Linear embeddings of nilpotent pro-$p$ groups

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 $$d$$.

• It can be embedded in $GL_n(\mathbb{Z}_p)$, for some $n$ (such a group is $p$-adic analytic). Aug 27, 2015 at 21:36
• 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". Mar 30, 2022 at 2:10
• @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. Dec 14, 2022 at 16:04
• @AlexanderBetts You are right, thank you. So I guess this is unresolved. Dec 14, 2022 at 17:37
• @DavidESpeyer I think I managed to get an argument going -- posted it as an answer below. Dec 14, 2022 at 19:05

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.

• 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. Dec 14, 2022 at 19:03
• 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 ---. Dec 14, 2022 at 20:01