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

1. May 17, 2012

### 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 im supposed to do with it, i know why S3 has only one element that commutes but i dont know how to prove it for all.

2. May 17, 2012

### micromass

Staff Emeritus
You could start by proving the following:

If f and g are in Sn, 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.

3. May 17, 2012

### 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.