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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.

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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 $=3+3=6$.

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  • $\begingroup$ I think the number of elements in an orbit should be less than or equal to $|Gal(\mathbb F_9/\mathbb F_3)|$. $\endgroup$
    – Shodharthi
    May 9 '15 at 6:47
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    $\begingroup$ @Chiranjeev Kumar can you please explain why | $Gal F_{9} / F_{3} $ | =3 ? $\endgroup$
    – James
    Aug 7 '20 at 10:49
  • $\begingroup$ @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. $\endgroup$
    – Why
    Mar 23 '21 at 16:35
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Hint: What is the order of the Galois group? What is the degree of this extension? How many groups are there with this order?

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