Projective special linear groups are co-hopfian? Is it known if $PSL(2,\, F)$, with $F$ a field of prime $p$ characteristic (maybe with all proper subfields of finite order), is co-hopfian?
I've searched everywhere but found nothing.
Definition
A group $G$ is called co-hopfian if $G$ has no proper subgroups isomorphic to $G$.
 A: It depends on the field $F$. For example, if $F$ is finite then $\operatorname{PSL}(2,F)$ is certainly co-hopfian, since it's finite. But if $F$ contains a proper subfield $F'$ isomorphic to $F$ then $\operatorname{PSL}(2,F')<\operatorname{PSL}(2,F)$, so it's not co-hopfian: for example, if $\mathbb{F}_p(x)$ is the field of rational functions, $\operatorname{PSL}\left(2,\mathbb{F}_p(x^2)\right)$ is a proper subgroup of $G=\operatorname{PSL}\left(2,\mathbb{F}_p(x)\right)$ that is isomorphic to $G$.
EDIT: In comments, the OP asked about fields such that all proper subfields are finite. I think that, more generally, if $F$ is algebraic then $\operatorname{PSL}(2,F)$ is co-hopfian. Here's at least the outline of a proof (I've not tried very hard to simplify it, but I suspect it can be simplified):
$F$ can be expressed as the union of a chain
$$F_1\leq F_2\leq \dots$$
of finite fields, where $F_i=\mathbb{F}_{p^{d_i}}$ (so $d_i$ divides $d_{i+1}$). A few of the facts I claim later on are false for some very small fields, but we can choose $F_1$ large enough that we avoid the counterexamples.
Suppose $\alpha:\operatorname{PSL}(2,F)\to\operatorname{PSL}(2,F)$ is an injective group homomorphism. We need to show that $\alpha$ is surjective.
Let $\alpha_i:\operatorname{PSL}(2,F_i)\to\operatorname{PGL}(2,\overline{\mathbb{F}}_p)$ be the composition of inclusion maps
$$\operatorname{PSL}(2,F_i)\to\operatorname{PSL}(2,F)\stackrel{\alpha}{\to}\operatorname{PSL}(2,F)\to\operatorname{PGL}(2,\overline{\mathbb{F}}_p).$$
If $p>2$, then by taking the pullback with the natural surjection $\operatorname{GL}(2,\overline{\mathbb{F}}_p)\to\operatorname{PGL}(2,\overline{\mathbb{F}}_p)$, we get an inclusion of a double cover of each $\operatorname{PSL}(2,F_i)$ into $\operatorname{GL}(2,\overline{\mathbb{F}}_p)$. Since $\operatorname{PSL}(2,F_i)$ has no non-trivial $2$-dimensional representations over $\overline{\mathbb{F}}_p$, this must be a non-trivial double cover. But the only such double cover is $\operatorname{SL}(2,F_i)$, so we get a compatible family of inclusions
$$\beta_i:\operatorname{SL}(2,F_i)\to \operatorname{GL}(2,\overline{\mathbb{F}}_p),$$
and so an inclusion
$$\beta:\operatorname{SL}(2,F)\to \operatorname{SL}(2,F),$$
that we need to show is surjective. (If $p=2$, then just take $\beta=\alpha$, $\beta_i=\alpha_i$.)
The only non-trivial $2$-dimensional representations of $\operatorname{SL}(2,F_i)$ are equivalent to Frobenius twists of the natural representation, so (letting $\varphi$ be the Frobenius map, raising all entries of matrices to the $p$th power) there are elements $g_i\in\operatorname{GL}(2,\overline{\mathbb{F}}_p)$ and non-negative integers $r_i$ such that
$$\beta_i=c_i\circ\varphi^{r_i},$$
where $c_i$ is conjugation by $g_i$.
Since the resriction of $\beta_{i+1}$ to $\operatorname{SL}(2,F_i)$ is equal to $\beta_i$, we must have $r_{i+1}\equiv r_i\pmod{d_i}$ and $g_{i+1}^{-1}g_i$ must be in the centralizer of $\operatorname{SL}(2,F_i)$ in $\operatorname{GL}(2,\overline{\mathbb{F}}_p)$. But this centralizer is the centre of $\operatorname{GL}(2,\overline{\mathbb{F}}_p)$, so without loss of generality we can take $g=g_1=g_2=\dots$.
But then the $\operatorname{im}(\beta)\leq\operatorname{SL}(2,F)$ is conjugate in $\operatorname{GL}(2,\overline{\mathbb{F}}_p)$, by $g$, to 
$$\operatorname{SL}(2,F)=\bigcup_i\operatorname{SL}(2,F_i).$$
Since $\operatorname{GL}(2,\overline{\mathbb{F}}_p)$ is the union of finite subgroups, $g$ has finite order, and so $\operatorname{im}(\beta)=\operatorname{SL}(2,F)$.
