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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.

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Hi Victor. I knew I must be missing something, but I can't believe it was something so obvious! I should have checked what you wrote more thoroughly. Anyway, I deleted my answer, since it contributes nothing and so it's better for the question to still have 0 answers. –  Tara Brough Mar 2 '12 at 10:54
    
Hi Tara. It is quite strange, but for groups this question is much harder to deal with than that for semigroups. Actually we proved the semigroup statement with Nik for the purposes of seeing how hopficity is preserved by finite Rees extensions, so we found this property on the go, by accident. This property is inetersting on its own. –  Victor Mar 2 '12 at 11:00
    
Perhaps you could give a reference for this fact in the semigroup case? –  HJRW Mar 2 '12 at 11:03
    
HW, actually so far it appeared only in my thesis. I could send you (and anybody else interested) a pdf with the thesis, just write to me to victor.maltcev@gmail.com for this. –  Victor Mar 2 '12 at 11:08

2 Answers 2

up vote 24 down vote accepted

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.

share|improve this answer
    
This is very neat! Thank you. –  Victor Mar 2 '12 at 13:56
    
And actually now I see that in the groups case, the argument is much easier :-) (see above comments). Well, "easier" only when you already know how to do it -- again, it is great to come up with such a beautiful proof! –  Victor Mar 2 '12 at 14:04
1  
Nice! Do you happen to know of an example if we remove the requirement that $H$ has finite index in $G$? I rather expect it's possible then, but I haven't thought of an example yet. –  Tara Brough Mar 2 '12 at 15:46
    
Victor - indeed, it is much easier when you know how! The proof is almost identical to the proof of Mal'cev's theorem. –  HJRW Mar 2 '12 at 16:26
3  
@tara, there are f.g. groups G isomorphic to GxG. So take the composition of such an iso with a projection to a factor. –  Benjamin Steinberg Mar 2 '12 at 17:26

Here is a variation on Henry's nice argument which uses Malcev's theorem. Let $N$ be the intersection of all finite index normal subgroups of $G$. Clearly $\phi(N)\subseteq N$ because a surjective endomorphism takes finite index normal subgroups to finite index normal subgroups. Thus $\phi$ induces a proper endomorphism of the finitely generated residually finite group $G/N$. By Malcev's theorem that f.g. residually finite groups are Hopfian, it follows $\phi$ induces an automorphism, which means $\ker \phi$ is contained in $N$. But then since each finite index subgroup contains a finite index normal subgroup, we have $\ker \phi\subseteq H$, which is a contradiction as Henry points out since in that case one would have $[G:H]=[\phi(G):\phi(H)]$.

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That's quite nice, too! –  Victor Mar 2 '12 at 14:34
3  
In fact, the proof of Mal'cev's Theorem really shows that the kernel of any self-epimorphism is contained in every finite-index subgroup. –  HJRW Mar 2 '12 at 20:56
    
Malcev's theorem is equivalent to the statement that the kernel of each self-epimorphism of a fg group is contained in the intersection of all finite index subgroup. –  Benjamin Steinberg Mar 2 '12 at 22:25
    
Malcev's theorem follows from the fact any surjective (weak) contraction of a compact metric space is an isometry. –  Benjamin Steinberg Mar 3 '12 at 1:16
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I prefer the second two proofs (which are essentially the same) since they clearly work for any algebraic structure. –  Benjamin Steinberg Mar 3 '12 at 11:45

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