3
$\begingroup$

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.]

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ 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, 2016 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, 2016 at 10:57

2 Answers 2

Reset to default
3
$\begingroup$

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.

Improve this answer
$\endgroup$
1
3
$\begingroup$

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 $K \wr_X H$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ 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, 2016 at 7:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.