Question

Show that $\mathit{\beta }$= (1236) (5910) (465) (5678) has order 21 in${\mathbf{\text{S}}}_{\mathbf{\text{n}}}\mathbf{\text{=n≥10}}$

Short answer

It is proved that$\beta$= (1236) (5910) (465) (5678) has order 21 in${\text{S}}_{\text{n}}=\text{n}\ge 10$

Step-by-step solution

Required Theorem

Theorem 7.25: The order of a permutation $\mathit{\tau }$in ${\text{S}}_{\text{n}}$is the least common multiple of the length of the disjoint cycles whose product is$\mathit{\tau }$.

Proving that β =(1236) (5910) (465) (5678) has order 21 in Sn (n : 10)

It is given that,

$\beta$= (1236) (5910) (465) (5678)

We can rewrite $\beta$as a product of its disjoint cycle:

$\beta$= (1236) (5910) (465) (5678)

= (1236784) (5910)

Since the length of the disjoint cycle are 7 and 3, so from the above theorem, the order should be the least common multiple of the length of disjoint cycles.

Therefore, the order is the LCM of 7 and 3 which is 21.

Hence, $\beta$ = (1236) (5910) (465) (5678) has an order 21 in (S_n$\ge$10).