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

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?

• 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
Commented May 28, 2023 at 13:52
• 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 mathoverflow.net/questions/282667, mathoverflow.net/questions/283336).
– YCor
Commented May 28, 2023 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 $\alpha (G)\neq \alpha (H)$). Commented May 28, 2023 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. Commented May 28, 2023 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? Commented May 28, 2023 at 14:48