Malnormal subgroups in solvable groups I have been reading through a book of Robinson where it is mentioned (informally!) that solvable groups have "many" subnormal subgroups (subgroups $H<G$ with $H=H_0 \lhd H_1 \lhd \ldots \lhd H_n = G$). There seems to be quite a body of work on this, but it brought me to ponder on:
$\textbf{Question:}$ Given an infinite solvable group $G$, are there "few"  malnormal subgroups (i.e. $H \cap gHg^{-1} = e$ for any $g \in G \setminus H$)?
I'm open to any interpretation of "few" (from "none" to "if there is one then it must be [insert property]" to ...) or any example.
[Edit: I forgot to put "finitely generated" in the question.]
 A: In general, if $G=N \rtimes H$ is a semidirect product in which all nontrivial elements of $H$ act fixed-point-freely on $N$, then $H$ is malnormal in $G$.
For example, if $K$ is any group and $H$ is any torsion-free  group, then $H$ is malnormal in the restricted wreath product $K \wr H$.
If we choose $K$ and $H$ to be finitely generated and solvable then $G$ will be too. We could take $G = {\mathbb Z} \wr {\mathbb Z}$, for example.
A: For any integer $n \in \mathbf Z$ with $\vert n \vert \ge 2$, 
consider the action of $\mathbf Z$ on $\mathbf Z[1/n]$
for which the generator $1 \in \mathbf Z$ acts by multiplication by $n$.
The solvable Baumslag-Solitar group
$\text{BS}(1,n) = \mathbf Z [1/n] \rtimes_n \mathbf Z 
= \begin{pmatrix} n^{\mathbf Z} & \mathbf Z [1/n]  \\ 0 & 1 \end{pmatrix}$
provides a nice particular case of Example 1 of Derek Holt.
His Example 2 carries over more generally to permutational wreath products:
let $H$ be a group acting on a set $X$ in such a way that $h^{\mathbf Z}x$
is infinite for all $h \in H$, $h \ne 1$, and $x \in X$, and let $K$ be any group with more than one element. Then $H$ is malnormal in $K \wr_X H$. 
If $H$ and $K$ are solvable, so is $K \wr_X H$.
