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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.

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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Would you mind terribly moving this answer to the other question (I link to it in the comment above)? That one includes a bit more and already has some answers. This is an exact duplicate (down to the wording) of the second half of that answer. – Arturo Magidin Sep 16 '11 at 2:41
Will do. Give me nine key strokes. – jspecter Sep 16 '11 at 2:46
If the downvote has to do with the logistics of the repeated question, I would ask the downvoter to remove it. If there is some other reason, kindly explain it. – Arturo Magidin Sep 16 '11 at 2:47
Downvote?? ${}$ – jspecter Sep 16 '11 at 2:47
I know. I hope it's not because of my comment... – Arturo Magidin Sep 16 '11 at 2:48

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