Question

Write each permutation in Exercise 3 as a product of transpositions.

(a)$\left(\begin{array}{cccccc}\mathbf{1}& \mathbf{2}& \mathbf{3}& \mathbf{4}& \mathbf{5}& \mathbf{6}\\ \mathbf{2}& \mathbf{1}& \mathbf{3}& \mathbf{5}& \mathbf{4}& \mathbf{7}\end{array}\begin{array}{c}\mathbf{7}\\ \mathbf{9}\end{array}\begin{array}{c}\mathbf{8}\\ \mathbf{8}\end{array}\begin{array}{c}\mathbf{9}\\ \mathbf{6}\end{array}\right)$ (b)$\left(\begin{array}{cccccc}\mathbf{1}& \mathbf{2}& \mathbf{3}& \mathbf{4}& \mathbf{5}& \mathbf{6}\\ \mathbf{3}& \mathbf{5}& \mathbf{1}& \mathbf{2}& \mathbf{4}& \mathbf{6}\end{array}\begin{array}{c}\mathbf{7}\\ \mathbf{8}\end{array}\begin{array}{c}\mathbf{8}\\ \mathbf{9}\end{array}\begin{array}{c}\mathbf{9}\\ \mathbf{7}\end{array}\right)$

(c) $\left(\begin{array}{cccccc}\mathbf{1}& \mathbf{2}& \mathbf{3}& \mathbf{4}& \mathbf{5}& \mathbf{6}\\ \mathbf{3}& \mathbf{5}& \mathbf{1}& \mathbf{2}& \mathbf{4}& \mathbf{9}\end{array}\begin{array}{c}\mathbf{7}\\ \mathbf{8}\end{array}\begin{array}{c}\mathbf{8}\\ \mathbf{7}\end{array}\begin{array}{c}\mathbf{9}\\ \mathbf{6}\end{array}\right)$ (c) $\left(\mathbf{14}\right)\left(\mathbf{27}\right)\left(\mathbf{523}\right)\left(\mathbf{34}\right)\left(\mathbf{1472}\right)$

(d)$\left(\mathbf{7236}\right)\left(\mathbf{85}\right)\left(\mathbf{571}\right)\left(\mathbf{1537}\right)\left(\mathbf{486}\right)$

Short answer

The product of transpositions is $\left(12\right)\left(45\right)\left(67\right)\left(79\right)$.

Step-by-step solution

To obtain the cycle notation

We have the definition of cycle that,

Let $a_1,a_2,\dots ,a_k$be the distinct elements of the set $\left\{1,2,3,\dots ,n\right\}$. Then, $\left(a_1{a}_2\dots a_k\right)$denotes the permutation in $S_n$that maps $a_1$to $a_2,a_2$, to $a_3,\dots ,a_{k-1}$to$a_k$ and$a_k$ to$a_1$ and every other elements of set maps to themselves.

$\left(a_1{a}_2\dots a_k\right)$is called a cycle of length k or k - cycle.

Two cycles which have no elements in common are called disjoint cycles.

A cycle of length 2 is called a transposition.

Now, the given permutation is,

$\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 2& 1& 3& 5& 4& 7\end{array}\begin{array}{c}7\\ 9\end{array}\begin{array}{c}8\\ 8\end{array}\begin{array}{c}9\\ 6\end{array}\right)$

Here, 1 is mapped to 2 and 2 is mapped to 1, 4 is mapped to 5 and 5 is mapped to 4, 6 is mapped to 7, 7 is mapped to 9 and 9 is mapped back to 6 and all other elements are mapped to themselves, that is, 3 is mapped to 3 and 8 is mapped to 8.

We have to find the product transpositions.

To find product of transpositions

By using Theorem 7.26 that,

‘Every permutation in$S_n$ is a product of (not necessarily disjoint) transpositions’, the product of transpositions is expressed as:

$\begin{array}{rcl}\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 2& 1& 3& 5& 4& 7\end{array}\begin{array}{c}7\\ 9\end{array}\begin{array}{c}8\\ 8\end{array}\begin{array}{c}9\\ 6\end{array}\right)& =& \left(12\right)\left(3\right)\left(45\right)\left(679\right)\left(8\right)\\ & =& \left(12\right)\left(45\right)\left(679\right)\\ & =& \left(12\right)\left(45\right)\left(67\right)\left(79\right)\end{array}$

Hence, the product of transpositions is $\left(12\right)\left(45\right)\left(67\right)\left(79\right)$.