Question

Express as a product of disjoint cycles:

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

Short answer

The product of disjoint cycles is$\left(13\right)\left(254\right)\left(69\right)\left(78\right)$

Step-by-step solution

Step by Step Solution Step 1: 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$\{1,2,3,\dots ,n\}$. 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,

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

Here, 1 is mapped to 3 and 3 is mapped to 1, 2, which is mapped to 5, 5, mapped to 4, and 4 is mapped back to 2, 7, which is mapped to 8, and 8 is mapped back to 7 and 6 is mapped to 9 and 9 is mapped back to 6.

We have to find the product of disjoint cycles.

To find product of disjoint cycles

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

$\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9\\ 3& 5& 1& 2& 4& 9& 8& 7& 6\end{array}\right)=\left(13\right)\left(254\right)\left(69\right)\left(78\right)$

Hence, the product of disjoint cycles is$\left(13\right)\left(254\right)\left(69\right)\left(78\right)$