Automorphism group of $\mathbb Q[\sqrt{13}, \sqrt[3]{7}]$. I want to calculate group of $\mathbb Q$-automorphisms of  $\mathbb Q[\sqrt{13}, \sqrt[3]{7}]$.
$\mathbb Q[\sqrt{13}, \sqrt[3]{7}]$ is separable as a separable extension of separable extension. Thus $|Aut| = [\mathbb Q[\sqrt{13}, \sqrt[3]{7}]: \mathbb Q] = 6$. So it is either $\mathbb Z_6$ or $S_3$.
I also know that for each $a \in Aut (a(\sqrt{13}) = \sqrt{13} \lor a(\sqrt{13}) = -\sqrt{13})$ and $a(\sqrt[3]{7}) = \sqrt[3]{7}$. 
But I still cant understand which group $Aut$ is isomorphic to and how this automorphisms are acting on the field.
Thanks!
 A: The extension $\Bbb Q(\sqrt{13},\sqrt[3]7)/\Bbb Q$ has degree six but is not Galois, so  it has fewer than $6$ automorphisms. It just has two, the non-trivial one taking $\sqrt{17}$ to $-\sqrt{17}$ and fixing $\sqrt[3]7$.
There is only one cube root of $17$ in this field, so it is fixed by
all automorphisms. The field is a quadratic extension of $\Bbb Q(\sqrt[3]7)$, so does have a non-trivial automorphism.
A: The Automorphism group is $\mathbb Z/2\mathbb Z $ because any such automorphism needs to fix $\mathbb Q$ and so must take a root of a polynomial with $\mathbb Q$ coefficients to another such root. Now $\mathbb Q(\sqrt[3]{7},\sqrt11)$ is real so does not contain the other two roots of $x^3-7$. So you have to take $\sqrt[3]{7}\mapsto\sqrt[3]{7}$ and all you can do is take $\sqrt{11}$ to itself or to $-\sqrt{11}$. So the automorphism group is $\mathbb Z/2\mathbb Z$.
This is an example of a non-Galois extension so the automorphism group having to be equal to the degree of the extension does not hold.
