$\mathbb{F}_2 \times \mathbb{F}_2$ is not subgroup separable I read that $\mathbb{F}_2 \times \mathbb{F}_2$ is not subgroup separable (ie. for every finitely generated subgroup $H$ and $g \notin H$, there exists a finite index subgroup $K$ such that $H \subset K$ and $g \notin K$). Do you know an argument justifying the assertion? 
We already know that $$\text{Subgroup separable} \Rightarrow \text{Residually finite} \Rightarrow \text{Hopfian},$$
is $\mathbb{F}_2 \times \mathbb{F}_2$ residually finite or Hopfian?
 A: Question 1: To prove that $F_2\times F_2$ is not subgroup seperable, the idea is as follows: take a surjective homomorphism $\phi:F_2\rightarrow G$ where $G$ is not residually finite (for example, take $G=\langle a, b; a^{-1}b^3ab^2\rangle$) and consider the diagonal subgroup $\Delta\leq G\times G$. Then $\Delta$ is not seperable in $G\times G$, and hence $Q=(\phi\times\phi)^{-1}(\Delta)$ is not seperable in $F_2\times F_2$ (this is not obvious, but it only uses surjections). Moreover, $Q$ is finitely generated (it is generated by the elements $(a, a)$, $(b, b)$, $(a^{-1}b^3ab^2, 1)$ and $(1, a^{-1}b^3ab^2)$).
A similar idea can be used to prove that the subgroup membership problem is insoluble for $F_2\times F_2$. The trick here is to take the group $G$ to be finitely presented with insoluble word problem. Also, I believe that the subgroup $Q=(\phi\times\phi)^{-1}$ is called a Mihailova subgroup, after K. A. Mihailova, The occurrence problem for direct products of groups Dokl. Acad. Nauk SSRR 119 (1958), 1103-1105.
Question 2: Recall that a group $G$ is residually finite if for every non-trivial $g\in G$ there exists a homomorphism $\phi_g: G\rightarrow Q_g$ where $Q_g$ is finite and where $\phi_g(g)\neq_{Q_g}1$. That is, for every element there exists a map onto a finite group where the element does not die.
Theorem: The group $F_2\times F_2$ is residually finite.
Proof: Suppose $1\neq(u, v)\in F_2\times F_2$, and without loss of generality suppose $u$ is non-trivial in $F_2$. Then $(u, v)$ does not die under the map $(u, v)\mapsto u$, which clearly extends to the following homomorphism
$$
\begin{align*}
\phi_{(u, v)}:
F_2\times F_2&\rightarrow F_2\\
(w_1, w_2)&\mapsto w_1
\end{align*}
$$
As $F_2$ is residually finite, there exists a map $\phi_u:F_2\rightarrow Q_u$ where $Q_u$ is some finite group and $\phi_u(u)$ is non-trivial. Then $\phi_{(u, v)}\circ\phi_u: F_2\times F_2\rightarrow Q_u$ is a homomorphism to a finite group and $\phi_{(u, v)}\circ\phi_u(u, v)$ is non-trivial. Hence, we can conclude that $F_2\times F_2$ is residually finite.
The only fact that we used here was that $F_2$ was residually finite. This means that the following, more general result holds (and the proof is essentially identical).
Theorem: If $H$ and $K$ are residually finite then so is $H\times K$.
