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.
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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 $C_{S_n}(a) = \langle a \rangle.$ |
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Hint: if $x$ is a fixed point of $a$, i.e. $a(x) = x$, then what can you say about $a(b(x))$? |
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