In my previous question: M Shahryari (mathoverflow.net/users/29488), Normal Subgroups of Free Products, Normal Subgroups of Free Products (version: 2012-11-28), I asked if a group $A$ has max-n property, is it true that the free product $A\ast \mathbb{Z}$ has also max-n? The answer was NO in that case. Now suppose $F$ is free group of finite rank and $A$ is a group having max-n (maximal condition on normal subgroups). A normal subgroup $N$ of $A\ast F$ is called an ideal if $N\cap A=1$. Is it true that $A\ast F$ has maximal property of ideals?

  • $\begingroup$ Looking at the comments on your previous question, I see that you are interested in groups which are not equationally Noetherian. You might be interested in this question of mine: mathoverflow.net/questions/75784/… . $\endgroup$ – HJRW Dec 6 '12 at 14:15
  • $\begingroup$ In particular, you might be interested in Denis Osin's answer to that question. $\endgroup$ – HJRW Dec 6 '12 at 14:16
  • $\begingroup$ @HW:Thank you for comment. You are right, I'm interested in versions of Hibert's basis theorem for groups. So, I'm trying to recognize groups $A$ having max-n such that A\ast F$ has maximal property on ideals. I need time to read your question and and D. Osin's answer. Thank you again. $\endgroup$ – M. Shahryari Dec 6 '12 at 20:15
  • $\begingroup$ I hope the answer for my question will be negative. $\endgroup$ – M. Shahryari Dec 6 '12 at 20:19
  • $\begingroup$ @M Shahryari, my answer worked for any group $A$, so no group $A$ has the property you want. However, are you just asking which groups are equationally Noetherian? I could write you a list of the current state of knowledge, if it would help. $\endgroup$ – HJRW Dec 8 '12 at 5:07

The answer to your question is 'no'. Consider any homomorphism $f:A*F\to G$ which is injective on $A$. Then $\ker f$ is an ideal in your sense (and this is necessary and sufficient). An infinite increasing chain of ideals is therefore equivalent to an infinite sequence of surjections

$A*F=G_0\to G_1\to G_2\to\ldots $

so that $A$ embeds into $G_n$ for all $n$. For a specific example, take $F=\langle b_1,b_2\rangle$ and construct $G_n$ from $G_{n-1}$ by adding a long small-cancellation relator in $b_1,b_2$.

Small-cancellation theory

At the OP's request, here are some references for the last sentence; they are all from Lyndon and Schupp's book Combinatorial group theory.

Take $r_1,r_2,\ldots$ to be an infinite sequence of elements of $F(b_1,b_2)$ such that, for every $n$, $R_n=\{r_1,\ldots,r_n\}$ satisfies condition $C'(1/6)$ (as defined on p. 240 of Lyndon and Schupp). It's a nice exercise to confirm that such sequences exist.

For each $n$, take $G_n=A*\langle b_1,b_2\mid R_n\rangle$. It's easy to check that there are infinitely many elements $g_i$ so that, for all distinct $i,j$, $g_ig_j^{-1}$ is $R$-reduced in the sense of p. 251 of Lyndon ad Schupp. Therefore, by Dehn's algorithm, $G_n$ is infinite for all $n$, as claimed.

Equational Noetherian groups

In response to a remark of the OP's in the comments, I want to point out that it does not follow that $G_0$, or any $G_n$, is not equationally Noetherian. Indeed, the $G_n$ constructed above are all word-hyperbolic and hence equationally Noetherian by a theorem of Sela. (Alternatively, for $C'(\lambda)$ for small enough $\lambda$, they are all linear by the work of Wise and friends, and hence are equationally Noetherian by Hilbert's Basis Theorem.)

To prove that $G_0$ (say) is not equationally Noetherian, you need an infinite sequence of proper epimorphisms

$L_0\to L_1\to L_2\to\cdots$

where each $L_n$ is residually $G_0$.

For more details, I suggest you look at the beginning of Bestvina and Feighn's paper Notes on Sela's work (here).

| cite | improve this answer | |
  • $\begingroup$ I just noticed that this is similar to a remark that Ashot Minasyan made on your previous question. $\endgroup$ – HJRW Dec 6 '12 at 14:00
  • $\begingroup$ @HW: I have few knowledge of small cancellation theory ( in the level of last chapter of Lyndon-Schupp). So, may I ask you to write some details? $\endgroup$ – M. Shahryari Dec 6 '12 at 20:24
  • $\begingroup$ I'll write some details when I'm next in the office and have access to Lyndon and Schupp---probably on Monday. $\endgroup$ – HJRW Dec 8 '12 at 20:40
  • $\begingroup$ @HW: Thank you so much, now I understand your argument. I will send my results to your email in next days. $\endgroup$ – M. Shahryari Dec 11 '12 at 20:51

You do not need to know small-cancellation thery. Take your favorite finitely generated but non-finitely presented group $$ \langle x_1,\dots,x_n\;|\;w_1=1,w_2=1,\dots\rangle. $$ Then $A*F(x_1,\dots,x_n)$ has an ascending chain of `ideals' $$ \langle\langle w_1\rangle\rangle \subset \langle\langle w_1,w_2\rangle\rangle \subset \dots. $$ Here, $\langle\langle\dots\rangle\rangle$ means the normal closure in $A*F(x_1,\dots,x_n)$.

The group $A$ plays no role here.

| cite | improve this answer | |
  • $\begingroup$ My favourite finitely-generated-but-not-finitely-presented group is provided by small-cancellation theory. :) $\endgroup$ – HJRW Dec 11 '12 at 15:51
  • 1
    $\begingroup$ And my is an amalgamated free product of two free groups. $\endgroup$ – Anton Klyachko Dec 11 '12 at 15:58
  • 1
    $\begingroup$ And mine is the wreath product of $\mathbb Z$ with $\mathbb Z$. ;-) $\endgroup$ – Ashot Minasyan Dec 12 '12 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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