Proving There are No Permutations of Order 18 in S_9: Permutation Question

[TTob]

Homework Statement

prove that there are not permutations of order 18 in S_9.

Homework Equations

The Attempt at a Solution

let t=c_1,...,c_k is cycle decomposition of such permutation. let r_1,...,r_k the orders of c_1,...,c_k.

then lcm(r_1,...,r_k) = 18 and r_1+...+r_k = 9.

What To Do Next ?

 

[morphism]

Why is r_1+...+r_k=9? There is definitely something to be said about the r_i's, but this is not it!

 

[TTob]

Why is r_1+...+r_k=9? There is definitely something to be said about the r_i's, but this is not it!

because this is cycle decomposition and hence c_1 is r_1-cycle,..., c_k is r_k cycle.

 

[sutupidmath]

well, here are my thoughts, but wait for morphism to confirm it.

Like you said, let

$$\alpha=\beta_1\beta_2...\beta_k-----(@)$$ be such a permutation written as a product of k disjoint cycles. Let $$o(\beta_i)=r_i,i\in\{1,2,...,k\}$$ be the orders of those cycles respectively.

Then we know that the order of that permutation is the least common multiple of the lengths(orders) of the cycles, that is

$$lcm[r_1,r_2,...,r_k]=18$$ (in here we are using proof by contradiction, that is we are assuming that indeed there is such a permutation in S_9 whose order is 18)

But this is not possible, why?

In order for $$lcm[r_1,r_2,...,r_k]=18$$ to be true there must be cycles in (@) with orders 9 and 6. But, such a thing is not possible, because say:

$$\beta_1=(a_1a_2a_3a_4a_5a_6), and \beta_2=(b_1b_2...b_9)$$ if

$$\beta=\beta_1\beta_2$$ there must be some $$a_i=b_j$$ for i=1,2...6 and j=1,2,...,9.

So, the contradiction derived, tells us that the assumption that $$lcm[r_1,r_2,...,r_k]=18$$ is not true.