Let $F$ be splitting Field of $x^3 -7 \in \mathbb{Q} [x]$ and let $G$ be the splitting field of $$ (x^3-7) (x^2-3) \in \mathbb{Q}[x]$$
Show that $Aut(G : \mathbb{Q})$ is the product of two normal subgroups which have trivial intersection
I know that the roots for $$ x^3-7$$ are $$\sqrt[3]{7}w_3^0,\sqrt[3]{7}w_3^1,\sqrt[3]{7}w_3^2 $$
and roots for $x^2-3$ are
$$ \sqrt{3},-\sqrt{3}$$
I think that that $G$ over $Q$ is of degree 6. F is normal since it splits the polynomial and call $H$ the splitting field of the deg 2 poly. They have nothing in common and generate $G$ over $Q$
Another way is to find what it is in $S_6$ and show that that group has two normal subgroups that generates it.
Appreciate Help. Reference where to read this subject clearly, how to solve this problem in sage, or good examples. Thanks