Finding the size of a Galois Group of a splitting field for a polynomial of degree 6. Let $f(x) = x^6 + ax^4 + bx^2 + c$ with a,b,c ∈ $\mathbb{Q}$ be an irreducible polynomial in $\mathbb{Q}$[x].  Let K be the splitting field of f(x) over $\mathbb{Q}$ and let G = Gal[K:$\mathbb{Q}$]. Then prove that |G| ≤ 48.
My attempt: 
I have that [$\mathbb{Q}$(α) : $\mathbb{Q}$] = 6.  I claim that one of the other roots must be a complex conjugate of α, which we can call β, and must be contained in $\mathbb{Q}$(α).
I then claim that [$\mathbb{Q}$(α, γ) : $\mathbb{Q}$(α)] = 4, with  the complex conjugate of γ.
I make a similar claim with [$\mathbb{Q}$(α, γ, ɛ) : $\mathbb{Q}$(α, γ)] = 2, which gives [K:$\mathbb{Q}$] = 48.
This answer was thrown together.  I was given a hint to note that all exponents of the polynomial are even, but I did not know what to gather from that.
 A: There is definitely some handwaving going on here regarding the mere existence of complex roots.  However, a similar argument to what you're using comes out of the observation that if $\alpha$ is a root, then so too is $-\alpha$, which is how the even exponents come into play. Therefore, you need only adjoin at most $3$ roots of $f$ to $\mathbb{Q}$ to yield the splitting field, since $-\alpha \in \mathbb{Q}[\alpha]$ by virtue of the field axioms.  Can you figure out a worst-case scenario for the degrees of each extension $\mathbb{Q} \subset \mathbb{Q}[\alpha_1] \subset \mathbb{Q}[\alpha_1, \alpha_2] \subset \mathbb{Q}[\alpha_1, \alpha_2, \alpha_3] = K$?  This information will give you a worst-case scenario for $[K:\mathbb{Q}]$, and of course $[K:\mathbb{Q}] = |\operatorname{Gal}(K/\mathbb{Q})|$.
A: Consider the splitting field $F$ of $g(x)=x^3+ax^2+bx+c$, it is a subgroup of $S_3$, with roots $u,v,w$, so degree at most 6. Now adjoin $\sqrt{u},\sqrt{v},\sqrt{w}$ to $F$ to get the splitting field of $f$, an extension of degree at most 8. So the splitting field is an extension of degree at most 48.
