Center of Symmetric Groups n>= 3 is trivial

  • #1

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.
 

Answers and Replies

  • #2
Office_Shredder
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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 Sn, then conjugating v by any other element doesn't change v. Do you know what the conjugates of an element of Sn look like?
 
  • #3
No I do not, I don't think we have covered conjugation.
 
  • #4
Dick
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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.
 

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