2
$\begingroup$

A finitely generated group $G$ is called limit group if, for any finite subset $S\subset G$, there exists a homomorphism $f:G\to F$ (where $F$ is a free group of finite rank) so that the restriction of $f$ to $S$ is injective.

Let $H$ be a proper retract of $G$, i.e., a subgroup of $H\lneq G$ for which there exists a homomorphism $r:G\to H$ such that $r(h)=h$, for all $h\in H$.

By Louder's theorem, there is a function $D(n)$ such that if $F_n \to L_1 \to \cdots \to L_k$ is a sequence of proper epimorphism of limit groups, where $F_n$ is a free group with $n$-generator, then $k\leq D(n)$. Let $G$ be a limit group. In particular, take $L_1$ equal to $G$ which is a finitely presented (as a limit group) then there exists an integer $k_G$ such that for every sequence with proper retrctions $G=L_1 \to G_2 \to \cdots \to G_k$, we have $k\leq k_G$. If $H$ is a proper retract of a limit group $G$, then is $k_H \lneq k_G$?

$\endgroup$
10
  • 1
    $\begingroup$ Once more, your question is ill-posed. The way you "define" $k_G$ is not unique, and you can increase arbitrarily the value of $k_G$ for one given $G$. So, for an arbitrary such assignment $G\mapsto k_G$, the answer is clearly "no". However, if you choose $k_G$ as the largest length of a chain of proper limit quotients of $G$, then the answer isa trivial yes. $\endgroup$
    – YCor
    Jun 8, 2023 at 15:35
  • $\begingroup$ @YCor That's right. You are right. What do you mean by proper limit quotients of $G$? You mean a sequence $G=G_0 \twoheadrightarrow‎ G_1 \twoheadrightarrow‎ \cdots \twoheadrightarrow‎ G_k$ of proper epimorphisms of limits groups $G_i$? $\endgroup$
    – Mahtab
    Jun 10, 2023 at 3:44
  • $\begingroup$ Yes, this is what I meant. $\endgroup$
    – YCor
    Jun 10, 2023 at 6:46
  • $\begingroup$ Dear @YCor , Is there any relation between $k_G$ and rank $F$? I mean can we write $k_G$ in terms of other group properties of limit group $G$? $\endgroup$
    – Mahtab
    Jun 10, 2023 at 10:34
  • 1
    $\begingroup$ @Mahtab: it's not possible to extract an exact formula for $k_G$ from Louder's proof. With some work one might be able to write down an explicit upper bound from Louder's proof, but it would be huge. $\endgroup$
    – HJRW
    Jun 10, 2023 at 11:02

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.