Proving one element in the symmetric group (s>=3) commutes with all element

[emath]

I am really stuck with how to prove that the only element in Sn (with n>=3) commuting with all the other elements of this group is the identity permutation id.

I have no idea what I am supposed to do with it, i know why S3 has only one element that commutes but i don't know how to prove it for all.

 

[micromass]

You could start by proving the following:

If f and g are in S_n, then the disjoint cycle representation of fgf^-1 is obtained by taking the disjoint cycle representation of g and by changing every number n in the representation by g(n).

Think how that would prove your claim.

 

[jgens]

I think micromass's solution is very good. If you have trouble with his solution, then you can also just do the following:

If \sigma \in S_n \setminus \{\mathrm{id}\}, then there exists an i \in \{1,\dots,n\} such that \sigma(i) = j and i \neq j. Now let k \in \{1,\dots,n\} be such that k \neq j,\sigma(j) and let \tau \in S_n be the transposition which switches j and k. The rest of the proof is trivial from here.