Galois group for $x^6-7x^2+7$ What is the galois group for $x^6-7x^2+7$?
My approach: first consider $x^3-7x+7$, it is irreducible by Esenstein, and its discriminant is a square, hence the galois group for this polynomial is$A_3$, and the splitting field for it is$Q(x_1)$ where $x_1$ is a root for $x^3-7x+7$, $[Q(x_1):Q]=3$
The problem I am having now is to find $[L:Q]$ where $L$ is the splitting field for $x^6-7x^2+7$ I know it is at least 6, and at most 24, since $[L:Q(x_1)]$ is at most 8 and at least 2.( Adjoin the square root of the three roots of $x^3-7x+7$ gives $L$ and $[L:Q(x_1)]=8$ when the three square roots are independent)
If I know $[L:Q]$ then I also know $|Gal(L/Q)|$ and moreover I know it is transitive and contains $A_3$ as a normal subgroup, then I should be able to compute $Gal(L/Q)$
 A: I gave a sketch of the argument above to show that this Galois group is $C_2\times A_4$, but it involved showing your polynomial is irreducible $\pmod{3}$, which I don't think is very easy (but I don't remember a lot of this Galois theory stuff!). So here is a different argument, which uses more group theory.
We will still use the fact that
$$ x^6-7x^2+7\equiv (x-4)(x+4)(x-5)(x+5)(x^2+2)\pmod{13},$$
which is easy enough to find by just checking roots.  So we still know our Galois group $G$ contains a 2-cycle.
Now you've already shown that $G$ has a cyclic quotient of order $3$ (so that the Sylow 2-subgroup is normal), and that $|G|$ is of the form $2^n\cdot3$, with $1\le n\le3$. Also, we know that the Sylow 2-subgroup is elementary abelian, since it is gotten by adjoining square roots.  Finally, we know $G$ has a subgroup of index $2$, since it contains an odd permutation (so $G\cap A_6$ works).
Thus $G$ looks like $P\rtimes C_3$, where $P$ is an elementary abelian group of order $2$, $4$, or $8$. Using this index 2 subgroup fact, we can see there are only the following possibilities for $G$:


*

*$C_6$

*$C_6\times C_2$

*$C_6\times C_2\times C_2$

*$C_2\times A_4$


[Note: we can avoid a lot of "cases" where $|P|=8$ by noting that $8\equiv2\pmod{3}$, so there's always one $C_2$ that is fixed by the $C_3$.]
So if we can show $G$ is not abelian, we will be done (case 4). But if $G$ was abelian, that 2-cycle from above would be central.  And the centralizer of a 2-cycle - like $(12)$ - in $S_6$ is of the form $S_2\times S_4$.  And that subgroup is not transitive. So we are done.
