## Compute the degree of the splitting field of $x^{12} – 2$ over $\mathbb{Q}$ and describe its Galois Group as a semidirect product

I have the polynomial $$f(x) = x^{12} – 2$$. I have to compute the degree of the splitting field over $$\mathbb{Q}$$ and describe its Galois Group as a semidirect product.

Clearly the splitting field is $$E= \mathbb{Q}(\zeta,\alpha$$), where $$\zeta$$ is a primitve $$12^{\text{th}}$$ root of unity and $$\alpha = \sqrt[12]{2}$$. The pertinent intermediate fields are $$L=\mathbb{Q}(\alpha)$$ and $$K=\mathbb{Q}(\zeta)$$. Since we have the degree of the splitting field as $$[E:\mathbb{Q}]=48$$, with $$[L:\mathbb{Q}]=12$$ and $$[K:\mathbb{Q}]=\varphi(12)=4$$

I look at $$\sigma:\zeta \mapsto \zeta,\ \alpha \mapsto\zeta \cdot \alpha$$, this automorphism fixes $$K$$ and is of order $$12$$, therefore we have $$N = \text{Gal}(E/K) = \langle \sigma \rangle \cong C_{12}$$, and we know $$N$$ is a normal subgroup of $$G = \text{Gal}(E/\mathbb{Q})$$, since $$K/\mathbb{Q}$$ is Galois.

Now, considering $$\tau:\zeta \mapsto \zeta^5,\ \alpha \mapsto \alpha$$. This automorphism fixes $$L$$ and is of order $$4$$, and therefore $$H = \text{Gal}(E/L) = \langle \tau \rangle \cong C_{4}$$, and we know $$H$$ is not normal, since $$L/\mathbb{Q}$$ is not Galois.

Am I correct? Is there more I can add in my reasonging? And how do I conclude that $$G= C_{12} \rtimes C_{4}$$? Any help is appreciated