# Problem from Armstrong's book, “Groups and Symmetry”

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. – Mark Bennet Sep 11 '11 at 19:22
@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. – Arturo Magidin Sep 11 '11 at 19:26
OR are we allowed the identity permutation on the points fixed by a, in which case I misread the question. – Mark Bennet Sep 11 '11 at 19:31
@Arturo Magidin: thanks for clarifying - I think I assumed an 'only' when there wasn't one. – Mark Bennet Sep 11 '11 at 19:32
Note that the second question has not been answered yet. – Arturo Magidin Sep 16 '11 at 2:39

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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