Center of Symmetric Groups n>= 3 is trivial

[Metahominid]

Homework Statement

The question is to show that the for symmetric groups, Sn with n>=3, the only permutation that is commutative is the identity permutation.

Homework Equations

I didn't know if it was necessary but this equates to saying the center is the trivial group.

The Attempt at a Solution

I was attempting to show that there will always exist a permutation that isn't commutative for a particular one, which I was figuring would be showing there can be permutations that move an element to different places so their compositions would be different.

 

[Office_Shredder]

If u and v are permutations, then uv=vu is the same thing as uvu^-1=v. So if v is in the center of S_n, then conjugating v by any other element doesn't change v. Do you know what the conjugates of an element of S_n look like?

 

[Metahominid]

No I do not, I don't think we have covered conjugation.

 

[Dick]

No I do not, I don't think we have covered conjugation.

You don't need it. Your first approach was fine. Sn contains a permutation cycle (x1,x2,x3) where x1,x2 and x3 are in {1,...n}. Write down a permutation that doesn't commute with it.