Showing equality of two groups.

Question

Consider $S_n$, the permutation group. Let $a\in S_n$. I want to show that if $a$ is an $n$ or an $n-1$ cycle, then $\langle a\rangle = \{c \in S_n : ca=ac\}$. Any help will be veru much appreciated.

Answer 1

Define an action $\,S_n\times S_n\to S_n\,$ by conjugation. Since there are $\,(n-1)!\,$ different $\,n-\,$cycles , we get that if $\,a\,$ is an $\,n-\,$ cycle then

$$(n-1)!=|\mathcal Orb(a)|=[S_n:C_{S_n}(a)]=\frac{|S_n|}{|C_{S_n}(a)|}\Longrightarrow |C_{S_n}{a}|=n$$

and since clearly every power of $\,a\,$ commutes with $\,a\,$ , these powers thus are the only elements in $\,S_n\,$ that do so.

Added on request of the OP: There are $\,(n-2)!\,n\,$ different $\,(n-1)-\,$ cycles in $\,S_n\,$ , so if $\,a\,$ is such a cycle:

$$|\mathcal Orb(a)|=[S_n:C_{S_n}(a)]\Longrightarrow |C_{S_n}(a)|=\frac{n!}{(n-2)!n}=n-1$$

and again we have that the only elements that commute with this cycle are its $\,n-1\,$ powers.