Galois Group of $(x^2-5)(x^2-7)$ over $ \mathbb{Q}$ I want to calculate the Galois Group of $p= (x^2-5)(x^2-7) \in \mathbb{Q} $. Because I saw a solution to this problem in a German textbook and think it is incorrect.
So: $p$ is irred. over $\mathbb{Q}$, and $\operatorname{char}( \mathbb{Q} )=0$ therefore the Galois group is a subgroup of $S_4$. Furthermore $ |\operatorname{Gal}(p)|=| \mathbb{Q} ( \sqrt{5}, \sqrt{7}): \mathbb{Q} | = | \mathbb{Q} ( \sqrt{5} ): \mathbb{Q} (\sqrt{7}) | \cdot| \mathbb{Q} ( \sqrt{7}): \mathbb{Q} |= 2 \cdot 2 = 4 $. Since $ \mathbb{Q}( \sqrt{5} ,\sqrt{7}) $ is the splitting field of $p$.
Now we have only $C_4$ (cyclic Group) or $V_4$ (Klein four-group) as possibilities.
If the discriminant of a polynomial is a square (over the the field $K = \mathbb{Q} $) then the Galois group is  subgroup of $A_n = A_4$. It is $\operatorname{disc}(p)= 8960 $ which is not a square of $\mathbb{Q}$ so we are left with $G= C_4$.
Does this look right to you? or did I miss something?
I apologize for my English and thank you in advance!

You have factored $p$, so it is apparently false that $p$ is irreducible over $\mathbb Q$. But your computation of $|\text{Gal}(p)|=4$ is more or less correct, except $|\mathbb Q(\sqrt 5, \sqrt 7):\mathbb Q(\sqrt 7)|=2$ (mistakenly as $\mathbb Q(\sqrt 5):\mathbb Q(\sqrt 7)|$) needs some justification.
It must be isomorphic to $V_4$, not $C_4$, as $C_4$ constains only one proper subgroup of order $2$, but the splitting field has at least two distinct intermediate subfields: $\mathbb Q(\sqrt 5)$ and $\mathbb Q(\sqrt 7)$.
