Question

Express as a product of disjoint cycles:

$$\begin{gathered} \left(a\right)\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9\\ 2& 1& 3& 5& 4& 7& 9& 8& 6\end{array}\right)\left(b\right)\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9\\ 3& 5& 1& 2& 4& 6& 8& 9& 7\end{array}\right) \\ \left(c\right)\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9\\ 3& 5& 1& 2& 4& 9& 8& 7& 6\end{array}\right)\left(d\right)\left(14\right)\left(27\right)\left(523\right)\left(34\right)\left(1472\right) \\ \left(e\right)\left(7236\right)\left(85\right)\left(571\right)\left(1537\right)\left(486\right) \end{gathered}$$

Short answer

The product of disjoint cycle is$\left(1573\right)\left(24\right)$

Step-by-step solution

Step by Step Solution Step 1: To obtain permutation by computing product of cycles

We have the definition of cyclethat,

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

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

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

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 the 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.

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

Here, 1 is mapped to 4, 4 is mapped to 7, 7 is mapped to 2. Moving to the 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 product of disjoint cycles

From step 1 we get,

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

Here, 1 is mapped to 5, 5, mapped to 7, 7, 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.

The above permutation can be expressed as a product of disjoint cycles as,

$$\begin{gathered} \left(14\right)\left(27\right)\left(523\right)\left(34\right)\left(1472\right)=\left(\begin{array}{ccccccc}1& 2& 3& 4& 5& 6& 7\\ 5& 4& 1& 2& 7& 6& 3\end{array}\right) \\ =\left(1573\right)\left(24\right) \end{gathered}$$

Hence, the product of disjoint cycles is $\left(1573\right)\left(24\right)$.