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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.

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I'm not sure I understand your last sentence. For instance, Rips and Sela proved that lots of torsion-free word-hyperbolic groups are co-Hopf. Of course, this doesn't immediately help you, as many groups don't embed into a word-hyperbolic group. But it makes a small-cancellation argument plausible. –  HJRW Nov 27 '12 at 10:39
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"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." Is this right? Or is it "does not contain elements of unbounded finite order"? I know little on this subject, but I do not understand why there's such a big difference. –  Jonathan Kiehlmann Nov 27 '12 at 16:15
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HW- Chalk that up to my ignorance. Perhaps I should have said "as this is usually accomplished in the proofs that I've seen by using..." Thanks for the pointer to a proof that works differently. Jonathan- More specifically, they showed that every countable group $G$ embeds into a complete hopfian quotient of $C_p * C_q$, with $p,q$ any natural numbers. Further, if $G$ contains no elements of order $p$ or if $G$ contains no element of order $q$, then the quotient can be made co-hopfian as well. So if there is some $n$ such that $G$ contains no elements of order $n$, ... –  Jay Williams Nov 27 '12 at 17:11
    
...then $G$ embeds into a complete hopfian and co-hopfian quotient of $C_n * C_3$, for example. –  Jay Williams Nov 27 '12 at 17:13

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up vote 7 down vote accepted

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 $H$).

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