Is there a finitely generated soluble group with uncountably many maximal subgroups? Any classification of such groups?

1$\begingroup$ We can't expect any classification... maybe it would sound more reasonable to ask about the class $C_1$ of f.g. solvable groups having only countably many maximal subgroups, its subclass $C_2$ of f.g. solvable groups all of whose maximal subgroups are finitely generated, its subclass $C_3$ of f.g. solvable groups all of whose maximal subgroups have finite index. All these classes are stable under quotients. Probably the inclusions $C_1\supset C_2\supset C_3$ are strict although I have no example in mind right now to prove it, and $C_3$ includes all nilpotentbypolycyclic f.g. groups. $\endgroup$– YCorCommented Sep 5, 2016 at 22:38

1$\begingroup$ And in turn to classify such groups (those f.g. solvable with countably many maximal subgroups), it is no restriction to assume that the intersection of all maximal subgroups is trivial. This yields some reduction: e.g., a group with this property has all its nilpotent normal subgroups abelian. $\endgroup$– YCorCommented Sep 6, 2016 at 9:53

$\begingroup$ PS: regarding my first comment, an example of group in the class $C_2\smallsetminus C_3$ is the wreath product $\mathrm{Sym}_3\wr\mathbf{Z}$. It contains $L=\mathrm{Sym}_2\wr\mathbf{Z}$ as a maximal subgroup, of infinite index so is not in $C_3$. That it belongs to $C_2$ is not hard: using that $V=\mathrm{Alt}_3^{(\mathbf{Z})}$ is a simple $L$module, one gets that if $M$ is a maximal subgroup, either it contains $V$ or is isomorphic to $L$ and hence is finitely generated. In the first case, we get a maximal subgroup of the quotient $L$, which is nilpotentbypolycyclic, hence of finite index. $\endgroup$– YCorCommented Feb 4, 2018 at 1:50
1 Answer
Yes, there is a finitely generated soluble group with uncountably many maximal subgroups.
Fix an odd prime $p$. Denote by $F_p$ the field on $p$ elements, $C_2$ the cyclic group on $2$ elements. Let $A$ be the set of functions from $\mathbf{Z}$ to $\{1,1\}$. Let $M$ be the free $F_p$module on the generators $(e_i)_{i\in\mathbf{Z}}$. Consider the lamplighter group $L=C_2\wr\mathbf{Z}$, with generators $u,t$ with $u^2=1$ and $u$ commuting with its conjugates.
For every $f\in A$, endow $M$ with a structure of an $L$module, where $t$ acts by the shift $e_i\mapsto e_{i+1}$ and $u$ acts diagonally by $u\cdot e_i=f(i)e_i$. We write it as $M_f$ to emphasize the dependency on $f$.
Each $f\in A$ can be viewed as a biinfinite word in the letters $\pm 1$. Let $A'$ be the set of "universal" sequences, that is, those sequences for which every finite word occurs at least once. Then if $f\in A'$, the module $M_f$ is a simple module. Indeed, consider any nonzero element $w=\sum_{i=k}^l a_ie_i$ with $a_k,a_l$ nonzero. Then by the assumption, we can find some conjugate $u'$ of $u$ mapping $e_i$ to $e_i$ for all $i = k, \dots, l  1$ and $e_l$ to $e_l$. Then $u'w + w = 2 a_l e_l$ generates $F_p e_l$. In turn using $t$ we get all the other basis elements.
Let $N=F_p[L]$ be the group algebra. Then mapping $1$ to $e_0$ defines a surjective $F_p[L]$module homomorphism $p_f:N\to M_f$. Let $I_f$ be its kernel. We claim that
Elements of $L$ can be written as $g=(t^n,P)$ with $P$ a finite subset of $\mathbf{Z}$. Let $\delta(g)$ be the corresponding basis element of $N$. Then $$p_f(\delta(t^n,P))=p_f((t^n,P)\delta(1,\emptyset))=(t^n,P)e_0$$ $$=(t^n,\emptyset)(\prod_{i\in P}f(i))e_0=(\prod_{i\in P}f(i))e_n.$$ In particular, $p_f(\delta(1,\{i\}))=f(i)e_0$, and $p_f(\delta(1,\{i\})\delta(1,\emptyset))=(f(i)1)e_0$. In particular, $\delta(1,\{i\})\delta(1,\emptyset)\in I_f$ iff $f(i)=1$. This shows that $f\mapsto I_f$ is injective.
Since $A'$ is clearly of continuum cardinality, this shows that $(I_f)_{f\in A'}$ is a family of continuum cardinality of maximal submodules of $N$.
Therefore, in the group $N\rtimes L$ (which is solvable of length 3, generated by 3 elements), we have a family $(I_f\rtimes L)_{f\in A'}$ of maximal subgroups, of continuum cardinality.
Of course we can pull this back to any group having $N\rtimes L$ as a quotient, such as the free solvable group of any length $\ge 3$ on $\ge 3$ generators.
Note: 3 is the minimal derived length. Indeed, finitely generated metabelian groups have maximal subgroups of finite index, hence there are at most countably many. This holds more generally in finitely generated nilpotentbypolycyclic groups.

$\begingroup$ Thank you for the nice construction. It is very helpful. $\endgroup$ Commented Sep 6, 2016 at 9:02
