# Number of orbits for the action of Galois group G. [duplicate]

Let G be the Galois group of a field with nine elements over its subfield with three elements. Then find the number of orbits for the action of G on the field with 9 elements.

Clearly $|Gal(\mathbb F_9/\mathbb F_3)|=2$. Since $3$ elements are fixed so there are $3$ singleton orbits. Number of elements in other orbits are divisors (other than $1$) of $|Gal(\mathbb F_9/\mathbb F_3)|$, i.e. $2$. So there are $3$ orbits containing $2$ elements.
Hence total number of orbits $=3+3=6$.
• I think the number of elements in an orbit should be less than or equal to $|Gal(\mathbb F_9/\mathbb F_3)|$. May 9 '15 at 6:47
• @Chiranjeev Kumar can you please explain why | $Gal F_{9} / F_{3}$ | =3 ? Aug 7 '20 at 10:49
• @Tim, Frobenious $\sigma$ is involution (i.e., $\sigma^2=$Identity) on $\mathbb{F}_9$. So every orbits on $\mathbb{F}_9 \setminus \mathbb{F}_3$ has the size $\{\sigma, \sigma^2=identity \}$ and hence the rest $6$ elements $(\bar 3, \bar 4, \bar 5,\bar 6,\bar 7, \bar 8$) consists of $3$ different orbits.