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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.]

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    $\begingroup$ If you just want an example, let $G=\langle A,t\rangle$ where $A$ is a normal abelian torsion group with all elements of odd order, $t$ has order $2$, and $t$ acts on $A$ by inverting every element. Then $\langle t \rangle$ is malnormal. $\endgroup$ – Derek Holt Jul 6 '16 at 10:43
  • $\begingroup$ thanks for the example! I forgot to add "finitely generated" in the question... My feeling was that malnormal subgroups should be finite or cyclic or "small" in some sense, but I was enable to make any pertinent progress... $\endgroup$ – ARG Jul 6 '16 at 10:57
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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.

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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 $K \wr_X H$.

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  • $\begingroup$ thanks! Is the following also an examlpe which is not a semi-direct product: take $G \simeq F_2 / F_2^{(2)}$ to be the free metabelian group (say generated by $x,y$) and look at $H = \langle x \rangle$? $\endgroup$ – ARG Jul 25 '16 at 7:50

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