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By a retract of a group $G$, we mean a subgroup $H$ of $G$ for which there is a homomorphism $r:G\to H$ such that $r\circ i=1_H$, where $i:H\to G$ in the inclusion. By a proper retract, I mean a retract $H$ such that $H\neq G$.

Let $\mathcal{X}$ be a class of groups closed under taking retracts. Is there a function $\alpha :\mathcal{X}\to \mathbb{R}$ satisfying:

(C1) If $K$ and $H$ are in $\mathcal{X}$, and $K$ is isomorphic to $H$, then $\alpha (K)=\alpha (H)$.

(C2) If $K$ is in $\mathcal{X}$, and $H$ is a proper retract of $K$, then $\alpha (H)\lneq \alpha (K)$.

We can easily check that:

(1) If $F$ is a free group of finite rank $r_F$ and $H$ be a proper retract of $F$, then we have $r_H\lneq r_F$.

(2) If $A$ is a finitely generated abelian group and $n_A$ denotes the number of direct summands in the canonical form of $A$ and $H$ be a proper retract of $A$, then we have $n_H\lneq n_A$.

Based on Lemma 1.3 of "Finiteness conditions on CW-complexes" by C.T.C. Wall, if $G$ is a finitely presented group, and $H$ a retract of $G$, then $H$ is finitely presented. Hence the class of finitely presented groups is closed under taking retracts.

Based on the above observations I have the following question:

Is there such a function for the class of finitely presented groups?

Mahtab
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  • Out of curiosity, why write $n\lneq m$ instead of $n<m$ for ordinary strict inequality of numbers? (the sign $\lneq$ then reads "$n$ is less or equal but not equal to $m$"...) – YCor May 28 '23 at 13:52
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    I would tend to guess that for a countable class $\mathcal{X}$, there is such a function iff no group in $\mathcal{X}$ is isomorphic to a proper retract of itself. Whether there's such a f.p. group seems to be open (see https://mathoverflow.net/questions/282667/, https://mathoverflow.net/questions/283336/). – YCor May 28 '23 at 13:57
  • @YCor What I really need is that the number $\alpha (H)$ is less or equal than $\alpha (G)$ when $H$ is a retract of $G$ but if $H$ is proper then $\alpha (H)$ is less than $\alpha (H)$ (I mean in this situation $\alpha (G)\neq \alpha (H)$). – Mahtab May 28 '23 at 14:19
  • @YCor Thanks for the comment and the links. That is right. Do you know such a countable class? I mean the the biggest well-known class in which no group is isomorphic to a proper retract of itself. – Mahtab May 28 '23 at 14:26
  • @YCor Trivially finite groups, finitely generated abelian groups and free groups of finite rank are examples of $\mathcal{X}$. Can $\mathcal{X}$ be the class polyclic-by-finite groups or poly-$\mathbb{Z}$-groups? – Mahtab May 28 '23 at 14:48
  • You need to explain what you mean by "example of $\mathcal{X}$". If you mean, of $\mathcal{X}$ that is stable under taking retracts and in which no group is isomorphic to a proper retract of itself, obviously the class of max-n groups work. Hence the class of virtually polycyclic groups works (and is countable). – YCor May 28 '23 at 16:44
  • @YCor Yes, I meant that. So could you please tell me how can I find the function $\alpha :\mathcal{X}\to \mathbb{R}$ with the properties mentioned in the question for the class of max-n groups? – Mahtab May 29 '23 at 03:18
  • @YCor For the class of polycyclic groups $G$, I tried to prove the Hirsch number of $G$, $h(G)$, plays the role of $\alpha (G)$ but I coundn't: Assume that $r:G\to H$ is a proper retraction. Assume, by the contrary, that $h(G)=h(H)$. We have $H\cong G/N$, where $N=ker(r)$. Then $h(H)=h(G/N)$. On the other hand, $h(G)=h(N)+h(G/N)$. This implies that $h(N)=0$. Now how can I use this to reach a contradiction? – Mahtab May 29 '23 at 03:23
  • No, you can have a polycyclic group of the form $G\ltimes F$ with $F$ finite. So the Hirsch number is not enough and you need to keep some information relative to size of the maximal finite normal subgroup, to define a function as you wish. – YCor May 29 '23 at 07:11
  • @YCor I got it. Thank you. Is there any proof for the fact that every polycyclic group is of of the form $G\ltimes F$ where $F$ is the maximal finite normal subgroup? Is that true for virtually polycyclic groups? – Mahtab Jun 01 '23 at 14:24
  • No, there is no proof, simply because it is false. A counterexample is the amalgam $C_4*_{C_2}C_4$. – YCor Jun 01 '23 at 14:29
  • Dear @YCor , about what you said "for a countable class $\mathcal{X}$, there is such a function iff no group in $\mathcal{X}$ is isomorphic to a proper retract of itself", I searched. These groups are called weakly Hopfian, i.e., groups which cannot be isomorphic to a proper retract of itself. It should be noted that there is not known at present any example of a finitely presented group that is not weakly Hopfian. – Mahtab Jun 08 '23 at 12:33

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