 3 \\ 4 \end{matrix}$$

Answer 2

In general, this is just an observation that any cycle can be decomposed into transpositions via the above method. Direct computation shows that they are the same permutation, and if we write

$(r_1r_2\cdots r_n)= (r_nr_{n-1})(r_nr_{n-2})\cdots (r_n)(r_1)$

we can show the equality holds by the same method.

For instance, choose $r_k$ with $k\neq n$. The first spot it appears (starting from the right) will be in the $k$th spot. It is in the transposition $(r_nr_k)$ so $r_k\mapsto r_n$. Then, $r_n$ will appear in the very next transposition so $r_n\mapsto r_{k+1}$ and $r_{k+1}$ is fixed. For $k=n$ we have that $r_n$ appears in the first transposition so $r_n\mapsto r_1$ and $r_1$ is fixed. But this is by definition the $n$-cycle we started with (i.e. $r_1\mapsto r_2 \mapsto \cdots \mapsto r_n \mapsto r_{1}$).