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?
In order for your argument to be acceptable, the formula you have for calculating conjugacy class size in terms of cycle type would have to be taken for granted as something you're allowed to use.
More elementary thoughts: Suppose $\sigma(12\cdots n)\sigma^{-1}=(\sigma(1)\sigma(2)\cdots\sigma(n))=(12\cdots n)$.
If two cycles are equal, then they are the same up to cycling the numbers in the notation.
So $\sigma(1)=i$, $\sigma(2)=i+1$, $\sigma(3)=i+2$, and so on. This means $\sigma=g^{\rm what~power}$?