A group $G$ acts on itself via the conjugation action


Find the orbit of $\sigma=\left(123\right)$ under the conjugation action of $S_{3}$. Then find the stabilizer of $\sigma$. Check that the Orbit-Stabilizer theorem holds in this case.

My attempt:

Suppose $\tau$ and $\sigma$ are cycles, so there are $n!$ orderings of the symbols and $n$ rearrangements for each $n$-cycle, so in total there are $\left(n − 1\right)!$ $n$-cycles. Therefore $\sigma$ has $\left(n − 1\right)!$ elements.

Then apply the orbit-stabilizer theorem. Let $G=S_{3}$. Let $G\sigma$ be the orbit of $\sigma$ and $G_{\sigma}$ its stabilizer. Then

$$\left|G\sigma\right|=\left|G:G_{\sigma}\right|\rightarrow\left(n − 1\right)!=\frac{n!}{\left|G_{\sigma}\right|}\rightarrow\left|G_{\sigma}\right|=n=3$$

Am I right? And is this all I need to answer the question?

Thanks for your attention, I really appreciate any help you can provide.

  • $\begingroup$ "Let Gσ be the orbit of σ and Gσ its stabilizer"? You should edit the text and make it more clear. Please also use MathJax math.meta.stackexchange.com/questions/5020/…. $\endgroup$ – eranreches Dec 7 '17 at 16:11
  • $\begingroup$ Thanks to point it out, I have changed it, does it look clearer? $\endgroup$ – BlackSky Dec 7 '17 at 16:15

You're not addressing the question. You are asked to find explicitly both the orbit and the stabilizer of $\sigma$. The idea is that $\tau\left(a_{0}\dots a_{n}\right)\tau^{-1}=\left(\tau\left(a_{0}\right)\dots\tau\left(a_{n}\right)\right)$ so the orbit of a $3$-cycle is the set of all $3$-cycles, which contains $2$ elements and is easy to find. Now find all $\tau$'s that keep $\sigma$ unchanged. It is not too hard to check all $6$ elements of $S_{3}$. Now you should use the orbit-stabilizer theorem and show that it is satisfied.


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