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$\left(13\right)\left(254\right)\left(789\right)$.

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 calleddisjoint cycles.

Now, the given permutation is,

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

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

We have to find the product of disjoint cycles.

To find the product of disjoint cycles

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

$\left(\begin{array}{ccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9\\ 3& 5& 1& 2& 4& 6& 8& 9& 7\end{array}\right)=\left(13\right)\left(254\right)\left(789\right)$

Hence, the product of disjoint cycles is$\left(13\right)\left(254\right)\left(789\right)$.