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!

Dealing with the confusion about the discriminant only as the others have handled the main question.

The group $S_4$ of permutations of $1,2,3,4$ contains several copies of $V_4$. The subgroup $H_1=\{id, (12)(34), (13)(24), (14)(23)\}$ consists of only even permutations whereas $H_2=\{id,(12),(34),(12)(34)\}$ contains both even and odd permutations. Therefore it is not possible to use the discriminant to rule out the Klein four as a Galois group.
Your polynomial has the latter variant given that we are looking at the four roots $x_1=\sqrt5$, $x_2=-\sqrt5$, $x_3=\sqrt7$, $x_4=-\sqrt7$. An automorphism will either swap $x_1$ and $x_2$, or swap $x_3$ and $x_4$, or swap both pairs or neither pair. Looks like $H_2$, doesn't it?

Mind you, if you, instead of $(x^2-5)(x^2-7)$, look at the polynomial
$$g(x)=x^4-24x^2+4$$
its splitting field is also $K=\Bbb{Q}(\sqrt5,\sqrt7)$. That's because the zeros of $g(x)$ are $\pm\sqrt5\pm\sqrt7$, all four sign combinations. This time the Galois group $G=Gal(K/\Bbb{Q})$ looks like $H_1$ instead. There is no contradiction because this time we study the action of $G$ on a different set of numbers. Also observe that the discriminant of $g(x)$
$$\Delta(g)=2^{14}\cdot5^2\cdot7^2$$
is a perfect square – all in line with the general result that you recalled.
