Dear All,


here is the question:


Does there exist a finitely generated group $G$ with a proper subgroup $H$ of finite index, and an (onto) homomorphism $\phi:G\to G$ such that $\phi(H)=G$?


My guess is "no", for the following reason (and this is basically where the question came from): in Semigroup Theory there is a notion of Rees index -- for a subsemigroup $T$ in a semigroup $S$, the Rees index is just $|S\setminus T|$. The thing is that group index and Rees index share the same features: say for almost all classical finiteness conditions $\mathcal{P}$, which make sense both for groups and semigroups, the passage of $\mathcal{P}$ to sub- or supergroups of finite index holds if and only if this passage holds for sub- or supersemigroups of finite Rees index. There are also some other cases of analogy between the indices. Now, the question from the post is "no" for Rees index in the semigroup case, so I wonder if the same is true for the groups.


Also, I beleive the answer to the question may shed some light on self-similar groups.



Mapping from a finite index subgroup onto the whole group
Here is a proof that there is no such finitely generated group. It's similar to Mal'cev's proof that finitely generated residually finite groups are non-Hopfian.


First, note that $\ker\phi$ is not contained in $H$---otherwise, $|\phi(G):\phi(H)|=|G:H|$. Let $k\in\ker\phi\smallsetminus H$. Because $\phi$ is surjective, there are elements $k\_n$ for each $n\in\mathbb{N}$ such that $\phi^n(k\_n)=k$.


Let $\eta:G\to\mathrm{Sym}(G/H)$ be the natural action by left translation. Then the homomorphisms $\eta\circ\phi^n$ are all distinct. Indeed,


$\eta\circ\phi^n(k\_n)=\eta(k)\neq 1$


because $k\notin H$, whereas


$\eta\circ\phi^{m}(k\_n)=\eta(1)=1$


for $m>n$. But there can only be finitely many distinct homomorphisms from a finitely generated group to a finite group.


