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?

  • $\begingroup$ 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 $g$. $\endgroup$ – Derek Holt Jun 5 '14 at 20:01
  • $\begingroup$ Would this work in the exam tomorrow? :) $\endgroup$ – lifin Jun 5 '14 at 20:04
  • $\begingroup$ 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! $\endgroup$ – Derek Holt Jun 5 '14 at 20:10

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 $\sigma=g^{\rm what~power}$?


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