Let $F(n)$ be free group of rank $n\geq 2$. Denote by $F\_d(n)$ the d-th derived subgroup, that is $F\_d(n)=[F\_{d-1}(n),F\_{d-1}(n)]$ where $F\_0(n)=F(n)$. The free solvable group of rank $n$ and solvable class $d$ is $S\_{n,d}=F(n)/F\_d(n)$.


By Corollary 2.14. in "FINITELY PRESENTED WREATH PRODUCTS AND DOUBLE
COSET DECOMPOSITIONS",(<http://arxiv.org/pdf/math/0509090.pdf>), the free solvable group of class 3 and finite rank is not a finitely presented group. Does this group contain a finitely presented subgroup of solvable class 3?



finitely presented subgroup and free solvable group of class 3
No, it does not exist.


It's not too hard to check that a fp subgroup of a free metabelian group is abelian. A consequence (since any quotient of a fp solvable group is fp, by Bieri-Strebel) is that any fp subgroup of a free 3-solvable group has abelian image in the metabelianization, and in particular, has to be metabelian (I'd guess that it even has to be abelian!).


I have to justify the first claim (fp subgroup of a free metabelian group $G$ is abelian). Write, $M=[G,G]$, $Q=G/M$. Then $M$ is a torsion-free $\mathbf{Z}Q$-module [edit: justification below].
Let $H$ a fg subgroup of $G$, and $Q\_H$ be the image of $H$ in $Q$.


If If $M\cap H$ or $Q\_H$ is trivial then clearly $H$ is abelian.
So assume otherwise, and let us show that $H$ is not finitely presented.
Then $M\cap H$ is also, by restriction, a torsion-free $\mathbf{Z}Q\_H$-module. So it is not finitely-generated over the subsemigroup $\{q\in Q\_H:\ell(q)\ge 0\}$ for any choice of nonzero homomorphism $Q\_H\to\mathbf{R}$. This means that the Bieri-Strebel invariant of the $Q\_H$-module $M\cap H$ is the whole space of homomorphism (in conventional literature, it will be rather called empty Bieri-Strebel invariant: I always define the Bieri-Strbel invariant as the closed complement rather than the open subset, in accordance to the idea that we define the spectrum of a matrix rather than the complement of its spectrum). Since a f.g. metabelian group $G$ is f.p. iff its Bieri-Strebel invariant (a closed subset of $\mathrm{Hom}(G,\mathbf{R})$, closed under positive scalar multiplication) contains a line, we deduce that $H$ is not finitely presented.




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[Edit:] That $M$ is a torsion-free $\mathbf{Z}Q$-module is a consequence of Magnus' embedding theorem. Assume, as we can, that $G$ is finitely generated free metabelian over generators $x\_1,\dots,x\_n$. Let $t\_1,\dots,t\_n,e\_1,\dots,e\_n$ be complex numbers, such that $(t\_1,\dots,t\_n)$ is algebraically independent over $\mathbf{Q}$, and $(e\_1,\dots,e\_n)$ are linearly independent over $\mathbf{Q}(t\_1,\dots,t\_n)$ (of course this holds if $(t\_1,\dots,t\_n,e\_1,\dots,e\_n)$ is algebraically independent).
Magnus says that mapping $x\_i\mapsto\begin{pmatrix}t\_i & e\_i\\ 0 & 1\end{pmatrix}$ yields an injective homomorphism from $G$ to $\mathrm{GL}\_2(\mathbf{C})$. Actually, it maps into the subgroup $H$ consisting of elements of the form $\begin{pmatrix}t & v\\ 0 & 1\end{pmatrix}$, where $t$ has the form $\prod t\_i^{m\_i}$, $(m\_1,\dots,m\_n)\in\mathbf{Z}^n$, and $v$ belongs to the $\mathbf{Z}[t\_1^{\pm 1},\dots,t\_n^{\pm 1}]$-submodule generated by $(e\_1,\dots,e\_n)$. We see $H$ is just a standard wreath product $$\mathbf{Z}^n\wr\mathbf{Z}^n=\mathbf{Z}[t\_1^{\pm 1},\dots,t\_n^{\pm 1}]^n\rtimes\mathbf{Z}^n=(\mathbf{Z}Q)^n\rtimes Q.$$ In particular, $M$ is isomorphic to a $\mathbf{Z}Q$-submodule of a free $\mathbf{Z}Q$-module of rank $n$, so is torsion-free. 


