Listing all the subfield and corresponding subgroup Here is an exercise in the book of Dummit Foote : 
Find the Galois group of the splitting field of $(x^2-2)(x^2-3)(x^2-5)$ over $\mathbb{Q}$. Then list all the subgroups and the corresponding subfield
Here is my argument :

It is not hard to see that $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})=K$ is the splitting field for the above polynomial. Since the degree of the extension $K/\mathbb{Q}$ is 8, so the order of the Galois group is 8. 
  The automorphism in $Gal(K/\mathbb{Q})$ are :
\begin{cases}
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}
  \begin{cases}
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}
  \begin{cases}
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\sqrt{2}\mapsto -\sqrt{2}\\
\end{cases}
  \begin{cases}
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}
  \begin{cases}
\sqrt{5}\mapsto \sqrt{5}\\
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\end{cases}
  \begin{cases}
\sqrt{5}\mapsto \sqrt{5}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\sqrt{2}\mapsto -\sqrt{2}\\
\end{cases}
  \begin{cases}
\sqrt{5}\mapsto -\sqrt{5}\\
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\end{cases}
  \begin{cases}
\sqrt{5}\mapsto \sqrt{5}\\
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\end{cases}

From there, we see that except the identity map(the last map) are of order 2. But I do not know any group of order 8 having that property. 
My first question is : what is the Galois group in this case ?
Here is the other argument :

Any element of $K$ can be represented in the form :
  $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}+e\sqrt{5}+m\sqrt{10}+n\sqrt{15}+p\sqrt{30}$
  Under the map : \begin{cases}
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}
  this element turns to : $a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6}-e\sqrt{5}+m\sqrt{10}+n\sqrt{15}-p\sqrt{30}$. 
This element is fixed under that map iff $b=c=e=p=0$ or it must has the form : $a+b\sqrt{6}+c\sqrt{10}+d\sqrt{15}$. 

But from that I can not deduce the corresponding subfield with the subgroup generated by the above map. My second question is : What is that corresponding subfield ?
 A: For the first question, consider $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Note this relationship for "independent" automorphisms.
Alright, since I don't know how to do the larger tables in latex here (usually I use xymatrix!) lets (diagrammatically) look at the example at the beginning of 14.2 in Dummit and Foote. The Galois group of $\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$ gives the table
$$ \begin{array}{cccccccc}
& \{1\} & \\
  \nearrow & \uparrow & \nwarrow  \\
\{1,\tau\} & \{1,\sigma\tau\} & \{1,\sigma\} \\
  \nwarrow & \uparrow & \nearrow \\
 & \{1,\sigma,\tau,\sigma\tau\} \end{array}$$
In particular, when both $\sqrt{2}$ and $\sqrt{3}$ are permuted, the fixed field is $\mathbb{Q}(\sqrt{6})$.
Now, in our larger example, if $\sqrt{6}$ is fixed, the automorphism $\sigma_2\sigma_3$ is being applied (where $\sigma_n:\sqrt{n}\mapsto -\sqrt{n}$). Similarly, when $\sqrt{10}$ and $\sqrt{15}$ are fixed, the respective automrophisms are $\sigma_2\sigma_5$ and $\sigma_3\sigma_5$. Here's where a complete description of the Galois group is useful:
$$ \operatorname{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})/\mathbb{Q}) \simeq \{1,\sigma_2,\sigma_3,\sigma_5,\sigma_2\sigma_3,\sigma_2\sigma_5,\sigma_3\sigma_5,\sigma_2\sigma_3\sigma_5\} $$
The subgroup you're interested in is then $\{1,\sigma_2\sigma_3,\sigma_2\sigma_5,\sigma_3\sigma_5\}$.
