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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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    I don't think the question can be quite right as stated. Take a 5-cycle $a$ in $S_7$ then $a^2$ commutes with $a$ and none of the conditions apply. I think it is possibly getting at something just a little more subtle.2011-09-11
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    @Mark Bennet: What do you mean? If $a$ is a $5$-cycle, say $(1,2,3,4,5)$, then it is *certainly* true that $a^2$ permutes the points fixed by $a$, namely $6$ and $7$: that is, it maps $\{6,7\}$ to itself. And the second part does not apply to that situation.2011-09-11
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    OR are we allowed the identity permutation on the points fixed by a, in which case I misread the question.2011-09-11
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    @Arturo Magidin: thanks for clarifying - I think I assumed an 'only' when there wasn't one.2011-09-11
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    Note that the second question has not been answered yet.2011-09-16

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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 $1\leq k_x\leq n$.

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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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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 $C_{S_n}(a) = \langle a \rangle.$