Let $K=\mathbb{Q}(\sqrt{2},i\sqrt{5})$ and $G={\rm Gal}(K,\mathbb{Q})$. Show that for every $n$-primary root of unity that ${\rm Gal}(K(\zeta),\mathbb{Q})$ is solvable.
My approach untill I got stuck:
I found that $G\simeq\mathbb{Z}_2\times \mathbb{Z}_2$. That means that every subgroup of $G$ is normal under $G$ (and of course every subfield of $K$ is a splitting field over $\mathbb{Q}$, including $K$.
We know that $K$ is a root extension since the series:$\mathbb{Q}\subseteq \mathbb{Q}(\sqrt{2})\subseteq \mathbb{Q}(\sqrt{2},i\sqrt{5})=K$ include every attribute needed.
Also $K(\zeta)$ is for sure a splitting field over $\mathbb{Q}$ since $\mathbb{Q}(\zeta)$ is a splitting field itself.
From the above we can conclude that $G$ is solvable. Yet the question implies that ${\rm Gal}(K(\zeta),Q)$. Well since $K(\zeta)$ is a splitting field over $\mathbb{Q}$ then $G$ is a normal subgroup of ${\rm Gal}(K(\zeta),Q)$. We know for a group to be solvable, other than every subgroup of a subgroup etc being normal under each other, we need ${\rm Gal}(K(\zeta),Q)/G$ be abelian or because we have finite groups, we $|{\rm Gal}(K(\zeta),Q)/G|=p$ (prime number).
From the fundamenta theorem of Galois theory we know that ${\rm Gal}(K(\zeta),Q)/G \simeq{\rm Gal}(K(\zeta),K)$.
And thus we need $|{\rm Gal}(K(\zeta),K)|=p$
Since $K(\zeta)$ is a splitting field $\mathbb{Q}$, then naturally $K(\zeta)$ is a splitting field over $K$ and for that reason $|{\rm Gal}(K(\zeta),K)|=[K(\zeta):K]\leq \phi(n).$
This is where I don't know how to follow. I'm aware of a lemma saying:
Let $F$ be a field that contains $n$-primary root of unity and $a \in \mathbb{C}$, for which $a^n \in F$. Then ${\rm Gal}(F(a),F)$ is solvable.
But was unable to use it somehow.
Any help would be appreciated. Thank you in advance