Determine the Galois group of the splitting field of $(x^3-1)(x^2-5)$ over $\mathbb{Q}$ 
Determine the Galois group of the splitting field of $(x^3-1)(x^2-5)$ over $\mathbb{Q}$.

I've been struggling with some of these Galois group questions.

Let $f=(x^3-1)(x^2-5)\in\mathbb{Q}[x]$. Note that
$$\begin{align}f&=(x-1)(x^2+x+1)(x^2-5)\\ &=(x-1)(x-\zeta_3)(x-\zeta_3^2)(x-\sqrt{5})(x+\sqrt{5})\end{align}$$
where $\zeta_3=\frac{-1+\sqrt{3}}{2}$ is a primitive cube root of unity. Thus, the splitting field of $f$ over $\mathbb{Q}$ is $$K=\mathbb{Q}(\zeta_3,\sqrt{5})=\{a+b\zeta_3+c\sqrt{5}+d\zeta_3\sqrt{5}\mid a,b,c,d\in\mathbb{Q}\}.$$
Suppose that $\phi:K\to K$ is an automorphism of $K$ fixing $\mathbb{Q}$. Then
$$\begin{align}\phi(a+b\zeta_3+c\sqrt{5}+d\zeta_3\sqrt{5})&=\phi(a)+\phi(b)\phi(\zeta_3)+\phi(c)\phi(\sqrt{5})+\phi(d)\phi(\zeta_3)\phi(\sqrt{5}) \\ &=a+b\phi(\zeta_3)+c\phi(\sqrt{5})+d\phi(\zeta_3)\phi(\sqrt{5})\end{align}$$
so that $\phi$ is determined entirely by the values of $\phi(\zeta_3)$ and $\phi(\sqrt{5})$; that is, if $\psi$ is an automorphism of $K$ such that $\psi(\zeta_3)=\phi(\zeta_3)$ and $\psi(\sqrt{5})=\phi(\sqrt{5})$, then in fact $\psi=\phi$.
Because $\zeta_3^2+\zeta_3+1=0$ and $(\sqrt{5})^2-5=0$, we must have that 
$$0=\phi(0)=\phi(\zeta_3^2+\zeta_3+1)=\phi(\zeta_3)^2+\phi(\zeta_3)+1$$
and
$$0=\phi(0)=\phi((\sqrt{5})^2-5)=\phi(\sqrt{5})^2-5$$
so that $\phi(\zeta_3)$ must be one of the two roots of $x^2+x+1$, namely $\zeta_3$ and $\zeta_3^2$, and $\phi(\sqrt{5})$ must be one of the two roots of $x^2-5$, namely $\sqrt{5}$ and $-\sqrt{5}$. 
Thus, there are at most four elements of $\operatorname{Gal}(K/\mathbb{Q})$ (there are two choices for where $\zeta_3$ goes, and two choices for where $\sqrt{5}$ goes):
$$\begin{align}
\operatorname{id}_K(\zeta_3)&=\zeta_3 &  \operatorname{id}_K(\sqrt{5})&=\sqrt{5}\\
\phi(\zeta_3) &= \zeta_3^2 & \phi(\sqrt{5})&=\sqrt{5}\\
\rho(\zeta_3) &= \zeta_3 & \rho(\sqrt{5})&=-\sqrt{5}\\
\sigma(\zeta_3) &= \zeta_3^2 & \sigma(\sqrt{5})&=-\sqrt{5}\\
\end{align}$$
 Check that all four do in fact define valid automorphisms, and see what the group structure on $\operatorname{Gal}(K/\mathbb{Q})$ is by composing them and seeing what comes up. For example, we have that $\phi\circ \rho=\sigma$, because
$$(\phi\circ\rho)(\zeta_3)=\phi(\rho(\zeta_3))=\phi(\zeta_3)=\zeta_3^2=\sigma(\zeta_3)$$
$$(\phi\circ\rho)(\sqrt{5})=\phi(\rho(\sqrt{5}))=\phi(-\sqrt{5})=-\phi(\sqrt{5})=-\sqrt{5}=\sigma(\sqrt{5})$$
Are there any elements of order 4?
