Question

Show that S₁₀ contains elements of orders 10, 20, and 30. Does it contain an element of order 40?

Short answer

It is proven that S₁₀ contains elements of orders 10, 20, and 30. There is no element with order 40.

Step-by-step solution

Statement of theorem 7.25

Theorem 7.25states that the order of a permutation$\mathrm{\zeta }$ in S_n would be the least common multiple of the lengths of the disjoint cycleswhose product would be ${\mathrm{\zeta }}^{\mathrm{t}}$.

Show that S10 contains elements of orders 10, 20, and 30

The order of $\left(12\right)\left(34567\right)$would be 10, the order of data-custom-editor="chemistry" $\left(1234\right)\left(56789\right)$would be 20, and the order of$\left(12\right)\left(345\right)\left(678910\right)$ is 30.

Hence, it is proven that S₁₀ contains elements of orders 10, 20, and 30.

Does S10 contain an element of order 40

In S₁₀, there have been no cycles of order 40. When$\mathrm{\zeta }$ was such an element of order 40 that was the product of disjoint cycles namely,$\mathrm{\zeta }=\prod _{\mathrm{i}=1}^{\mathrm{k}}{\mathrm{\sigma }}_{\mathrm{i}}$ the least common multiples of the lengths of the${\mathrm{\sigma }}_{\mathrm{i}}$ would be 40. The only possibility will be for$\mathrm{\zeta }={\mathrm{\sigma }}_1{\mathrm{\sigma }}_2$ having${\mathrm{\sigma }}_1$ of length 8 and${\mathrm{\sigma }}_2$ of length 5 because $40=2^3.5$. However, this will entail a permutation of total 13 elements.

Therefore, it is not possible in S₁₀.

Thus, there is no element with order 40.