5
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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 $=3+3=6$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

Hint: What is the order of the Galois group? What is the degree of this extension? How many groups are there with this order?

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.