A group $G$ is co-hopfian if every injection $f\colon G \rightarrow G$ is an automorphism, or equivalently if $G$ is not isomorphic to any of its proper subgroups. Miller and Schupp, using small cancellation theory, showed that every countable group that does not contain elements of every finite order can be embedded in a 2-generated co-hopfian group. I can't find a more general result.


Is it known if every countable group embeds into a finitely generated co-hopfian group? At least, does every finitely generated group embed into a finitely generated co-hopfian group? My instinct is to use small cancellation theory, but if a group contains elements of every finite order it doesn't seem to give enough control over the embeddings $f\colon G\rightarrow G$, as this is usually accomplished using torsion elements.



Does every group embed into a co-hopfian group?
Like with many problems solvable by small cancellation methods, the answer can be found by looking at A.Yu. Ol'shanskii's papers. 


Indeed, take any countable group $H$. Without loss of generality we can assume that $H$ is not virtually cyclic and has some non-trivial element of finite order. Now, let $K$ be a finitely generated torsion-free group that is not embeddable in $H$ (such $K$ exists as $H$ has only countably many f.g. subgroups, but there are uncountably many f.g. pairwise non-isomorphic torsion-free groups). 


In Theorem 2 of the paper [``Efficient embeddings of countable groups.'', Vestnik
Mosk. Univ., Ser. Matem. (1989), N 2, 28-34 (in Russian); English translation in Moscow Univ. Math. Bull. 44 (1989), no. 2, 39-49] Ol'shanskii proves a powerful embedding theorem which implies that there is a $2$-generated simple group $G$, containing both $H$ and $K$, in which every proper subgroup is either infinite cyclic or infinite dihedral or is conjugate inside $H$ or $K$ in $G$. 


Thus $H$ embeds in $G$, and it's not hard to see that $G$ is co-hopfian. Indeed, let $A < G$ be a proper subgroup such that $G \cong A$. Since $G$ is not virtually cyclic, $A$ must be contained in a conjugate of $H$ or $K$ in $G$. Since $K$ does not embed in $H$, $A$ must be conjugate inside of $K$, but $K$ is torsion-free and $H$ has torsion, giving a contradiction.


Of course this construction is flexible. For example, if $H$ is torsion-free one can take $G$ to be torsion-free, by using some other invariants instead of torsion (to ensure that $H$ is not embeddable in $K$ and $K$ is not embeddable in $H$).


