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Possible Duplicate:
Problem from Armstrong's book, “Groups and Symmetry”

If $a$, $b$ are members of the permutation group $S_n$, and $ab=ba$. Show that $b$ must be a power of $a$ when $a$ is an $n$-cycle.

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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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    I know. I hope it's not because of my comment...2011-09-16