# Number of Galois conjugates

Let $L/\mathbb Q$ be a (finite) Galois extension of degree $n$ with Galois group $\Gamma$. We know that there is a primitive element or generator $\alpha$ of this extension.

My question:

Is the number of Galois conjugates of $\alpha$ (i.e. the length of the orbit of $\alpha$ under $\Gamma$) smaller than or equal to the order of the Galois group?

And under which conditions are they the same?

• If $\alpha$ is a generator, then it necessarily has $n$ distinct conjugates. For if $\alpha$ is fixed by an element $\sigma\in\Gamma$ then $\alpha$ belongs to a proper subfield, namely $\operatorname{Inv}(\langle\sigma\rangle)$, and hence cannot be a generator. – Jyrki Lahtonen Jun 22 '15 at 15:56

What does the Orbit-Stabilizer theorem tell us? If $\Gamma = \textrm{Gal}(L/\mathbb{Q})$, then because the extension is finite, $$|\Gamma | = |\Gamma \cdot \alpha| \cdot | \textrm{Stab}_{\Gamma}(\alpha)|.$$ In particular, the number $|\Gamma \cdot \alpha|$ of Galois conjugates of $\alpha$ is less than or equal to the order $|\Gamma|$ of the Galois group. (It will, in fact, be an equality: the stabilizer of $\alpha$ must be trivial, else $\alpha$ would not be a generator.)
• let $\alpha$ be an algebraic element over $\mathbb Q$, and $f$ is the irreducible polynomial of $\alpha$, with $deg(f)=n$. What can we say about other roots of $f$ in the field $\mathbb Q(\alpha)$? Are they necessarily conjungate with $\alpha$? – Yilong Zhang Jun 23 '15 at 8:16
• Yes all of the roots of $f$ will be Galois conjugate - the key point is that the extension $\mathbb{Q}(\alpha) / \mathbb{Q}$ is Galois. – msteve Jun 23 '15 at 15:23
• Now all of the roots of $f$ are Galois conjugate, then can we say every permuatation $\sigma \in S_n$ act on these $n$ roots induce a $\mathbb Q$-automorphism of the whole field $\mathbb Q(\alpha)$? If yes, then the order of $ord(Gal(\mathbb Q(\alpha)/\mathbb Q)) =ord(S_n)=n!$. Is there anything wrong with my reasoning? – Yilong Zhang Jun 23 '15 at 15:32
• Just because the Galois group acts transitively on the roots does not mean it is equal to $S_n$, or more generally, there are groups that act transitively that are not symmetric groups e.g. $\mathbb{Z}/n\mathbb{Z}$ acts transitively on itself but is not equal to $S_n$. – msteve Jun 23 '15 at 15:40
• For a Galois theory example, consider the polynomial $x^3-3x-1$, which is irreducible over $\mathbb{Q}$. The splitting field of this polynomial has Galois group $A_3$. – msteve Jun 23 '15 at 15:41