# Galois group of degree 7 polynomial

From Bosch‘s Algebra (p. 368):

Determine the galois group of $$X^7 - 8X^5 - 4X^4 + 2X^3 - 4X^2 + 2 \in \mathbb Q[X]$$ and decide if it‘s solvable or not.

How does one find the galois group of that polynomial?

My usual way is to find the roots $$\alpha_1, \ldots, \alpha_n$$, find a splitting field $$L = L(\alpha_1, \ldots, \alpha_n)$$ and then study the structure of the galois group $$\operatorname{Gal}(L/\mathbb Q) = \operatorname{Gal}(f)$$ as a subgroup of $$S_n$$.

However that does not seem to work in this case since already finding roots seems difficult. What is a useful approach in this case?

Hint

1. Use Eisenstein's Criterion to show that $$f$$ is irreducible.
2. Show that $$f$$ has exactly $$5$$ real roots. (Since $$f$$ has negative discriminant, $$f$$ has either exactly $$1$$ or exactly $$5$$ real roots, hence it suffices to show that $$f$$ has at least $$2$$ real roots.)

What kind of permutations do (1), (2) (separately) imply occur in $$\operatorname{Gal}(f)$$?

1. So, $$\operatorname{Gal}(f)$$ contains a $$7$$-cycle.

2. Complex conjugation exchanges the $$2$$ nonreal roots of $$f$$, hence it determines a transposition in $$\operatorname{Gal}(f)$$.

• Thank you! Why does (1) imply that $\operatorname{Gal}(f)$ contains a $7$-cycle? But then, since the Galois group contains a $7$-cycle and a transposition, it is $S_7$, right?
– ATW
Commented Feb 13 at 17:45
• @ATW By Cauchy's theorem, it is sufficient to show that the order of $\text{Gal}(f)$ is divisible by $7$. Show this by using irreducibility.
– Mark
Commented Feb 13 at 18:01
• @ATW The Galois group has an element of order $7$, namely a $7$-cycle, and the subgroup generated by it has order $7$ and divides the group order by Lagrange. Commented Feb 13 at 20:06
• @ATW Take any root $\alpha$. Then $\mathbb{Q}(\alpha)$ is a subfield of the splitting field. What is $[\mathbb{Q}(\alpha):\mathbb{Q}]$?
– Mark
Commented Feb 13 at 20:06
• @ATW Since $f$ is irreducible, it itself must be the minimal polynomial.
– Mark
Commented Feb 13 at 20:38