1. ## Abstract Algebra

Express (7 3 2 6) (5 7 1) (1 3 5 7) (4 6 8) as a product of transpositions

2. Originally Posted by mathlg

Express (7 3 2 6) (5 7 1) (1 3 5 7) (4 6 8) as a product of transpositions
The rule of transpositions is,
That a cycle,
$\displaystyle (a_n,a_{n-1},a_{n-2},...,a_2,a_1)$
Can be expressed as,
$\displaystyle (a_n,a_{n-1})(a_n,a_{n-2})...(a_n,a_1)$

I am confused by your question, are you asking to express that product as a transposition. Or is each cycle an individual problem. Your question is not well-defined.

3. Yes I am asking how to express that product as a transposition. Not each individual cycle.

4. That big cycle which you mentioned is actually a premutation. I assume you are working in $\displaystyle S_8$
Let us express,
$\displaystyle (7,3,2,6) (5,7,1) (1,3,5,7) (4,6,8)$
As a premutation.
Since this is a function composition I will start with the inner function first, i.e. with (4,6,8). Then evaluate it by the second by the right and so one.

To find what premutation this is, be write out the mappings of each element from $\displaystyle S_8$.
Thus,
$\displaystyle \left\{ \begin{array}{c}1\to 1\to 3\to 3\to 2\\ 2\to 2\to 2 \to 2 \to 6\\ 3\to 3\to 5\to 7\to 3 \\ 4\to 6\to 6\to 6\to 7\\ 5\to 5\to 7\to 1\to 1\\ 6\to 8\to 8\to 8\to 8\\ 7\to 7\to 1\to 5\to 5\\ 8\to 4\to 4\to 4\to 4$
As a result we have the following premutation,
$\displaystyle \left( \begin{array}{cccccccc}1&2&3&4&5&6&7&8\\ 2&6&3&7&1&8&5&4 \end{array} \right)$
To express this premutation as a product of cycles, we need to find its orbits,
$\displaystyle 1\to 2\to 6\to 8\to 4\to 7\to 5\to 1$
Hence, the two equivalence classes are,
$\displaystyle \{1,2,6,8,4,7,5\} \mbox{ and } \{3\}$
Thus, this premutation is itself a cycle, since it has at most one orbit containing more than one element.
Thus, it can be expressed as,
$\displaystyle (1,2,6,8,4,7,5)$
Using the formula I stated above we can express it as a product of transpositions as,
$\displaystyle (1,2)(2,6)(6,8)(8,4)(4,7)(7,5)$