For a field $K$ let denote by $Tr_1(d,K)$ the nilpotent group of all upper triangular $d\times d$-matrices over $K$ with each diagonal entry equal to 1. Let $\mathbb{Q}_p$ the field of $p$-adic numbers and consider $Tr_1(d,\mathbb{Q}_p)$ with the $p$-adic topology (observe that this is a $p$-adic analytic group). Then $Tr_1(d,\mathbb{Q})$ is a dense subgroup of $Tr_1(d,\mathbb{Q}_p)$. The question is: If $H$ is a closed subgroup of $Tr_1(d,\mathbb{Q}_p)$ is it true that $H\cap Tr_1(d,\mathbb{Q})$ is dense in $H$?.

Now $Tr_1(d,\mathbb{Z}_p)$ is an open compact subgroup of $Tr_1(d,\mathbb{Q}_p)$ and in fact it is the pro-$p$ completion of $Tr_1(d,\mathbb{Z})$. It is a fact that for every closed subgroup $H$ of $Tr_1(d,\mathbb{Z}_p)$ we have that $H\cap Tr_1(d,\mathbb{Z})$ is dense in $H$. Using this it is easy to see that for any closed subgroup $H$ of $Tr(d,\mathbb{Q}_p)$ we have that $\overline{H\cap Tr(d,\mathbb{Q})}$ is a an open subgroup of $H$. But I can't prove that it is equal to $H$.


1 Answer 1


Pick $\lambda \in \mathbb{Z}_p \backslash \mathbb{Q}$ and define

$H = \left\lbrace\begin{pmatrix} 1 & 0 & a \cr 0 & 1 & \lambda a \cr 0 & 0 & 1 \end{pmatrix} : a \in \mathbb{Z}_p\right\rbrace$.

Then $H$ is a closed subgroup of $Tr_1(3,\mathbb{Q}_p)$ but $H \cap Tr_1(3,\mathbb{Q})$ is the trivial group, hence certainly not dense in $H$.


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