Given $K=\mathbb{Q}(x,y)$ where $x$ and $y$ are independent transcendentals over $\mathbb{Q}$ and $G=\langle \sigma, \tau \rangle$ where $\sigma$ and $\tau$ are automorphisms of $K$ with $\sigma(x)=y$, $\sigma(y)=-x$, $\tau(x)=x$ and $\tau(y)=-y$ I need to show that the fixed field is $K^G = \mathbb{Q}(x^2+y^2,x^2 y^2)$.
What I've tried so far: it's very quick to check that $x^2+y^2$ and $x^2 y^2$ are fixed by the generators of $G$ (which is all that needs to be checked to show $G$ fixes them, and can be checked immediately). I've used an earlier result that states that $|G| \leq [K:K^G]$ and shown that $|G|=8$, so using the tower law we have $$[K:K^G] [K^G:\mathbb{Q}]= [K:\mathbb{Q}]$$ so we have $$8 [K^G:\mathbb{Q}] \leq [K:\mathbb{Q}].$$ It might also be worth noting that $[K:\mathbb{Q}]$ would be the order of the Galois group, but I can't see how you would find this directly from the information we have. How can we show the fixed field is as given?