Galois groups of rational function fields extensions Consider the subfields $$ K_{1}:=\Bbb{C}\big(4x(1-x),4y(1-y)\big) $$
$$ K_{2}:=\Bbb{C}\Big(\frac{4x(1-x)(1-2y)}{(1-2xy)^{2}},\frac{4y(1-y)(1-2x)}{(1-2xy)^{2}}\Big) $$ of $ K:=\Bbb{C}(x,y) $. 
I want to compute the Galois groups $ G_{1}:=Gal(K/K_{1}),G_{2}:=Gal(K/K_{2}) $ as well as the intersection $ K_{3}:=K_{1} \cap K_{2} $ and $ G_{3}=Gal(K/K_{3}) $.
First, I claim that $G_{1} \simeq C_{2} \times C_{2} $. To this end, define $$ \sigma_{1}: x \mapsto  1-x, y \mapsto y $$ and $$ \sigma_{2}: x \mapsto x, y \mapsto 1-y $$ Notice that both $ \sigma_{1} $ and $ \sigma_{2} $ are their own inverses and map the generators $ 4x(1-x) $ and $ 4y(1-y) $ of $ K_{1} $ to themselves so the $ \sigma_{i}'s $ induce $ K_{1} $-automorphisms of $ K $. 
Let $$ p_{1}(T):=(T-x)(T-\sigma_{1}(x))=T^{2}-(x+\sigma_{1}(x))T+x\sigma_{1}(x)=T^{2}-T+x(1-x) $$
and $$ p_{2}(T):=(T-y)(T-\sigma_{2}(y))=T^{2}-(y+\sigma_{2}(y))T+y\sigma_{2}(y)=T^{2}-T+y(1-y) $$
Observe that $ p_{1}(T),p_{2}(T) \in K_{1}[T] $ and that $ p_{1}(x)=p_{2}(y)=0 $ so $ [K_{1}(x):K_{1}],[K_{1}(y):K_{1}] \leq 2 $.
On the other hand, we can't have that $ [K_{1}(x):K]=1 $ or that $ [K_{1}(y):K]=1 $ as that would mean that $ x $ or $ y $ are in $ K $ which is false because of degree considerations. So $$ [K_{1}(x):K_{1}]=[K_{1}(y):K_{1}]=2 $$ Furthermore, we have that $$ [K_{1}(x,y):K_{1}]=[K_{1}(x,y):K_{1}(x)][K_{1}(x):K_{1}]=2[K_{1}(x,y):K_{1}(x)]=4 $$ as $ y \notin K_{1}(x) $ because of the same degree considerations. So we obtain that $ K $ is a degree $ 4 $ extension of $ K_{1} $ whose Galois group is generated by $ \sigma_{1} $ and $ \sigma_{2} $, each being non-trivial $ K_{1} $-automorphisms of $ K $ of order $ 2 $. Thus $ G_{1} \simeq C_{2} \times C_{2} $.
I don't know how to go about computing $ G_{2},G_{3} $ and finding $ K_{3} $. Any help would be much appreciated. Thank you!

Looking at this over finite fields $\Bbb F_q$ with large $q$, one finds experimentally that the map $\phi : (x,y) \mapsto (u(x,y),v(x,y))$ (where $u$ and $v$ are the two generators of $K_2$) is basically $8$-to-$1$. For a choice of a pair $(u,v)$, either there is no solution to $(u,v) = \phi(x,y)$ (approximately $7/8$ of the time, in which case the solutions are all in $\Bbb F_{q^2}$), either there are $8$ of them (approximately $1/8$ of the time).
This is very strong evidence (and would be a proof if I knew how to get some nice effective bounds from the Weil conjectures)  that $K$ is Galois over $K_2$, that the group is isomorphic to $C_2^3$, and that it can be given by a bunch of explicit transformation formulas.
Those are (after quite a bit of work), 
$id : (x,y) \mapsto (x,y) \\
 f : (x,y) \mapsto  (1-x,-\frac y{1-2y}) \\
 g : (x,y) \mapsto  (- \frac x{1-2x},1-y) \\
fg : (x,y) \mapsto (\frac{1-x}{1-2x}, \frac {1-y}{1-2y}) \\
h : (x,y) \mapsto (\frac 1{2y}, \frac 1{2x}) \\
fh : (x,y) \mapsto (-\frac{1-2y}{2y}, \frac 1{2(1-x)}) \\
gh : (x,y) \mapsto (\frac 1{2(1-y)}, -\frac{1-2x}{2x}) \\
fgh : (x,y) \mapsto (\frac{1-2y}{2(1-y)},\frac{1-2x}{2(1-x)})
$

Furthermore, it looks like the group $G_3$ generated by the two Galois groups is finite (noncommutative, of order 32, looking at the size of most orbits in finite fields, maybe it's a skewed product), so if this is the case, $K$ should also be Galois over $K_3$.
