Permutations, Fixed symbols in $S_n$ Prove that if $\rho \in S_n$ leaves just one symbol fixed, then every element of $S_n$ that commutes with $\rho$ must leave the same symbol fixed?  Here is a definition: A permutation $\rho \in S_n$ is completely factored when it a written as a product of disjoint cycles and 1-cycles of all fixed elements.  Here is an example: The permutation $\rho \in S_n$ defined by $\rho$=(1,3,6,2)(4 5) is factored into disjoint cycles, but the factorization is not complete. When it is presented as $\rho$=(1,3,6,2)(45)(7)(8)(9) it is completely factored. Here the fixed elements of $\rho$ were 7, 8 and 9.  Giving this information, how can I prove the giving question? I am having trouble with this
 A: Call the symbol that $\rho$ leaves fixed $\mathrm{a}$. Let $\chi \in S_n$ be any permutation such that $\chi(\mathrm{a}) \ne \mathrm{a}$.
Then $(\chi\rho)(\mathrm{a}) = \chi(\rho(\mathrm{a})) = \chi(\mathrm{a})$, but $(\rho\chi)(\mathrm{a}) = \rho(\chi(\mathrm{a}))$.
Now, since $\rho$ leaves only one symbol fixed, and since $\chi(\mathrm{a}) \ne \mathrm{a}$, we know then that $\chi(\mathrm{a}) \ne \rho(\chi(\mathrm{a}))$.
Therefore, $(\chi\rho)(\mathrm{a}) \ne (\rho\chi)(\mathrm{a})$, and therefore $\chi\rho \ne \rho\chi$.
Therefore, any permutation that does commute with $\rho$ must leave $\mathrm{a}$ fixed.
A: WLOG, let $\rho (1)=1$ and $\rho(k)\neq k$ for $k=2,3,\dots, n$. Take $\tau \in S_n$ arbitrarily with $\rho \tau =\tau \rho$. Now we observe how $\rho \tau$ or $\tau \rho$ sends $1$.
Note that $\tau (1) = \tau (\rho (1)) = \rho (\tau(1))$. Now as $\rho(k)\neq k$ unless $k=1$ and  $ \rho (\tau(1)) = \tau (1)$, it follows that $\tau (1) =1$.
A: Without loss of generality we can assume $S_n$ act on the set $\{1,\ldots,n\}$ and that the only point fixed by $\chi$ be $n$, then $\chi$ moves all the points of $\{1,\ldots,n-1\}$. The only elements of $S_{n-1}$ that commute with $\chi$ are the elements of the group generated by $\chi$ so they leave $n$ fixed.
