1. ## order 171

This question is part of a larger question, but I don't understand the solution I have been given here.

Show that there is no element of order 171 in the symmetric group $S_{20}$.

Solution: The smallest n such that $S_n$ contains an element of order $171 = 3^2 \cdot 19$ is n = 9 + 19 = 28.

I don't understand at all why that statement is true. The question comes at the end of a section about the Sylow theorems. Any help with this would be appreciated

2. Originally Posted by slevvio
This question is part of a larger question, but I don't understand the solution I have been given here.

Show that there is no element of order 171 in the symmetric group $S_{20}$.

Solution: The smallest n such that $S_n$ contains an element of order $171 = 3^2 \cdot 19$ is n = 9 + 19 = 28.

I don't understand at all why that statement is true. The question comes at the end of a section about the Sylow theorems. Any help with this would be appreciated

Facts you must check you know/you can prove:

1) Every permutation can be written as a product of disjoint cycles

2) The order of a permutation is the lowest common multiple of the orders (=lengths) of the cycles appearing in its decomposition in disjoint cycles.

From here the solution follows at once.

Tonio

3. Suppose there exists an element of order 171 in $S_n$. We want to find the different ways to write this element as the product of disjoint cycles, and then look at the smallest possible one.

The factors of 171 are 1,3,9,19,57, and 171.

The smallest one we can get is a cycle of 9 elements times a cycle with 19 elements (lcm(19,9) = 171) , so there must be at least 19 + 9 = 28 elements to choose from to put into those cycles