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

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

Short answer

The product of transpositions is$\left(15\right)\left(37\right)\left(73\right)\left(24\right)$.

Step-by-step solution

To obtain the permutation by computing the product of cycles

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 role="math" localid="1653543244747" $a_3,\dots ,a_{k-1}$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,

$\left(14\right)\left(27\right)\left(523\right)\left(34\right)\left(1472\right)$

First, we compute the product of these cycles to obtain a permutation.

Here, the given product is a composition of functions. So, the right-hand permutation is performed first. Thus, we will start from 1 in the rightmost cycle.

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

Here, 1 is mapped to 4, 4 is mapped to 7, 7 is mapped to 2. Moving to left-hand permutation, 2 is mapped to 3 and 3 is mapped to 5. So, 1 is mapped to 5.

Next, 2 is mapped to 1 and 1 is mapped to 4. So , 2 is mapped to 4.

Next, 3 is mapped to 4 and 4 is mapped to 1. So, 3 is mapped to 1.

Next, 4 is mapped to 7, 7 is mapped to 2. So, 4 is mapped to 2.

Next, 5 is mapped to 2, 2 is mapped to 7. So, 5 is mapped to 7.

Next, 6 is mapped to 6 and 7 is mapped to 2, 2 is mapped 3. So, 7 is mapped to 3.

To find the product of transposition

From step 1 we get,

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

Here, 1 is mapped to 5, 5 is mapped to 7, 7 is mapped to 3 and 3 is mapped back to 1; 2 is mapped to 4 and 4 is mapped back to 2; 6 is mapped to itself.

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,

$\begin{array}{rcl}\left(14\right)\left(27\right)\left(523\right)\left(34\right)\left(1472\right)& =& \left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 5& 4& 1& 2& 7& 6\end{array}\begin{array}{c}7\\ 3\end{array}\right)\\ & =& \left(1573\right)\left(24\right)\\ & =& \left(15\right)\left(57\right)\left(73\right)\left(24\right)\end{array}$

Hence, the product of transposition is $\left(15\right)\left(57\right)\left(73\right)\left(24\right)$.