I haven't gotten all that far with this:
If $a$, $b$ are members of the permutation group $S_n$, and $ab=ba$, prove that $b$ permutes those integers which are left fixed by $a$.
Show that $b$ must be a power of $a$ when $a$ is an $n$-cycle.
I haven't gotten all that far with this:
If $a$, $b$ are members of the permutation group $S_n$, and $ab=ba$, prove that $b$ permutes those integers which are left fixed by $a$.
Show that $b$ must be a power of $a$ when $a$ is an $n$-cycle.
Here is my attempt: For the second part, I showed that if $\beta(x)=x$, then $\beta(k)=k$ for all $1\leq k\leq n$. Then, it follows that if $\beta\neq i$ where $i$ is the identity mapping, then $\beta$ is a $n$-cycle. By writing $\alpha=(a_1\ a_2\ \dots\ a_n)\ \text{and }\beta=(123\dots n),$
I found that there is a $1\leq k\leq n$ such that $\alpha=(k\ k+1\ \dots k+n)$ where addition is modulo $n$. To do this, I looked at diagrams similar to what follows. For $1 \leq x\leq n$, $\beta\alpha: x\rightarrow k_x \rightarrow k_x+1$ for some $1\leq k_x\leq n$.
Hint: if $x$ is a fixed point of $a$, i.e. $a(x) = x$, then what can you say about $a(b(x))$?
Here's the solution to the second part.
Recall the following facts.
(1) Two elements of $S_n$ are conjugate if and only if they have the same cycle structure. Hence, the orbit of $a$ under the action of conjugation by elements of $S_n$ is exactly the $n$-cycles.
(2) There are (n-1)! $n$-cycles in $S_n$
(3) The size of the orbit of $a$ under the action of conjugation in $S_n$ times the size of the centralizer of $a$ is equal to the order of $S_n.$
It follows there are exactly $n$ elements that commute with $a.$ As every power of $a$ is such an element, we conclude the $C_{S_n}(a) = \langle a \rangle.$