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