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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

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    Argue the following: If two maps $\tau$ and $\rho$ from a set $X$ to itself commute, then $\tau$ sends each $\rho$-fixed point to a $\rho$-fixed point. – 2017-02-07
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    I am not using torsion for the problem. I see what you are trying to say but the question is hard to me – 2017-02-07

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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.

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    This makes sense but do have to make up a cycle like (1423)(65)(7)(8)(9) to show that this proof holds? – 2017-02-07
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    No; why would you? (also, you need a permutation that holds only a single symbol fixed, and that holds three symbols fixed) – 2017-02-07
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    Oh you are right but my suggestion works right? Using just 1 fixed symbol – 2017-02-07
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    For a proof, you need to show that it works for every permutation that matches the condition (leaves one symbol fixed). Giving a single example won't really help you prove it for *all* permutations that leave one symbol fixed. – 2017-02-07
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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 $\tau (1) =1$.

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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.