Is the system of subgroups whose quotient is finitely generated, an inverse system?

Question

In a similar fashion to the construction of the profinite completion, let $G$ be a group and let $$\mathfrak{M}=\{H\trianglelefteq G\;|\;G/H\text{ is finitely generated}\}$$ ordered by reverse inclusion. Is $\mathfrak{M}$ an inverse system? i.e, if $G/H_1$ and $G/H_2$ are finitely generated, is $G/(H_1\cap H_2)$ also finitely generated? (or the quotient by some subgroup contained in $H_1\cap H_2 $). Since a subgroup of a finitely generated group need not be finitely generated, it is not obvious that this should be true, but I can't think of a counterexample.

Answer 1

This may be overkill... (from “Structure and finiteness properties of subdirect products of groups” by Bridson and Miller, Example 3. Because the obvious thing to try is to send $G/(N_1\cap N_2)\hookrightarrow (G/N_1)\times (G/N_2)$, giving a subdirect product...)

Let $K=\langle a,b\mid R\rangle$ be a $2$-generated group that is not finitely presented; let $F_1=\langle x,y\rangle$ and $F_2=\langle z,w\rangle$ be two free groups. Let $\phi_1\colon F_1\to K$ be the quotient map induced by $x\mapsto a$ and $y\mapsto b$, and let $\phi_2\colon F_2\to K$ be induced by $z\mapsto a$ and $w\mapsto b$.

Let $G\leq F_1\times F_2$ be the subdirect product $$ G = \{(r,s)\in F_1\times F_2\mid \phi_1(r)=\phi_2(s)\}.$$ The groups $N_1=\ker(\phi_1)\times\{e\}$ and $N_2=\{e\}\times\ker(\phi_2)$ are normal in $G$, and $G/N_1\cong G/N_2\cong F$, so both quotients are finitely generated. And $N_1\cap N_2$ is trivial.

$G$ is generated by the elements $(x,z)$, $(y,w)$, and all elements of the form $(n,e)$ with $n\in N_1$; in fact, it suffices to take $n$ in the set $R$ of the description of $K$ (after replacing $a$ and $b$ with $x$ and $y$). No finite subset of these elements can generate $G$: we need both $(x,z)$ and $(y,w)$; if we had finitely many elements $(n_1,e),\ldots,(n_m,e)$ that together with $(x,z)$ and $(y,w)$ generate $G$, then $K$ would be presented by $\langle x,y\mid n_1,\ldots,n_m\rangle$, which contradicts the choice of $K$.

This means $G$ cannot be finitely generated, for you would be able to express each of the finitely many generators in terms of finitely many of these special generators, but no such finite set generates $G$.

So $G/N_1$ and $G/N_2$ are finitely generated, but $G/(N_1\cap N_2)= G/\{e\}$ is not.