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Recall Thompson's group $F$, acting by piecewise-affine homeomorphisms on the interval. I wish to make the following conjectures on all subgroups of $F$, which originate from this question on MO.

Recall also that, for a group $H$, a composition series is a family of subgroups $(H_\alpha)$ indexed by ordinals, with $H_0=H$ and $H_{\alpha+1}\triangleleft H_\alpha$ with $H_\alpha/H_{\alpha+1}$ simple and $H_\lambda=\bigcap_{\alpha<\lambda}H_\alpha$ for limit ordinals $\lambda$.

Conjecture 1. Every subgroup of $F$ admits a composition series.

Conjecture 2. In every such composition series, all composition factors are $C_p$ or $F'$.

I know that there are lots of complicated subgroups of $F$, see e.g. arXiv:1711.10998, but all these subgroups seem to satisfy the conjecture.

Question: is any of this reasonable for the $F$ experts out there?

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  • $\begingroup$ You mean $H_\lambda = \bigcap_{\alpha<\lambda} H_\alpha$ for limit ordinals, right? $\endgroup$
    – Lior Silberman
    Commented Aug 30, 2019 at 6:51
  • $\begingroup$ "Recall that a composition series is...": is this in any way standard? Anyway, assuming that the definition is fixed with $\bigcap$ in the limit case, Conjecture 1 is equivalent to "Every nontrivial subgroup of $F$ has a simple quotient". The latter is true for finitely generated subgroups, so is quite orthogonal to the previous question on finite subquotients. $\endgroup$
    – YCor
    Commented Aug 30, 2019 at 8:12
  • $\begingroup$ ... and conjecture 2 could be restated in a more robust way as "Every simple subquotient of $F$ is cyclic or isomorphic to $F'$." (which technically is a bit stronger since one could imagine a subgroup with another simple quotient, but not part of any composition series). $\endgroup$
    – YCor
    Commented Aug 30, 2019 at 9:02
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    $\begingroup$ As a particular instance of a subgroup, one has the subgroup $F_4$ of elements with only slopes powers of $4$. Its derived subgroup $F'_4$ might be an infinite simple subgroup of $F$ not isomorphic to $F'$. $\endgroup$
    – YCor
    Commented Aug 30, 2019 at 9:54
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    $\begingroup$ There is a conjecture (still open) by Guba and myself that every simple subgroup of $F $ is isomorphic to $F_n'$ for some $n$. Your conjecture 1 with $F $ replaced by $F_n $ is stronger than our conjecture. I do not know (seems unlikely) if all these derived subgroups are isomorphic. $\endgroup$
    – user6976
    Commented Aug 31, 2019 at 18:19

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