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A subgroup $H$ of a group $G$ is malnormal if $gHg^{-1}\cap H=\{e\}$ for all $g\in G$ with $g\notin H$. It is almost malnormal if we merely require $gHg^{-1}\cap H$ to be finite.

I am wondering whether $\mathrm{SL}(n,\mathbb{Z})$, or $\mathrm{PSL}(n,\mathbb{Z})$, for $n\geq 2$, have (almost) malnormal non-abelian free subgroups. And why or why not?

More generally, I'd be happy to know a bit more about when countably infinite groups contain malnormal non-abelian free subgroups. Thanks!

Edit: As pointed out in Moishe's answer, this really should be a question for $\mathrm{PSL}(n,\mathbb{Z})$, not $\mathrm{SL}(n,\mathbb{Z})$, to avoid the possibility of (trivially) conjugating with $-I$.

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    $\begingroup$ The answer is yes for n= 2 (take any maximal virtually cyclic subgroup). I suspect that the answer is no for $n>2$, the reason being that most elements in $SL(n,Z)$ will have large (rank >2) centralizers. Hence the intersection of a conjugate of a free subgroup by a centralizer of an element in it will have infinite intersection with the subgoup. $\endgroup$
    – Ian Agol
    Commented May 10, 2023 at 17:07
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    $\begingroup$ @IanAgol the question requires "non-abelian free". If one asks about free subgroups, well $\{1\}$ is malnormal, period. If one asks about non-trivial free subgroups, in a sense the infinite cyclic case is a special story. Note anyway that $\begin{pmatrix}2&1&0&0\\1&1&0&0\\0&0&0&-1\\0&0&1&0\end{pmatrix}$ has a centralizer of rank 1 (although not cyclic, only virtually cyclic). $\endgroup$
    – YCor
    Commented May 10, 2023 at 18:35
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    $\begingroup$ @YCor In this context, any restriction on the number of generators is irrelevant: malnormality is transitive, and any non-abelian countable free group contains malnormal subgroups of arbitrary rank (including infinite). $\endgroup$
    – ADL
    Commented May 11, 2023 at 8:34
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    $\begingroup$ @IanAgol if you refer to the matrix in my previous comment: my 4-dimensional example is not great since it has a finite power (its square) with centralizer containing $\mathbf{Z}^2$. But the matrix $\begin{pmatrix}2& 1&0\\1&1&0\\0&0&1\end{pmatrix}$ has virtually infinite cyclic centralizer as well as all its powers (but not cyclic: there's the diagonal matrix $(-1,-1,1)$). Yet it might be that some finite index subgroup of $\mathrm{SL}_3(\mathbf{Z})$ has a malnormal infinite cyclic subgroup. $\endgroup$
    – YCor
    Commented May 11, 2023 at 16:25
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    $\begingroup$ The normalizer in $\mathrm{SL}_3(\mathbf{Z})$ of the cyclic subgroup generated by the matrix $\begin{pmatrix}9 & 7 & 0\\14 & 11 & 0\\ 0&0&1\end{pmatrix}$, or by any of its positive powers is infinite cyclic (not just virtually cyclic). So $\mathrm{SL}_3(\mathbf{Z})$ has a malnormal infinite cyclic subgroup. Same for $\begin{pmatrix}5 & 4 & 0\\9 & 7 & 0\\ 0&0&-1\end{pmatrix}$. $\endgroup$
    – YCor
    Commented May 12, 2023 at 7:50

1 Answer 1

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First of all, you have to work with $G=PSL(2, {\mathbb Z})$ and not $SL(2, {\mathbb Z})$, for otherwise the claim is clearly false. Then $G$ is a nonelementary hyperbolic group with trivial maximal finite normal subgroup. According to Lemma 8 in

Minasyan, Ashot; Olshanskii, Alexander Yu.; Sonkin, Dmitriy, Periodic quotients of hyperbolic and large groups., Groups Geom. Dyn. 3, No. 3, 423-452 (2009). ZBL1234.20051.

the group $G$ contains a malnormal subgroup which is also free of rank 2.

As for the group $PSL(n, {\mathbb Z})$, $n\ge 3$, the situation is unclear. I do know that it is impossible to find an almost malnormal Zariski dense subgroup because such subgroups always contain real-regular elements [2] and the latter have centralizers which are not virtually cyclic [1].

The references are

[1] Prasad, Gopal; Raghunathan, M. S., Cartan subgroups and lattices in semi-simple groups, Ann. Math. (2) 96, 296-317 (1972). ZBL0245.22013.

[2] Prasad, Gobal, $\mathbb{R}$-regular elements in Zariski-dense subgroups, Q. J. Math., Oxf. II. Ser. 45, No. 180, 541-545 (1994). ZBL0828.22010.

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    $\begingroup$ For $n\ge 4$ I'm pretty sure that each element has a power whose centralizer contains $\mathbf{Z}^2$. I'll double check. $\endgroup$
    – YCor
    Commented May 12, 2023 at 18:34

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