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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: