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 cycles is$\left(123\right)\left(456\right)\left(78\right)$.

Step-by-step solution

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

We have the definition of cycle that,

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,

$\left(7236\right)\left(85\right)\left(571\right)\left(1537\right)\left(486\right)$

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 second cycle from right-hand.

$\left(7236\right)\left(85\right)\left(571\right)\left(1537\right)\left(486\right)=\left(\begin{array}{cccccccc}1& 2& 3& 4& 5& 6& 7& 8\\ 2& 3& 1& 5& 6& 4& 8& 7\end{array}\right)$

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

Next, 3 is mapped to 7 and 7 is mapped to 1. So , 3 is mapped to 1.

Next, 4 is mapped to 8 and 8 is mapped to 5. So, 4 is mapped to 5.

Next, 5 is mapped to 3, 3 is mapped to 6. So, 5 is mapped to 6.

Next, 6 is mapped to 4.

Next, 7 is mapped to 1, 1 is mapped to 5, 5 is mapped to 8. So, 7 is mapped to 8.

Next, 8 is mapped to 6, 6 is mapped 7. So, 8 is mapped to 7.

To find the product of disjoint cycles 

From step 1 we get,

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

Here, 1 is mapped to 2, 2 is mapped to 3, 3 is mapped back to 1; 4 is mapped to 5 and 5 is mapped to 6 and 6 is mapped back to 4; 7 is mapped to 8 and 8 is mapped back to 7.

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

$$\begin{gathered} \left(7236\right)\left(85\right)\left(571\right)\left(1537\right)\left(486\right)=\left(\begin{array}{cccccccc}1& 2& 3& 4& 5& 6& 7& 8\\ 2& 3& 1& 5& 6& 4& 8& 7\end{array}\right) \\ =\left(123\right)\left(456\right)\left(78\right) \end{gathered}$$

Hence, the product of disjoint cycles is$\left(123\right)\left(456\right)\left(78\right)$