Non- simplicity of $\text{Bir}(\mathbb{P}_\mathbb{R}^2)$. It was proven by Serge Cantat and Stéphane Lamy that the Cremona group $\text{Bir}(\mathbb{P}_\mathbb{C}^2)$ is not simple. It is later proven separately that the real Cremona group $\text{Bir}(\mathbb{P}_\mathbb{R}^2)$ is also not simple, albeit using slightly different techniques. However in their paper: https://perso.univ-rennes1.fr/serge.cantat/Articles/nsgc-acta-c.pdf the authors said that their result 'directly implied' that  $\text{Bir}(\mathbb{P}_\mathbb{R}^2)$ is not simple as well. How can we infer the non-simplicity of  $\text{Bir}(\mathbb{P}_\mathbb{R}^2)$ just from the fact that  $\text{Bir}(\mathbb{P}_\mathbb{C}^2)$ is not simple?
I am thinking of choosing a proper normal subgroup, say $N$, of $\text{Bir}(\mathbb{P}_\mathbb{C}^2)$ and intersecting it with $\text{Bir}(\mathbb{P}_\mathbb{R}^2)$. But there is no guarantee that $\text{Bir}(\mathbb{P}_\mathbb{N}^2)\cap N$ is non-trivial. How do I justify that $\text{Bir}(\mathbb{P}_\mathbb{N}^2)\cap N$ is nontrivial?
 A: This is to convert my comment into a proper answer. 
In the main result of the linked paper, Theorem A, they prove that there exists an integer $k$ such that if $g\in G=Bir({\mathbb C}P^2)$ is a "general element" of degree $d\ge 2$ then for any $n\ge k$, the normal closure in $G$ of the cyclic subgroup $\langle g^n\rangle$ is a proper subgroup of $G$. Moreover, this subgroup does not contain any element $f\in G-\{1\}$ whose degree is $<nd$. 
Here an element of an (irreducible) algebraic variety $V$ (in this case, the variety is $V_d$ consisting of degree $d$ rational maps of ${\mathbb C}P^2$ ) is general if it belongs to the complement of a countable union of proper Zariski closed subsets of $V$. Hence, a more careful statement of their Theorem A is that for each $d\ge 2$ there exists a countable union $Z$ of (Zariski) closed proper subsets $Z_i\subset V_d$  such that for every element $g\in G- Z$ the normal closure of  $\langle g^n\rangle$ does not contain any  $f\in G-\{1\}$ whose degree is $<nd$. 
Now, since the set $V_d({\mathbb R})$ of real points of $V_d$  is Zariski dense in $V_d$, it follows that every intersection $Z_i\cap V_d({\mathbb R})$ has empty interior, hence, $Z\cap  V_d({\mathbb R})$  is meagre (in the classical topology!), hence, by Baire theorem, its complement in $V_d({\mathbb R})$ is nonempty. Now, take any $g$ in this complement. Since the normal closure of $\langle g^n\rangle$ in $G$ does not contain any real birational transformation $f\ne 1$ whose degree is $<nd$, it follows that the normal closure of $\langle g^n\rangle$ in the real Cremona group $Bir({\mathbb R}P^2)$ is a proper normal subgroup. Hence, $Bir({\mathbb R}P^2)$ is not simple. 
