Generating Fixed-Point-Free Permutations in Sn

[banana112]

Homework Statement

What subgroup is generated by the fixed-point-free permutations?

Homework Equations

The Attempt at a Solution

I know that the elements that have no fixed points are the ones whose cycle type adds up to n (i.e. all the numbers in {1,...,n} have to be used). I don't know what to do other than just start computing examples and multiplying out every single combination. EDIT I think they generate all of Sn. I know this is true up to 5, and whenever we have n odd. (We can generate the transpositions, which in turn generate all of Sn.) For odd n this works by: (1,2)=[(1,2)(3,4,...,n)]^(n-2). I haven't yet found something similar for the case when n is even.

 

[Dick]

Homework Statement

What subgroup is generated by the fixed-point-free permutations?

Homework Equations

The Attempt at a Solution

I know that the elements that have no fixed points are the ones whose cycle type adds up to n (i.e. all the numbers in {1,...,n} have to be used). I don't know what to do other than just start computing examples and multiplying out every single combination. EDIT I think they generate all of Sn. I know this is true up to 5, and whenever we have n odd. (We can generate the transpositions, which in turn generate all of Sn.) For odd n this works by: (1,2)=[(1,2)(3,4,...,n)]^(n-2). I haven't yet found something similar for the case when n is even.

Look for other patterns that will let you prove all transpostions are generated. Like for S8, (1,2)(3,4,5)(6,7,8) to the third power. What's a good one for S10? You might have to get a little more creative for S6 but you can certainly do it without multiplying out every single combination.