A group $G$ is called co-Hopfian if it is not isomorphic to a proper subgroup of itself; equivalently, every injective group homomorphism $\varphi : G \to G$ is surjective and hence an isomorphism. Examples of co-Hopfian groups include finite groups, $\mathbb{Q}$, $\mathbb{Q}/\mathbb{Z}$, and fundamental groups of closed hyperbolic manifolds. Examples of groups which are not co-Hopfian include free groups, free abelian groups, $\mathbb{R}$, and $\mathbb{Q}^*$ (take $\varphi(x) = x^3$).
In this MathOverflow answer, Ian Agol shows that if a group $C$ is freely indecomposable (i.e. not a non-trivial free product) and co-Hopfian, then $A\ast C \cong B\ast C$ implies $A\cong B$. As all the examples of co-Hopfian groups I have been able to find are freely indecomposable, I wonder if they all must be, in which case the first condition on $C$ is superfluous.
Is a co-Hopfian group necessarily freely indecomposable?
Note, answering this question is equivalent to answering the question in the title.
I've seen it claimed in a couple of papers that the answer is yes, but references weren't provided which makes me think that this should be fairly elementary. If $G \cong A\ast B$ is co-Hopfian, then there is a proper subgroup $H$ with $H\cong G$. By the Kurosh subgroup theorem, there is a set $X \subseteq G$, a family of subgroups $(A_i)_{i\in I}$ of $A$, a family of subgroups $(B_j)_{j\in J}$ of $B$, and families of elements $(g_i)_{i\in I}$, $(f_j)_{j\in J}$ of $G$ such that $H = F(X)\ast\left(\ast_{i\in I}g_iA_ig_i^{-1}\right)\ast\left(\ast_{j\in J}f_jB_jf_j^{-1}\right)$. I don't know where to go from this point. In particular, I don't know if the fact that $H$ is a proper subgroup yields any information about the cardinalities of $X$, $I$, and $J$.
Best Answer
$A*B$ is never co-hopfian, as it contains as a proper (isomorphic) subgroup $A*(ba)B{(ba)}^{-1}$, where $a, b$ are two arbitrary non trivial elements in $A$ and $B$ respectively.