Is it true that every finitely generated (topologically) torsion-free nilpotent pro-$p$ group is isomorphic to a subgroup of $U_d(\mathbb{Z}_p)$, the group of $d\times d$-upper triangular matrices with 1's in the diagonal, for some $d$?.

This question is the analogous of this well known result: every finitely generated torsion-free nilpotent group is isomorphic to a subgroup of $U_d(\mathbb{Z})$ for some $d$.

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    $\begingroup$ It can be embedded in $GL_n(\mathbb{Z}_p)$, for some $n$ (such a group is $p$-adic analytic). $\endgroup$ Aug 27, 2015 at 21:36
  • $\begingroup$ There is a theory of sylow groups for profinite groups groupprops.subwiki.org/wiki/… and I'm pretty sure that $U_d(\mathbb{Z}_p)$ is the $p$-Sylow of $GL_n(\mathbb{Z}_p)$. Any pro-$p$-subgroup is conjugate into the $p$-Sylow, so Yassine's comment implies that the answer is "yes". $\endgroup$ Mar 30, 2022 at 2:10
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    $\begingroup$ @DavidESpeyer Actually the $p$-Sylow of $GL_n(\mathbb Z_p)$ is bigger. Even when $n=1$, the $p$-Sylow subgroup in $GL_1(\mathbb Z_p)=\mathbb Z_p^\times$ is $1+p\mathbb Z_p$. And in general a $p$-Sylow subgroup of $GL_n(\mathbb Z_p)$ is the preimage of $U_n(\mathbb F_p)$ under the map $GL_n(\mathbb Z_p) \twoheadrightarrow GL_n(\mathbb F_p)$, since the kernel of that map is a pro-$p$ group. $\endgroup$ Dec 14, 2022 at 16:04
  • $\begingroup$ @AlexanderBetts You are right, thank you. So I guess this is unresolved. $\endgroup$ Dec 14, 2022 at 17:37
  • $\begingroup$ @DavidESpeyer I think I managed to get an argument going -- posted it as an answer below. $\endgroup$ Dec 14, 2022 at 19:05

1 Answer 1


Yes, this is true. In the argument below, all references are to the book Analytic Pro-$p$ Groups by Dixon–Du Sautoy–Mann–Segal.

Let $G$ be a finitely generated torsion-free nilpotent pro-$p$ group. We'll deal first with the special case that $G$ is uniform, meaning here that the quotient of $G$ by the closed subgroup $G^p$ generated by $p$th powers is abelian [Theorem 4.5]. In this case, one can endow $G$ with the structure of a Lie algebra over $\mathbb Z_p$, as explained in [Sections 4.3 & 4.5]. That is, for any elements $g,h\in G$, the element $g^{p^n}h^{p^n}$ is the $p^n$th power of a unique element of $G$, and the element $g^{p^n}h^{p^n}g^{-p^n}h^{-p^n}$ is the $p^{2n}$th power of a unique element of $G$. So we can define an addition and Lie bracket on $G$ by $$ g + h = \lim_{n\to\infty} (g^{p^n}h^{p^n})^{1/p^n} \quad\text{and}\quad [g,h] = \lim_{n\to\infty} (g^{p^n}h^{p^n}g^{-p^n}h^{-p^n})^{1/p^{2n}} \,. $$ These make $G$ into a Lie algebra $L(G)$ over $\mathbb Z_p$, which is finitely generated free as a $\mathbb Z_p$-module and for which the pro$-$p topology on $G$ coincides with the natural $\mathbb Z_p$-module topology [Proposition 4.16, Theorems 4.17 & 4.30]. Moreover, the Baker–Campbell–Hausdorff power series converges on $L(G)$ and recovers the group law on $G$ (follows from [Lemma 7.12]).

Now the fact that $G$ is nilpotent implies that $L(G)$ is a nilpotent Lie algebra. (This can be proved inductively by writing $G$ as a central extension of a finitely generated torsion-free nilpotent pro-$p$ group $G'$ — which is also uniform — by $\mathbb Z_p$ and noting that the induced sequence on Lie algebras is also a central extension.) So $\mathbb Q_p\otimes_{\mathbb Z_p}L(G)$ is a finite-dimensional nilpotent Lie algebra over $\mathbb Q_p$, and the Baker–Campbell–Hausdorff power series makes it into the $\mathbb Q_p$-points of a unipotent group $U(G)$ over $\mathbb Q_p$.

The upshot of all this is that $G$ admits a continuous embedding in the $\mathbb Q_p$-points of a unipotent group $U(G)$, namely via the composite $$ G \cong L(G) \subset \mathbb Q_p\otimes_{\mathbb Z_p}L(G) = U(G)(\mathbb Q_p) \,. $$ But every unipotent group embeds as a closed algebraic subgroup of the group $U_n$ of upper triangular matrices, and hence we get a continuous embedding of $G$ as a subgroup of $U_n(\mathbb Q_p)$, i.e. $G$ has a faithful unipotent continuous representation defined over $\mathbb Q_p$. It is now not hard to show that we can rescale coordinates on this representation so as to make the representation take values in $U_n(\mathbb Z_p)$, and we are done.

We've now dealt with the case that $G$ is uniform. In the general case, we know that $G$ has a uniform open normal subgroup $H$ [Corollary 4.3], and this subgroup will again be finitely generated, torsion-free and nilpotent. So we know from the above that $H$ admits a continuous embedding $f\colon H\hookrightarrow U(H)(\mathbb Q_p)$ into the $\mathbb Q_p$-points of a unipotent group. We want to show that $f$ extends uniquely to a continuous group homomorphism $f'\colon G\to U(H)(\mathbb Q_p)$ — such an $f'$ would automatically be an embedding since $G$ is torsion-free and we'd be done.

Since $U(H)(\mathbb Q_p)$ is a uniquely divisible group, there is only one possibility for $f'$: we must take $f'(x) = f(x^n)^{1/n}$ where $n$ is chosen so that $x^n\in H$. We want to show that this $f'$ is a group homomorphism, for which we use the Hall–Petresco identity [Appendix A]. This states that for any two elements $x,y$ in a group $G$, there is a sequence of elements $c_1=xy,c_2,c_3,\dotsc$ of $G$, with each $c_i$ lying in the $i$th step of the descending central series of $G$, such that $$ x^ny^n = (xy)^nc_2^{n\choose2}c_3^{n\choose3}\dots c_{n-1}^nc_n = (xy)^n\prod_{i=2}^{\infty}c_i^{n\choose i} $$ for all $n\geq0$ (the product on the right-hand side has only finitely many terms different from $1$). In our case, since $G$ is nilpotent, we only need to take the product on the right up to some fixed $N$. So if we choose $n$ divisible by a sufficiently large power of $p$, then all of the terms $x^n$, $y^n$, $(xy)^n$ and all $c_i^{n\choose i}$ in the Hall–Petrescu identity lie in the subgroup $H$. So we obtain $$ f'(x)^nf'(y)^n = f'(xy)^n \prod_{i=2}^Nf'(c_i)^{n\choose i} $$ for all such $n$. Now since $U(H)$ is a unipotent group, we know that the multiplication and exponentiation in $U(H)$ are given coordinatewise by polynomials with coefficients in $\mathbb Q_p$. So both sides of the above equality are given coordinatewise by polynomials in $n$. Since the equality holds for infinitely many $n$, it in fact must hold for all $n$. Setting $n=1$ we recover $$ f'(x)f'(y)=f'(xy) $$ so that $f'$ is a group homomorphism and we are done.

  • $\begingroup$ By the way, the map $G\to U(G)(\mathbb Q_p)$ in the first part, and the map $G\to U(H)(\mathbb Q_p)$ in the second part exhibit this group $U(G)$ or $U(H)$ as the $\mathbb Q_p$-Malcev completion (or pro-unipotent completion) of the topological group $G$, meaning that it is the initial continuous group homomorphism from $G$ to the $\mathbb Q_p$-points of a pro-unipotent group. $\endgroup$ Dec 14, 2022 at 19:03
  • $\begingroup$ I edited your TeX dashes to Unicode, but TeX note: in TeX, -- is an en-dash, as in Baker–Campbell–Hausdorff; but in other contexts—such as here—you want an em-dash, which is entered in TeX as ---. $\endgroup$
    – LSpice
    Dec 14, 2022 at 20:01

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