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Find the Galois group of $(x^5)-1$ ∈ Q[x], its subgroup diagram and the corresponding subfield diagram of the splitting field of the polynomial $x^5 − 1$. Is it isomorphic to the Galois group of $x^5 − 2$∈ Q[x]?

I know that the zeros of $x^5-1$ are the 5th roots of unity and am having trouble where to go from there

Sara Badenhausen
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  • How about factorising $x^5-1$ into irreducibles? Also please see https://math.meta.stackexchange.com/questions/5020/ – Angina Seng May 10 '18 at 05:45
  • @LordSharktheUnknown Sorry! this is my first stack exchange question! I know that (x^5 -1) factors into (x-1)(x^4 + x^3 + x^2 + x + 1). Im really lost – Sara Badenhausen May 10 '18 at 05:54

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The Galois group of $x^n-a$ is known:

Proposition [Jacobson, Velez]: One has $G\cong \mathbb{Z}/n \rtimes (\mathbb{Z}/n)^{\times}$ if and only if $n$ is odd, or $n$ is even and $\sqrt{a}\not\in \mathbb{Q}(\zeta_n)$.

Reference: Computing the Galois group of polynomials $x^n-a \in \mathbb{Q}[x]$

For $n=5$ we obtain $G=\Bbb{Z}/5\rtimes \mathbb{Z}/4$, both for $a=1$ and $a=2$.

For the subfields, see for example here:

Subfield of the Galois Group of $x^5 - 1$

Dietrich Burde
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