Automorphisms of a smooth quadric surface $Q\subset\mathbb{P}^{3}$ Let $Q\cong\mathbb{P^{1}_{1}}\times\mathbb{P^{1}_{2}}\subset\mathbb{P}^{3}$ be a smooth quadric surface. We have the following two actions on $Q$:
$$S_2\times Q\rightarrow Q,\; (\sigma,(x,y))\mapsto\sigma(x,y)$$
where $S_2$ is the symmetric group of order $2$, and
$$Aut(\mathbb{P^{1}_{1}})\times Aut(\mathbb{P^{1}_{2}})\times Q\rightarrow Q,\;((f,g),(x,y))\mapsto (f(x),g(y)).$$
These two actions do not commute. Furthermore $h^{0}(Q,T_{Q}) = 6$, that is $dim(Aut(Q)) = 6$.
Is it true that $Aut(Q)$ is the semi-direct product
$$Aut(Q)\cong(Aut(\mathbb{P^{1}_{1}})\times Aut(\mathbb{P^{1}_{2}}))\rtimes S_2 ?$$
 A: Yes, it is true. This is easily seen by looking for example at the Picard group, generated by $C$ and $D$, the two fibres of the projections. Since $C^2=D^2=0$ and$C\cdot D=1$, the only curves of self intersection $0$ are multiple of $C$ or $D$, and the irreducible ones are equivalent to $C$ or $D$. Composing an automorphism by the exchange of coordinates you fix then $C$ and $D$ and thus act on each factor via an automorphism. This gives the result.
A: Here is another way to see this. The automorphism group of any quadric hypersuface $$Q(x) = 0 \subset \mathbb{P}^n,$$ is exactly the projective orthogonal group $\textrm{PO}(Q)$ of $Q$. 
The key point now is that we have the coincidence that $\textrm{SL}(2) \times \textrm{SL}(2)$ is a double cover of $\textrm{SO}(4)$. On taking suitable projective quotients and carefully keeping track of the $2$-torsion, one arrives at the isomorphism
$$\textrm{PO}(4) \cong \textrm{PGL}(2) \times \textrm{PGL}(2) \rtimes S_2,$$
and the result is proved on noticing that $\textrm{Aut}(\mathbb{P}^1) \cong \textrm{PGL}(2).$
A: You can see the question in this other way. Take an automorphism $\phi:Q\rightarrow Q$ and compose it with one of the two projections $\pi_i\circ \phi$. Then 
$$\pi_i\circ\phi:Q\rightarrow\mathbb{P}^1$$
is a fibration with connected fibers. Therefore $\pi_i\circ\phi$ necessarily factorizes through $\pi_{j_{i}}$ with $j_i\in\{1,2\}$. Associating to $\phi$ the permutation $\{i\mapsto j_i\}$ we get a surjactive morphism of groups
$$f:Aut(Q)\rightarrow S_2.$$ 
On the other hand if $\pi_i\circ\phi$ factorized through $\pi_i$ this means than $\phi$ comes from an automorphism in $Aut(\mathbb{P}^1)\times Aut(\mathbb{P}^1)$. Therefore we have an exact sequence
$$0\mapsto Aut(\mathbb{P}^1)\times Aut(\mathbb{P}^1)\rightarrow Aut(Q)\rightarrow S_2\mapsto 0.$$ 
Now, it is enough to obeserve that $f$ admits a sections to conclude that $Aut(Q)\cong (Aut(\mathbb{P}^1)\times Aut(\mathbb{P}^1))\rtimes S_2$.  
