Orbit-Stabiliser Theorem Application

Question Let $G$ be the symmetric group $S_n$ acting on the $n$ points $\lbrace 1, 2, 3, . . . , n\rbrace$, let $g\in S_n$ be the n-cycle $(1,2,3,. . . , n)$. By applying the Orbit-Stabiliser Theorem or otherwise, prove that $C_G(g) = <g>$ (where $C_G(g)$ is the centraliser of $g$).

My attempt Let $g = (1,2,3,. . . , n) \in G=S_n$. Since $gg^i = g^ig = g^{i+l}$, we have $<g>$ contained in $G$, and since $<g> = |g| = n$, it is enough to show that $|C_G(g)| = n$.

It can be shown that $|Cl_G(g)=|G|/|C_G(g)|$ (1) by application of the Orbit-Stabiliser Theorem.

It is known that $|S_n|=n!$, and the order of a conjugacy class $|Cl_G(g)|=\frac{n!}{\Pi r^{n_r}*n_r!}$ where $r$ is the cycle length and $n_r$ is number of those cycles. Since $g=(1,..,n)$ then its cycle length is $n$ which occurs once, therefore $|Cl_G(g)|=\frac{n!}{n}=(n-1)!$.

So by (1), $|C_G(g)|=\frac{|G|}{|Cl_G(g)|}=\frac{n!}{(n-1)!}=n$.

Is there something wrong with this argument?

• There's nothing wrong, but by assuming the formula for order of ${\rm Cl}_G(g)$, you are really assuming something much stronger than the result you are trying to prove. You really ought to try and prove directly that $|{\rm Cl}_G(g)| \ge (n-1)!$, which you can do simply by writing down $(n-1)!$ distinct conjugates of $g$. – Derek Holt Jun 5 '14 at 20:01
• Would this work in the exam tomorrow? :) – lifin Jun 5 '14 at 20:04
• Oh I see! No, it wouldn't because the formula for $|{\rm Cl}_G(g)|$ was not proved in the course, so you can't use it. Good luck, anyway! – Derek Holt Jun 5 '14 at 20:10

More elementary thoughts: Suppose $\sigma(12\cdots n)\sigma^{-1}=(\sigma(1)\sigma(2)\cdots\sigma(n))=(12\cdots n)$.
So $\sigma(1)=i$, $\sigma(2)=i+1$, $\sigma(3)=i+2$, and so on. This means $\sigma=g^{\rm what~power}$?