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Let $p$ be a prime number, $S = C_p$ a cyclic group of order $p$, $G = \mathbb{Z}_p$ the profinite additive group of $p$-adic integers. It is well known that all the closed nontrivial subgroups of $G$ are open (equivalently, of finite index).

Motivated by this example, we call an infinite profinite group $H$ thin if all its closed nontrivial subgroups are open.

What is known about thin profinite groups? Is there a complete list?

Is there an analogue of $G$ for a nonabelian finite simple group $T$? That is:

Is there a thin profinite group $M$ such that the composition factors of every nontrivial continuous finite image of $M$ are all isomorphic to $T$?

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    $\begingroup$ Take a non-trivial element and look at the subgroup it generates. This results in: in a profinite group is thin, then it is either finite, or virtually $\mathbf{Z}_p$ for some prime $p$ (maybe some little effort shows it's then isomorphic to $\mathbf{Z}_p$). In any case, this means that the class of thin profinite groups is minute, and not of much interest. $\endgroup$
    – YCor
    Oct 26, 2014 at 15:50
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    $\begingroup$ Voting to close as answered in the comments. $\endgroup$ Oct 27, 2014 at 2:03
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    $\begingroup$ @BenjaminSteinberg: the question is whether the conclusion can be strengthened from "is virtually isomorphic to $\mathbf{Z}_p$" to "is isomorphic to $\mathbf{Z}_p$". $\endgroup$
    – YCor
    Oct 27, 2014 at 22:44
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    $\begingroup$ ... and this is true. We need to show that if a pro-$p$-group $G$ is torsion-free and virtually $\mathbf{Z}_p$ then it's isomorphic to $\mathbf{Z}_p$. By induction it's enough to suppose $G$ has an open normal subgroup $N\simeq\mathbf{Z}_p$ of index $p$. Pick $x\notin N$. The action of $x$ on $N$ is by multiplication some element $t$ of $\mathbf{Z}_p^*$. Since $x^p\in N$, this shows that either $x^p=1$, or $t=1$. In the first case, we contradict that $G$ is torsion-free. In the second case, being central-by-cyclic, $G$ is abelian and the conclusion easily follows (e.g. by Pontryagin duality). $\endgroup$
    – YCor
    Oct 27, 2014 at 22:58
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    $\begingroup$ There do exist a continuum of hereditarily just infinite profinite groups, all of whose composition factors are of a given non-abelian isomorphism type. In fact, in constructing an h.j.i. profinite group, you are free to choose independently for each non-abelian finite simple group whether you want zero, finitely many or infinitely many composition factors of that type. This is the closest non-abelian analogue of 'thin' that I have heard of. $\endgroup$
    – Colin Reid
    Oct 28, 2014 at 1:01

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