Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In this question, I asked whether there existed groups $G$ with finitely presentable subgroups $H$ such that $gHg^{-1}$ is a proper subgroup of $H$ for some $g \in G$. Robin Chapman pointed out that the group of affine automorphisms of $\mathbb{Q}$ contains examples where $H \cong \mathbb{Z}$.

This leads me to the following more general question. A group $\Gamma$ is "coHopfian" if any injection $\Gamma \hookrightarrow \Gamma$ is an isomorphism. To put it another way, $\Gamma$ does not contain any proper subgroup isomorphic to itself. The canonical example of a non-coHopfian group is a free group $F_n$ on $n$ letters. Chapman's example exploits the fact that $F_1 \cong \mathbb{Z}$ contains proper subgroups $k \mathbb{Z}$ isomorphic to $\mathbb{Z}$.

Now let $\Gamma$ be a non-coHopfian group and let $\Gamma' \subset \Gamma$ be a proper subgroup with $\Gamma' \cong \Gamma$. Question : does there exist a group $\Gamma''$ such that $\Gamma \subset \Gamma''$ and an automorphism $\phi$ of $\Gamma''$ such that $\phi(\Gamma) = \Gamma'$? How about if we restrict ourselves to the cases where $\Gamma$ and $\Gamma''$ are finitely presentable? I expect that the answer is "no", and I'd be interested in conditions that would assure that it is "yes".

If such a $\Gamma''$ existed, then we could construct an example answering my linked-to question above by taking $G$ to be the semidirect product of $\Gamma''$ and $\mathbb{Z}$ with $\mathbb{Z}$ acting on $\Gamma''$ via $\phi$. This question thus can be viewed as asking whether Chapman's answer really used something special about $\mathbb{Z}$.

share|improve this question

2 Answers 2

up vote 10 down vote accepted

Let $\alpha: \Gamma\to\Gamma$ be an injection sending $\Gamma$ to $\Gamma'$. Then the $\Gamma''$ you're looking for is the infinite amalgamated product

$\cdots *_{\Gamma}\Gamma *_{\Gamma}*\Gamma*_{\Gamma}\cdots$

where, at each stage, $\Gamma$ maps to the left by the identity and to the right by $\alpha$. Now the 'shift' automorphism has the property that you want, and the semidirect product with $\mathbb{Z}$ that you suggest is just the ascending HNN extension of $\Gamma$ via $\alpha$.

share|improve this answer
6  
This was the original purpose of H,N&N defining HNN extensions. (They asked the question Given a group G and a pair of isomorphic subgroups H and H' of G, can we embed G into a group G' so that H and H' are conjugate in G'?'. Then they answered this (and a bunch of other embedding questions) by defining HNN extensions. [The paper is Embedding theorems for groups' J.LMS 24 (1949), 247-254.] –  Daniel Groves Apr 6 '10 at 0:56

Effectively you have a group $\Gamma$ and a monomorphism $\phi:\Gamma\to\Gamma$ which is not a surjection. Take the direct limit of the sequence $(\Gamma_n)$ where each $\Gamma_n=\Gamma$ and each map from $\Gamma_n$ to $\Gamma_{n+1}$ is $\phi$. I think this direct limit is the group $\Gamma''$ you want.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.