Question

Find the order of ${\mathbf{\sigma }}^{\mathbf{1000}}$, where$\mathbf{\sigma }$ is the permutation$\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 3& 7& 8& 9& 4& 2\end{array}\begin{array}{c}7\\ 2\end{array}\begin{array}{cc}8& 9\\ 1& 6\end{array}\right)$ is the identity subgroup.

Short answer

The order of${\mathrm{\sigma }}^{1000}$ would be 3.

Step-by-step solution

Statement of Exercise 17 and 18

Exercise 17states that k-cycle in the group S_n contains order k.

Exercise 18states that when $\mathrm{\sigma }=\left({\mathrm{a}}_1{\mathrm{a}}_2....{\mathrm{a}}_{\mathrm{k}}\right)$and $\mathrm{T}=\left({\mathrm{b}}_1{\mathrm{b}}_2.....{\mathrm{b}}_{\mathrm{r}}\right)$ can be disjoint cycles in S_n then, $\mathrm{\sigma T}=\mathrm{T\sigma }$. Let the three cases be: i is one of the a’s, i is one of the b’s, i would be neither.

Determine the order of σ1000

Here, $\sigma =\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 3& 7& 8& 9& 4& 5\end{array}\begin{array}{ccc}7& 8& 9\\ 2& 1& 6\end{array}\right)=\left(138\right)\left(27\right)\left(4965\right)$.

Therefore, according to Exercises 17 and 18,

$\begin{array}{rcl}{\mathrm{\sigma }}^{1000}& =& {\left(138\right)}^{1000}{\left(27\right)}^{1000}{\left(4965\right)}^{1000}\\ & =& \left(138\right)\end{array}$

Since there are ${\left(138\right)}^3=1$,${\left(27\right)}^3=1$and ${\left(4965\right)}^3=1$. As a result, the order of${\mathrm{\sigma }}^{1000}$ would be 3.

Hence, the order of${\mathrm{\sigma }}^{1000}$ would be 3.