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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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marked as duplicate by Arturo Magidin, Srivatsan, Amitesh Datta, anon, Zhen Lin Sep 16 '11 at 2:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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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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