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

I'm having some trouble with this question. I know the $|S_n| = n! $ and if we need to prove that $C_G(g) = <g>$ then the order of $C_G(g) = n$ and by the Orbit-Stabiliser Theorem, $|C_G(g)||Cl_G(g)| = |G|$ so I need to show that $|Cl_G(g)| = (n-1)!$ where $Cl_G(g)$ denote the conjugacy class containing g, i.e. the orbit. $C_G(g)$ denotes the centraliser of G i.e. the stabiliser.

My understanding of conjugacy classes isn't great, and i assume that its not totally obvious that the order of the conjugacy class is $(n-1)!$, but i can't seem to show that this indeed is the order.

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    $\begingroup$ Are you aware of the fact that two elements of $S_n$ are conjugate if and only if their cycle types are the same? $\endgroup$ – Servaes May 28 '14 at 11:28
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It is not too hard to prove that $|{\rm Cl}_G(g)| \ge (n-1)!$, but here is another way to solve the problem.

You know that $\langle g \rangle \le C_G(g)$, and $\langle g \rangle$ acts transitively on $\{1,2,\ldots,n\}$ and hence so does $C_G(g)$.

So, by the Orbit-Stabilizer Theorem, $|C_G(g)| = n|S|$, where $S = {\rm Stab}(C_G(g),1)$. So the problem is equivalent to proving that $S$ is trivial.

To do that, let $s \in S$. Then, $g^{k}sg^{-k} = s$ for any $k$ with $1 \le k \le n$, so $s(k) = g^{k}sg^{-k}(k) = k$ and hence $s=1$.

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