# 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 $Gal(L/Q)$

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Welcome to Math.SE! +1 for showing your thoughts and the plan of attacking the problem. –  Jyrki Lahtonen Mar 18 '13 at 18:27
Sage says $C_2 \times A_4$. –  Myself Mar 18 '13 at 20:30
Reduction mod 3 shows there is a 6-cycle in your Galois group. Reduction mod 13 shows there is a 2-cycle, so the group has order at least 12. But a 2-cycle can't normalize a 6-cycle in $S_6$. so the group must have maximal order, namely 24. –  user641 Mar 18 '13 at 20:50
As for the structure of the Galois group itself, since $[\mathbb{Q}(x_1):\mathbb{Q}]=3$ and is a splitting field, your Galois group has a normal subgroup of order $8$, and it is elementary abelian (it is gotten by adjoining square roots). So your group is the semidirect product $(C_2\times C_2\times C_2)\rtimes C_3$. Since there is a 6-cycle, one of those $C_2$ is centralized by $C_3$, and your group is $C_2\times (C_2\times C_2)\rtimes C_3$. This last semidirect product is non-trivial, since the 6-cycle wasn't central, so we get $C_2\times A_4$. –  user641 Mar 18 '13 at 20:52

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 $G$:

1. $C_6$
2. $C_6\times C_2$
3. $C_6\times C_2\times C_2$
4. $C_2\times A_4$

[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.

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Thanks very much for your answer, really appreciate that. –  Gogwaben Mar 19 '13 at 1:05
Irreducibility mod 3 can be seen (without CAS) as follows. The polynomial $p(x)=x^3-x+1$ is well known to be irreducible ($x^p-x-a$). Let $\alpha\in GF(27)$ be any one of its roots. Then $\alpha^{13}=\alpha\cdot\alpha^3\cdot\alpha^9$ is the norm of $\alpha$ (down to the prime field). But this is equal to $-p(0)=-1$. Therefore $\alpha$ is of multiplicative order $26.$ Consider the relevant polynomial $f(x)=p(x^2)$. The above calculation shows that the roots of $f$ are of multiplicative order $52=2\cdot 26$. Therefore they are all in $GF(3^6)$, i.e. $f(x)$ is irreducible in $\mathbb{Z}_3[x]$. –  Jyrki Lahtonen Mar 19 '13 at 10:36