Compute Galois Group of splitting field of $x^p-a$ over $\mathbb Q$ I am having trouble computing the Galois Group of the splitting field $E$ of $x^p-a$ (where $p$ and $a$ are prime) over $\mathbb{Q}$. 
Let $w$ be a $p^\text{th}$ root of unity, and $\alpha$ a root of $x^p-a$. $E$ should be isomorphic to $\mathbb{Q}(w,\alpha)$. The extension $\mathbb{Q}(w,\alpha)/\mathbb{Q}(\alpha)$ has basis $\{w,\ldots, w^p\}$ and $\mathbb{Q}(\alpha)/\mathbb{Q}$ has basis $\{1,\alpha,\ldots, \alpha^{p-1}\} \implies [E:\mathbb{Q}] = p(p-1)$.
But I am not sure what the Galois Group should be. Is it $C_p \times C_{p-1}$, and how do I justify this? There can only be one (cyclic) group of order $p$, but there could be other groups of order $p-1$ that are not cyclic. 
 A: You should not expect the Galois group to be abelian, even though the correspondence gives you two subgroups $\operatorname{Gal}(\Bbb Q(\omega, \sqrt[n](\alpha))/\Bbb Q(\omega))$ and $\operatorname{Gal}(\Bbb Q(\omega, \sqrt[n](\alpha))/\Bbb Q(\sqrt[n]{a}))$ that are cyclic of order $p$ and $p-1$, respectively.  These two subgroups generate the whole group, and the one of order $p$ must be normal, but this only implies that $G$ is a semidirect product of these groups.  It remains to determine the action of $C_{p-1}$ on $C_p$.
Note that this action will in general not be trivial.  An example is the splitting field for $x^3-2$.  This extension is degree six, but we also know that the Galois group is a subgroup of $S_3$.  Therefore, it is the nonabelian group $S_3$.   In any case, since we know that $\Bbb Q(\sqrt[n]{a})$ is not a normal extension of $\Bbb Q$, that implies that the corresponding subgroup $C_{p-1}$ will not be a normal supgroup.
My advice, if you are having trouble determining the action, is to write down generators for your two cyclic subgroups explicitly as field automorphisms, and compute the conjugation action by hand.
