How to find the number of orbits in $S_3$ under $H=\langle(123)\rangle$. By conjugation action ?
2026-03-30 09:52:32.1774864352
How To find number of orbits in S3 by using burnside's Formula
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Each element of $A_3$ is in its own orbit. Then by direct calculation you can show that the remaining elements, which are all of order $2$, are all in the same orbit. Thus the number of orbits is $4$.
If we want to use Burnside's formula, note that each of $(123)$ and $(132)$ have $3$ fixed points, and the identity has $6$, so the result is $(3+3+6)/3=4$.