field extension with Galois group Q8

Let α=(2+2)(3+3),E=(α). We will show that E is Galois over and that G=Gal(E/)Q8 (the group of quaternions).

We begin by showing that [E:]=8. Let F=(2,3)=(2+3). Claim that FE. To show that they are not equal, we show that αF, i.e. that (2+2)(3+3) is not a square in F. If it were, say (2+2)(3+3)=c2,cF, take σGal(F/) to be the element


Then (2+2)(3+3)σ((2+2)(3+3))=(cσ(c))2, so


But cσ(c)=TrF/(2)(c)(2), and thus 6=(cσ(c)2+2)2, so 6(2), a contradictionMathworldPlanetmathPlanetmath. Thus FE. We show that FE by showing that 2+3E.

(2+2)(3+3)=6+32+23+6E, so
32+23+6E, so
(32+23+6)2=36+12(2+3+6)E, so
2+3+6E, so
32+23+6-2-3-6=3+22E, so
(3+22)2=11+46E, so
6E, so

So FE and thus [E:F]=2. Then [E:]=[E:F][F:]=8.

Now, the irreducible polynomialMathworldPlanetmath f(x) for (2+2)(3+3) over is the productMathworldPlanetmathPlanetmathPlanetmath of x-τ((2+2)(3+3)) as τ ranges over Gal(F/):


so that f(x2) is a degree 8 polynomialMathworldPlanetmathPlanetmathPlanetmath with α as a root. In fact,


This polynomial must be irreduciblePlanetmathPlanetmath since α is of degree 8, so f(x2) is the minimal polynomial for α over . The roots of f(x2) are obviously


Furthermore, it is easy to see that each of these roots lies in E, for

α(2-2)(3+3) =2(3+3)F
α(2+2)(3-3) =6(2+2)F
α(2-2)(3-3) =26=23F

so dividing through by α we see that


Thus E is in fact Galois over , is the splitting fieldMathworldPlanetmath for f(x2), and has Galois groupMathworldPlanetmath G=Gal(E/) of order ( 8.

G acts transitively on the roots of f(x2), and E=(α), so an element of G is determined by the image of α. Thus the elements of G are the automorphismsPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of E that map α to any of the eight roots of f(x2). Let

α=(2+2)(3+3) β=(2-2)(3+3)
γ=(2+2)(3-3) δ=(2-2)(3-3)

and let σ:αβ,τ:αγ be elements of G.

σ(α2)=β2, so σ(2+2)σ(3+3)=(2-2)(3+3). This is an equation in F, so regarding σ as an automorphism of F/, it must be the automorphism 2-2,33. Since αβ=2(3+3), we have σ(αβ)=-αβ and thus that σ(β)=-α. It follows that σ is an element of order ( 4 in G.

Similarly, τ(α2)=γ2, so τ(2+2)τ(3+3)=(2+2)(3-3), so that τ, regarded as an automorphism of F/, must be 22,3-3. Since αγ=6(2+2), we have τ(αγ)=-αγ, so that τ(γ)=-α, and τ is also an element of order ( 4 in G. Note also that σ2(α)=-α=τ2(α), so that σ2=τ21.

Looking at στ,




and thus τσ3(α)=τσσ2(α)=-τσ(α)=-δ=στ(α). So στ=τσ3.

Putting this all together, we see that G is generated by σ,τ, and that the generatorsPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath satisfy the relationsMathworldPlanetmathPlanetmath


Define φ:GQ8 by φ(σ)=i,φ(τ)=j. This is easily seen to be a homorphism, and φ(στ)=ij=k, so φ is surjectivePlanetmathPlanetmath and is thus an isomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath since both groups have order ( 8. Thus Gal(E/)Q8.

Title field extension with Galois group Q8
Canonical name FieldExtensionWithGaloisGroupQ8
Date of creation 2013-03-22 17:44:28
Last modified on 2013-03-22 17:44:28
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 7
Author rm50 (10146)
Entry type Example
Classification msc 12F10