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A uniform pro-$p$ group is an inverse limit of finite $p$-groups, a Lie group, and (the restriction to $\mathbb{Z}_p$ of) a finite-dimensional Lie algebra all at the same time. I think you'd be hard- …

Not in general. Consider for instance the case that $G$ is the same abstract group as $P_n$, but with the discrete topology. If $G$ is finitely generated, the answer is yes. See this article of Nik …

"Is there an example of a group of td-type which is not locally profinite?" No. This was proved by D. van Dantzig in the 1930s: Van Dantzig, D.: Zur topologischen Algebra. III. Brouwersche und Cant …

No. Let $K_n = K_n(G)$ be the intersection of all open normal subgroups of $G$ of index at most $n$. Then $\alpha(K_n) = K_n$. If we replace $G$ with $G/K_n$, we still have $\alpha$ acting as an au …

I am looking for an example of the following situation: $G$ is an infinite profinite group, with a dense normal subgroup $N$. However $N$ does not contain any non-trivial closed normal subgroup of $ …

Yves has answered your first question, so I'll address your more general aim. One can regard the Frattini subgroup of a profinite group $G$ as the set of 'non-generators': a non-generator is an eleme …

A pro-$p$ group of finite subgroup rank has an open subgroup $P$ that is uniformly powerful, meaning that $[P,P]$ is contained in the group generated by $2p$-th powers in $P$, and raising elements to …

Let $P$ be a finitely generated pro-$p$ group and let $G$ be a semidirect product $P \rtimes A$, where $A$ is a finite group of order coprime to $p$ that acts faithfully on $P$. Then one can show tha …

It will only happen if $m=1$. See this paper: http://mlarsen.math.indiana.edu/~larsen/papers/2gen.pdf Indeed, the pro-$p$ groups that are linear over local fields of characteristic $0$ are just the …