Why is every nilpotent-by-finite finitely generated pro-p-group always $p$-adic analytic

Question

I'm studying the paper "On the verbal width of finitely generated pro-p groups" by Andrei Jaikin-Zapirain (link at ProjectEuclid) and I cannot see a claim made in a proof. I don't know if the my question is appropriate to this site, I apologize if not.

Theorem 1.3. Let $G$ be a compact $p$-adic analytic group. Then any word $w$ of a free group $F$ has finite width in $G$.

Definition. We say that $w$ is a $\cal{N}_p$-word if for every finitely generated pro-$p$ group $H$, $H/\overline{\omega(H)}$ is nilpotent-by-finite, where $\overline{\omega(H)}$ denotes the closure of $\omega(H)$ in $H$.

Theorem 3.1. Let $\omega$ be a $\cal{N}_p$-word and $G$ a finitely generated pro-$p$ group. Then $\omega(G)$ is closed.

Set $d = d(G)$ and $H$ a free pro-$p$ group on generators $x_1,...,x_d,z$. We have that $\gamma_n(H^{p^t}) \leq \overline{\omega(H)}$ since $\omega$ is a $\cal{N}_p$-word. The generators $y_1,...,y_s$ of $\overline{\langle x_1,...,x_d \rangle^{p^t}}$ are pro-$p$ words in $x_i$. So he applies the theorem 1.3 to get $k$ such that, for every $i_1,...,i_n \in \{1,...,s\}$, $$[z,y_{i_1},...,y_{i_n}] \equiv v_{i_1,...,i_n} \pmod{\gamma_{n+2}(H^{p^t})}$$ where $v_{i_1,...,i_n}$ is a product of at most $k$ $\omega$-values in $H$.

I suppose that he works on the quotient group, but I cannot see the conditions to apply the theorem 1.3. Whence the question:

Why is every nilpotent-by-finite finitely generated pro-p-group always $p$-adic analytic?

Answer 1

This is indeed true: every finitely generated nilpotent-by-finite (= virtually nilpotent) pro-$p$-group is $p$-adic analytic.

Since every finitely generated nilpotent group has all its subgroup finitely generated, a finitely generated nilpotent profinite group $G$ has a composition series by closed subgroups in which each successive quotient is procyclic. If $G$ is moreover pro-$p$, it follows that each successive quotient is either isomorphic to $\mathbf{Z}_p$, or a quotient thereof (which is finite). So these successive quotients are $p$-adic analytic. Since being $p$-adic analytic is stable under taking extensions, it follows that every finitely generated nilpotent pro-$p$-group is $p$-adic analytic, and the same follows with "nilpotent-by-finite".

Answer 2

If you work with finite rank all of these questions are quite trivial. In particular, you will notice that finite rank by finite rank is finite rank.