$\DeclareMathOperator{\Sub}{\operatorname{Sub}}$ Let $G$ be a profinite group and consider the space $\Sub(G)$ of closed subgroups of $G$ equipped with the profinite topology. That is, we have $G = \underleftarrow\lim(G\_i)$ for finite groups $G\_i$, and we construct $\Sub(G)$ as $\lim(\Sub(G\_i))$. The space $\Sub(G)$ admits an action of $G$ by conjugation, therefore we may also consider the quotient space $\Sub(G)/G$.


In a pair of papers [1,2], Gartside—Smith consider various topological properties of the space $\Sub(G)$. I am interested in some of the corresponding statements for $\Sub(G)/G$.


Explicitly, assume that $G$ is second countable (which is equivalent to $\Sub(G)$ or $\Sub(G)/G$ being second countable). Is is true that if $\Sub(G)$ is uncountable then $\Sub(G)/G$ is also uncountable? Note that this is clearly true for abelian groups.


This question is aiming towards having a classification of those profinite $G$ such that $\Sub(G)/G$ is scattered, in analogy to the Gartside—Smith results for $\Sub(G)$.


[1] Counting the closed subgroups of profinite groups J. Group Theory (2010)


[2] Classifying the closed subgroups of profinite groups J. Group Theory (2010)



Spaces of closed subgroups of a profinite group up to conjugacy
Just to restate the question concisely:



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> Let $G$ be a second-countable profinite group. If $G$ has uncountably many subgroups, does it have uncountably many closed subgroups modulo conjugacy?
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The answer is no. A counterexample is the $p$-adic group $\mathrm{SL}\_2(\mathbf{Z}\_p)$ for $p$ prime.


It clearly has uncountably many closed subgroups (mapping a line of $\mathbf{Q}\_p^2$ to its stabilizer in $G$ is injective). So the main claim is



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> $\mathrm{SL}\_2(\mathbf{Z}\_p)$ has countably many conjugacy classes of closed subgroups.
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Since $\mathrm{SL}\_2(\mathbf{Z}\_p)$ has countable index in $H=\mathrm{SL}\_2(\mathbf{Q}\_p)$, it is enough to check that the latter has only countably many compact subgroups up to conjugacy.


Note that every closed subgroup of $H$ is $p$-adic analytic, so is locally determined by its Lie algebra. The action of $\mathbf{SL}\_2(\mathbf{Q}\_p)$ on its Lie algebra has countably many orbit on the Grassmanian, so it is enough to check that for a given Lie subalgebra $\mathfrak{k}$, there are only countably many possible subgroups $K$.


Write $d=\dim(\mathfrak{k})=\dim(K)$.


If $d=3$ $K$ is a compact open subgroup and there are only countably many.


If $d=0$, $K$ is finite and it is known that there are only finitely many conjugacy classes of finite subgroups (e.g., as every finite group has only finitely many representations on $\mathbf{Q}\_p^2$ modulo $\mathrm{SL}\_2(\mathbf{Q}\_p)$-conjugation).


If $d=2$, the Lie algebra is a line stabilizer, isomorphic to $\mathbf{Q}\_p^\*\ltimes\mathbf{Q}\_p$ (with action $t\cdot x=t^2x$); a compact open subgroup has to lie in $\mathbf{Z}\_p^\*\ltimes\mathbf{Q}\_p$. It intersects $\mathbf{Q}\_p$ in $p^n\mathbf{Z}\_p$ for some $n\in\mathbf{Z}$. The intersection being given, the subgroup is determined by some compact open subgroup of the quotient $\mathbf{Z}\_p^\*\ltimes \mathbf{Q}\_p/p^n\mathbf{Z}\_p$, necessarily contained in $\mathbf{Z}\_p^\*\ltimes p^m\mathbf{Z}\_p/p^n\mathbf{Z}\_p$ for some $m\le n$. This is a semidirect product $\mathbf{Z}\_p^\*\ltimes\mathbf{Z}/p^k\mathbf{Z}$ for some $k$. In turn, this has to contain some congruence subgroup in $\mathbf{Z}\_p$, and then there are only finitely many possibilities.


If $d=1$, and the Lie algebra is diagonalizable over an extension, then the normalizer of the Lie algebra is a one-dimensional group $M$, with a compact open normal subgroup isomorphic to $\mathbf{Z}\_p$. The intersection with $\mathbf{Z}\_p$ is some $p^n\mathbf{Z}\_p$, the quotient being finite or virtually isomorphic to $\mathbf{Z}$, hence has countably many subgroups.


Finally if $d=1$ and the Lie algebra is nilpotent, it is conjugate to the Lie algebra of strictly upper triangular matrices. So it's a compact subgroup of its normalizer, which is the group of upper triangular matrices of determinant 1 (as when $d=2$). So as when $d=2$, $K\subset\mathbf{Z}\_p^\*\rtimes\mathbf{Q}\_p$. Then $K$ contains $e\_{12}(p^n\mathbf{Z}\_p)=I\_2+p^n\mathbf{Z}\_pE\_{12}$ for some $n$, so is determined by its image in the quotient, a finite subgroup of $\mathbf{Z}\_p^\*\ltimes p^m\mathbf{Z}\_p/p^n\mathbf{Z}\_p$ for some $m\le n$. Since $\mathbf{Z}\_p^\*\simeq \mathbf{Z}\_p\times F$ with $F$ finite and $\mathbf{Z}\_p$ is torsion-free, the subgroup has to lie in the finite group $F\ltimes p^m\mathbf{Z}\_p/p^n\mathbf{Z}\_p$. So for given $m,n$, this leaves only finitely many possibilities.




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Remark 1: it can be shown along the same lines that $\mathrm{SL}\_2(\mathbf{Q}\_p)$ has countably many conjugacy classes of closed subgroups that are not (infinite discrete). While it has uncountably many conjugacy classes of discrete infinite cyclic subgroups (e.g., those generated by the diagonal matrices $\mathrm{diag}(t,t^{-1})$ when $|t|>1$ are pairwise non-conjugate).


Remark 2: in $\mathrm{SL}\_3(\mathbf{Z}\_p)$ there are uncountably many conjugacy classes of closed subgroups, just because there is a closed subgroup isomorphic to $\mathbf{Z}\_p^2$ (in which the subgroups $L\cap\mathbf{Z}\_p^2$, when $L$ ranges among lines in $\mathbf{Q}\_p^2$, are pairwise distinct).


