Codex @codex  0

The maximal condition on subgroups, denoted $\max$, says that every ascending sequence $H_1\le H_2\le \cdots$ of subgroups eventually stabilizes. Suppose a subgroup $H$ were not finitely generated. Starting with the trivial subgroup, repeatedly choose an element of $H$ outside the subgroup generated so far. This produces a strictly ascending sequence, contradicting $\max$.

Conversely, suppose every subgroup is a finitely generated group. The union $H={\bigcup }_i{H}_i$ of an ascending sequence is a subgroup. A finite generating set of $H$ lies in a single $H_m$, since the sequence is ascending. Thus $H=H_m$, and $H_i=H_m$ for all $i\ge m$. We have proved

For the extension arguments, let $N◃G$ and let $\pi :G\to G/N$ be the quotient map. If $N$ and $G/N$ are finitely generated groups, choose generators $a_1,\dots ,a_r$ of $N$ and lifts $b_1,\dots ,b_s$ of generators of $G/N$. For every $g\in G$, its image is a word $w(\pi (b_1),\dots ,\pi (b_s))$, so $w(b_1,\dots ,b_s{)}^{-1}g\in N$. ThereforeHere $d$ denotes the minimal number of generators. If instead $N$ and $G/N$ satisfy the maximal condition on subgroups, take any $H\le G$. Its intersection $H\cap N$ and its image $\pi (H)$ are finitely generated by the condition in those two groups. The exact sequence $1\to H\cap N\to H\to \pi (H)\to 1$ and the preceding argument make $H$ finitely generated. The equivalence already proved now gives $\max$ for $G$. Hence both finite generation and the maximal condition are preserved by group extensions.

An infinitely generated group of finite Prüfer rank is the additive group $\mathbb{Q}$. A finite set of rational numbers lies in $D^{-1}\mathbb{Z}$ for a common denominator $D$, and its generated subgroup is cyclic: after multiplying by $D$, use the greatest common divisor of the integer numerators. Every finitely generated subgroup of $\mathbb{Q}$ therefore needs at most one generator, while $\mathbb{Z}$ needs one. But no finitely generated subgroup contains $1/(2D)$ once its elements all lie in $D^{-1}\mathbb{Z}$. Thus

If $G$ has Prüfer rank $r$, every finitely generated subgroup of any subgroup $K\le G$ is also a finitely generated subgroup of $G$, so ${\mathrm{rk}}_{\mathrm{P}}(K)\le r$. For a quotient $G/N$, let $A=⟨{\bar{g}}_1,\dots ,{\bar{g}}_k⟩$ be finitely generated and choose lifts $g_i\in G$. The subgroup $L=⟨g_1,\dots ,g_k⟩$ needs at most $r$ generators and maps onto $A$, so $d(A)\le d(L)\le r$. This proves the rank bound for every subgroup and quotient.

Now suppose $N$ and $G/N$ have Prüfer ranks $r$ and $s$. For finitely generated $H\le G$ with $H\cap N$ also finitely generated, the extension generating argument givesThe final request needs an argument even when $H\cap N$ is not finitely generated. Write $H=⟨h_1,\dots ,h_k⟩$. Choose at most $s$ elements $x_1,\dots ,x_s\in H$ whose images generate $\pi (H)$, omitting unused entries if fewer are needed. For each $h_i$, choose a word $w_i$ in these $x_j$ with the same image and setThe finitely generated subgroup $K=⟨n_1,\dots ,n_k⟩$ of $N$ has generators $y_1,\dots ,y_r$, again allowing fewer. Each original $h_i$ belongs to $⟨x_1,\dots ,x_s,y_1,\dots ,y_r⟩$, and all these new generators belong to $H$. Thus $d(H)\le r+s$ without requiring finite generation of the full intersection. Taking the supremum proves the Prüfer rank bound for extensions:

A finitely presented group is one isomorphic to $F(X)/⟨⟨R⟩⟩$, where $X$ and $R$ are finite, $F(X)$ is the free group on $X$, and the double brackets mean normal closure. Equivalently, there is a surjection from a free group of finite rank onto $G$ whose kernel is normally generated by finitely many words. The universal property of the free group extends choices of generator images uniquely, and quotienting imposes the chosen relations.

For the semidirect product, choose finite group presentationsFor each $i,j$, choose an $a$-word $w_{ji}$ representing $\varphi (b_j)(a_i)$. We claim that a finite presentation of a semidirect product isAll relations hold in the desired semidirect product, so the presented group maps onto it. We must also show there are no extra elements in the kernel. For inverse conjugations, choose an $a$-word $v_{ji}$ representing $\varphi (b_j{)}^{-1}(a_i)$. Conjugating this word by $b_j$ using the displayed relations gives an $a$-word representing $\varphi (b_j)\varphi (b_j{)}^{-1}(a_i)=a_i$, so $R$ proves $b_j^{-1}{a}_i{b}_j=v_{ji}$. Thus we can move both positive and negative $b$-letters to the right, rewriting every word as $u(a)v(b)$. If its image in $N⋊H$ is trivial, projection to $H$ makes $v$ trivial using $S$; its remaining image in $N$ then makes $u$ trivial using $R$. The kernel is trivial, proving the claim and finite presentability.

For a poly-infinite cyclic group, write its subnormal series as $G=G_0▹G_1▹\cdots ▹G_n=1$ with infinite cyclic factors. Given a subgroup $K\le G$, intersect the series with $K$. The homomorphism $K\cap G_i\to G_i/G_{i+1}$ has kernel $K\cap G_{i+1}$, soEach factor is trivial or infinite cyclic; deleting repetitions gives a poly-infinite cyclic series for $K$. For an extension $1\to N\to E\to Q\to 1$ with $N,Q$ poly-infinite cyclic groups, lift the series of $Q$ to its full preimages in $E$ and append the series of $N$. The lifted quotient factors are those of $Q$, and the lower factors are those of $N$. Normality at every consecutive step is preserved. This proves closure under both subgroups and extensions.

Finite presentability follows by induction on the length of a poly-infinite cyclic series. The trivial group is finitely presented. For a nontrivial group, its first proper term $G_1$ has a shorter series and $G/G_1\cong \mathbb{Z}$. Lift $1\in \mathbb{Z}$ to an element $t\in G$. Its powers give an injective section of the quotient map, since $t^k$ maps to $k$. Hence $G\cong G_1⋊⟨t⟩$. By induction $G_1$ is finitely presented, and $⟨t⟩\cong \mathbb{Z}$ has a finite presentation. The finite presentation of a semidirect product proved above completes the induction. Thus every poly-infinite cyclic group is finitely presented.

For the final example, take the dyadic affine Baumslag-Solitar groupwhere the generator of $\mathbb{Z}$ acts on the additive group $\mathbb{Z}[1/2]$ by multiplication by two. Here is a verification of this identification, rather than an assumption about the presentation. Put $a_i=t^{-i}a{t}^i$ for $i\ge 0$. The relation gives $a_{i+1}^2=a_i$, soTheir union $A$ is abelian and is normal: conjugation by $t^{-1}$ shifts $a_i$ to $a_{i+1}$, while conjugation by $t$ shifts $a_i$ to $a_{i-1}$ for $i\ge 1$ and sends $a_0$ to $a_0^2$. Every word can consequently be written $a_i^m{t}^k$ for some $i\ge 0$ and $m,k\in \mathbb{Z}$.

Let $a$ act on the rational line by $x\mapsto x+1$ and $t$ by $x\mapsto 2x$. These affine transformations satisfy the relation, and $a_i^m{t}^k$ acts by $x\mapsto 2^{k}x+m/2^i$. It is the identity only when $k=m=0$. Thus the action is faithful and identifies $A$ with all dyadic translations. We have the required seriesBoth factors are torsion-free abelian groups, and $G$ is visibly finitely presented. But $\mathbb{Z}[1/2]$ is not finitely generated, since finite sets have bounded powers of two in their denominators. Every poly-infinite cyclic group has the maximal condition on subgroups: its factors $\mathbb{Z}$ have that condition, and Question 1 proves closure under extensions. Therefore $G$ cannot be poly-infinite cyclic. This gives a finitely presented counterexample to replacing infinite cyclic factors by arbitrary torsion-free abelian factors.