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



Finitely generated soluble groups with uncountably many maximal subgroups
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 bi-infinite 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.




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**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 nilpotent-by-polycyclic groups.


