The conjugation action in completed group rings

TL;DR

This paper investigates the conjugation action in completed group rings of p-valuable groups, establishing structural links between automorphisms of the algebra and the underlying group, and calculating fixed rings under subgroup actions.

Contribution

It proves that automorphisms of the completed group algebra induced by conjugation correspond to automorphisms of the group itself and computes fixed rings for subgroup conjugation actions.

Findings

Inner automorphisms of $kG$ correspond to those of $G$.

Calculated $ ext{Gamma}$-fixed rings for certain ideals.

Established structural links between algebra automorphisms and group automorphisms.

Abstract

Let $k = \mathbb{F}_p$ or $\mathbb{Z}_p$ (or finite extensions of these). Let $G$ be a $p$-valuable group, and form its completed group algebra $kG$. By analysing the conjugation action of $G$ on itself, we prove two structural results. Firstly, we show that all inner automorphisms of $kG$ that preserve $G$ are induced from inner automorphisms of $G$. Secondly, for a closed subgroup $\Gamma$ of $G$, we calculate the $\Gamma$-fixed ring of $kG/I$ under the conjugation action of $\Gamma$, for certain ideals $I$ induced from the $G$-centraliser of $\Gamma$.

Equations

$$\underset{i\in I}{\underset{⟵}{\lim }}{A}_i:=\left\{(a_i{)}_{i\in I}\in \prod _{i\in I}{A}_i:{\pi }_{jk}(a_j)=a_k\text{ for all }j\ge k\right\}.$$

$$R/I\to \underset{J\in \mathcal{J}}{\underset{⟵}{\lim }}\left(R/(J+I)\right)$$

$$kG:=\underset{U}{\underset{⟵}{\lim }}k[G/U],$$

$$[G,H_i]=[G,G_i\cap H]\le [G,G_i]\cap [G,H]\le G_{i+1}\cap H=H_{i+1}$$

$$a=\sum _{x\in T}{a}_{x}x=\sum _{x\in T}\sigma (h)(a_{x}x)h^{-1}=\sum _{x\in T}(\sigma (h)a_{x}x{h}^{-1}{x}^{-1})x.$$

$$a=\sum _{x\in T}{a}_{xN}\cdot x,$$

$$a=\sum _{x\in T}{a}_{xN}\cdot t_x=\sum _{y\in T^{\prime }}{a}_{yN}\cdot t_y.$$

$$\sum _{x\in T}{a}_{xN}\cdot x=\sum _{x\in T}\sigma (g)a_{xN}\cdot x{g}^{-1}=\sum _{x\in T}{a}_{xN}^{\sigma (g{)}^{-1}}{n}_{g\text{hashsym}x}\cdot t_{g\text{hashsym}x}.$$

$$\sum _{x\in T}{a}_{xN}\cdot x=\sum _{g\text{hashsym}x\in g\text{hashsym}T}{a}_{g\text{hashsym}xN}\cdot t_{g\text{hashsym}x}.$$

$$a_{xN}^{\sigma (g{)}^{-1}}{n}_{g\text{hashsym}x}=a_{g\text{hashsym}xN}$$

$$\sum _{xN\in G\text{hashsym}yN}\epsilon (a_{xN})=p^r\epsilon (a_{yN}).$$

$$\epsilon (a)=\epsilon \left(\sum _{x\in T}{a}_{xN}x\right)=\sum _{xN\in G/N}\epsilon (a_{xN})=\sum _{\begin{array}{c}\text{distinct}\\ \text{orbits}\\ G\text{hashsym}yN\end{array}}\left(\sum _{xN\in G\text{hashsym}yN}\epsilon (a_{xN})\right),$$

$$\theta (x)\theta (y)=\boxed{x^{-1}gx{g}^{-1}}\boxed{y^{-1}}gy{g}^{-1},$$

$$h^{-1}\theta (x)h=\boxed{h^{-1}}\boxed{x^{-1}}gx{g}^{-1}h,$$

$$h^{-1}\theta (x)h=x^{-1}g{h}^{\prime -1}x{h}^{\prime }g.$$

$$(H{g}^{-1}xg{)}^{p^f}=H{g}^{-1}{x}^{p^f}g=H{x}^{p^f},$$

$$ag=x^{-1}(ag)x=zag,$$

$${\tau }_n{\rho }_n{\tau }_n$$

$$=\sigma {\tau }_{n-1}{\rho }_{n-1}{\pi }_n\sigma {\tau }_{n-1}{\overline{\pi }}_n$$

$$=\sigma {\tau }_{n-1}{\rho }_{n-1}{\tau }_{n-1}{\overline{\pi }}_n$$

$$=\sigma {\tau }_{n-1}{\overline{\pi }}_n$$

$$={\tau }_n$$

$${\hat{x}}_1=\sum _{\beta \in T}{x}_1^{\beta }=\sum _{\beta \in T}a{x}_2^{h\beta }=\sum _{h\beta \in hT}a{x}_2^{h\beta }=a{\hat{x}}_2,$$

$$f(\hat{y})=\sum _{\gamma \in T}f(y{)}^{\gamma }=\frac{|T|}{|S|}\sum _{\gamma \in S}{x}^{\gamma }=\frac{|T|}{|S|}\hat{x},$$

$$c=b_1{\hat{y}}_1+\cdots +b_s{\hat{y}}_s,f(c)=a_1{\hat{x}}_1+\cdots +a_r{\hat{x}}_r,$$

$${\mathrm{pr}}_{M(x)}(f(b_1{\hat{y}}_1+\cdots +b_s{\hat{y}}_s))=f(b_1)n_1\cdot {\mathrm{pr}}_{M(x)}({\hat{z}}_1)+\cdots +f(b_s)n_s\cdot {\mathrm{pr}}_{M(x)}({\hat{z}}_s),$$

$$R=\underset{t\in T}{⨁}(kH/I)t.$$

$$R_{n}=k[G/G_{n}]/\theta(IkG)\cong(k[H/H_{n}]*F){\big{/}}(\theta(I)*F)\cong A_{n}*F.$$

$$(P/Q)\cap Z(kG/Q)=(P/Q)\cap (Q+kZ)/Q=(Q+(P\cap kZ))/Q=Q/Q,$$

$$\theta (ga{g}^{-1})=(1-ga{g}^{-1})+\mathfrak{ga}=g(\eta +\mathfrak{ga})g^{-1}=\eta +\mathfrak{ga}=\theta (a),$$

Full text

1. 0.1 Inner automorphisms

2. 0.2 Fixed rings

3. 0.3 Ideals and normal subgroups

4. 1 Preliminaries

5. 1.1 $p$-valuable groups

6. 1.2 Profinite spaces, groups and rings

7. 1.3 Topological rings

8. 1.4 Completed group algebras

9. 1.5 Ideals

10. 2 Inner automorphisms

11. 2.1 Uniform subgroups

12. 2.2 The general result

13. 3 The action of $\Gamma$

14. 3.1 Centralised group elements

15. 3.2 Choosing a compatible family of transversals

16. 4 The lifting procedure

17. 4.1 Twisted permutation modules

18. 4.2 Lifting orbits

19. 5 Applications and open questions

20. 5.1 Ideal separation conditions

21. 5.2 Some parallels

Introduction

Throughout this paper, $p$ will denote a fixed prime number, and $\mathbb{Z}_{p}$ will be the ring (or additive group) of $p$-adic integers.

Our main objects of study are the completed group rings $kG$ of compact $p$-adic analytic groups $G$ over certain profinite rings $k$ (usually $\mathbb{F}_{p}$ or $\mathbb{Z}_{p}$). The rings $kG$ are often known as Iwasawa algebras in the literature, and play an important role in modern number theory.

Compact $p$-adic analytic groups (also known as compact $p$-adic Lie groups) may be most simply defined as closed subgroups of the topological group $GL_{n}(\mathbb{Z}_{p})$: see e.g. [6, 13] for references. In studying these groups, it is common first to study a smaller subclass consisting of particularly nice groups: common choices include uniform [6, §4] groups, or the slightly more general $p$-valuable [13, III, 3.1.6–7] groups.

Both $p$-valuable and uniform groups are torsion-free pro-$p$ groups of finite rank with well-behaved filtrations, which makes them easier to study. Moreover, arbitrary compact $p$-adic analytic groups are “almost” $p$-valuable (or uniform) in several important senses: for instance, given a compact $p$-adic analytic group $G$, we can always find an open (i.e. closed and finite-index) subgroup $G_{0}$ which is both $p$-valuable (or uniform) and normal in $G$ [6, Corollary 8.34]. Having chosen such a $G_{0}$, we can form the ring $kG$ as a crossed product of the ring $kG_{0}$ with the finite group $G/G_{0}$: by doing this, we can often pass information from $kG_{0}$ to $kG$. Such descriptions have been concretely applied with fruitful results in previous work: see, for instance, [24, 26, 27, 4].

For this reason, in the current paper, we work primarily with $p$-valuable and uniform groups $G$. The most important properties of these groups are recalled for the convenience of the reader below in §1.1 and §2.1.

There is an ongoing project to understand the (prime) ideals in these completed group rings $kG$: various precise questions were posed in the survey paper [4, Questions G–O], and despite significant work on this front in the last decade, most of these questions still remain partly or entirely open. The most substantial results in this direction have been those of the paper [3], which answers special cases of [4, Questions N and O]:

Let $G$ be a nilpotent uniform group with centre $Z$.

• •

Is every prime ideal of $\mathbb{F}_{p}G$ completely prime?

• •

Is $\mathrm{Spec}(\mathbb{F}_{p}G)$ the disjoint union of finitely many commutative strata?

• •

Is every faithful prime ideal of $\mathbb{F}_{p}G$ controlled by $Z$?

More recently, the results of the thesis [11] achieve positive results towards the more general case of soluble uniform $G$.

Many of the questions in [4] were posed exclusively in the (“modular”) case when $k=\mathbb{F}_{p}$. The (“integral”) case $k=\mathbb{Z}_{p}$ is often of even greater interest in areas of number theory, where Iwasawa algebras originated; and the study of the ideals $I$ of $\mathbb{F}_{p}G$ may be viewed as the study of the ideals $J$ of $\mathbb{Z}_{p}G$ which contain $p$, so that the modular case is a special case of the integral case. However, the integral case appears to be considerably more difficult, and currently very little is known about ideals of $\mathbb{Z}_{p}G$ which do not contain $p$. For some recent developments, see [11, Theorems C and E] and [12].

In all of the below results, $k$ will denote either a finite field of characteristic $p$ or a complete discrete valuation ring whose residue field has characteristic $p$.

0.1 Inner automorphisms

Our first results concern automorphisms of $kG$.

Theorem A. Let $G$ be a $p$-valuable group, and let $\sigma$ be an automorphism of $kG$ that preserves the naturally embedded copy of $G$. Then $\sigma$ is the (unique) extension of an inner automorphism of $G$.∎

The importance of this result stems from an analysis of the behaviour of skew power series rings $R[[x;\sigma,\delta]]$ (see [21, 15, 28] for the definition and basic properties of these rings in the Iwasawa context). Forthcoming joint work of the current author and Adam Jones shows that the behaviour of $R[[x;\sigma,\delta]]$ varies depending on whether $\sigma$ is inner or outer. A similar dichotomy has also been observed in the world of skew polynomial rings $R[x;\sigma,\delta]$, e.g. [14, Theorem 3.6].

If $G$ is a $p$-valuable group and $H$ is a closed normal subgroup such that $G/H\cong\mathbb{Z}_{p}$, choose an arbitrary topological generator $gH$ for $\mathbb{Z}_{p}$: that is, choose an element $g\in G$ such that $G=\overline{\langle H,g\rangle}$. Then $kG$ can be written as a skew power series ring $kH[[x;\sigma,\sigma-\mathrm{id}]]$, where $x=g-1$ and $\sigma$ is conjugation by $g$ [21, §4]. Hence $\sigma$ is an automorphism of $kH$ which preserves the naturally embedded copy of $H$, and so if it is inner, we can apply Theorem A to obtain the following simplification.

Corollary B. Let $G$ be a $p$-valuable group and $H$ a closed normal subgroup such that $G/H\cong\mathbb{Z}_{p}$. Choose an element $g\in G$ such that $G=\overline{\langle H,g\rangle}$. If the conjugation action of $g$ on $kH$ is inner, then $G\cong H\times\mathbb{Z}_{p}$.∎

0.2 Fixed rings

Next, we turn our attention to subrings of $kG$ fixed under the conjugation action of a closed subgroup $\Gamma\leq G$.

Theorem C. Let $G$ be a $p$-valuable group, $\Gamma$ an arbitrary closed subgroup which we consider acting on $G$ by conjugation, and $H$ the centraliser in $G$ of $\Gamma$. If $I$ is a faithful $G$-prime ideal of $kH$, then the set of $\Gamma$-invariant elements of $kG/IkG$ is precisely $(kH+IkG)/IkG$.∎

Note that, under this setup, $kG/IkG$ is a ring, as $H$ is a normal subgroup: see Lemma 1.3 below. Note also that we may identify the subring $(kH+IkG)/IkG\leq kG/IkG$ with the ring $kH/I$. (Indeed, $kG$ is a flat $kH$-module by [5, Lemma 4.5], and is faithfully flat by [16, Proposition 7.2.3] as the maximal ideal of $kG$ contains the maximal ideal of $kH$. Now it follows that $IkG\cap kH=I$, as in [3, Lemma 5.1].)

This result, proved below as Theorem 4.14, is an analogue of Roseblade’s [19, Lemma 10], which was used to answer similar questions about group algebras of polycyclic groups. While Theorem A is not quite as general as [19, Lemma 10], in practice it is general enough to be used in an analogous context:

Corollary D. Let $G$ be a supersoluble $p$-valuable group, and $Z$ its centre. Then $kG$ is centrally separated if and only if all its faithful prime ideals are controlled by $Z$.∎

Here, by centrally separated, we mean the following: if $P\lneq Q$ are two elements of $\mathrm{Spec}(kG)$, then $Q/P$ contains a nonzero central element of $kG/P$. An ideal $I\lhd kG$ is faithful if $G$ embeds naturally into $kG/I$, i.e. if the kernel of the natural map $G\to(kG)^{\times}\to(kG/I)^{\times}$ is trivial.

This result is proved in §5.1. Of course, the question of when $kG$ satisfies either condition remains open in general: to our knowledge, the only case that is known is that of [3]. Still, we explore the potential significance of this result in the larger context, and give some conjectures.

0.3 Ideals and normal subgroups

Finally, in §5.2, we provide a first result linking the ideal structure of $\mathbb{Z}_{p}G$ with the normal subgroup structure of $G$:

Theorem E. Let $G$ be a soluble $p$-valuable group. If $\mathbb{Z}_{p}G$ is polycentral, than $G$ is in fact nilpotent. ∎

This mirrors similar results for group algebras of polycyclic-by-finite groups, as well as finite-dimensional Lie algebras: see §5.2 for a discussion.

1 Preliminaries

1.1 $p$-valuable groups

As mentioned in the introduction, every compact $p$-adic analytic group $G$ contains an open $p$-valuable normal subgroup $G_{0}$, and $p$-valuable groups are in some sense much better behaved. We recall some of the basic properties of $p$-valuable groups.

Definition 1.1**.**

If $H$ is a closed subgroup of $G$, then $H$ is called orbital if it has finitely many $G$-conjugates. An orbital closed subgroup is called isolated if, whenever $H^{\prime}$ is an orbital closed subgroup of $G$ strictly containing $H$, we have $[H^{\prime}:H]=\infty$. [25, Definition 1.4].

Properties 1.2**.**

Let $G$ be a $p$-valuable group, as defined in [13, III, 3.1.6].

1. (i)

$G$* is a torsion-free pro-$p$ group of finite rank. [13, III, 2.1.2]* 2. (ii)

$G$* is orbitally sound [3, Proposition 5.9], i.e. if $H$ is any orbital closed subgroup of $G$, then its normal core $H^{\circ}:=\bigcap_{g\in G}H^{g}$ has finite index in $H$. In particular, this implies that its isolated orbital closed subgroups are normal in $G$ [25, Lemma 1.10]. Hence an orbital closed subgroup $H$ of $G$ is isolated if and only if it is normal and $G/H$ is torsion-free (compare [3, §5.3]), and by (i), this means that an orbital closed subgroup $H$ is isolated if and only if it is normal and $x\in G$, $x^{p}\in H\implies x\in H$.* 3. (iii)

Closed subgroups of $G$ are $p$-valuable **[13, III, 2.1.2]**. 4. (iv)

If $H$ is an isolated normal closed subgroup of $G$, then $G/H$ is $p$-valuable. **[13, IV, 3.4.2]** 5. (v)

If $x,y\in G$ satisfy $x^{p}=y^{p}$, then $x=y$. **[13, III, 2.1.4]**

Lemma 1.3**.**

Let $G$ be a $p$-valuable group, $\Gamma$ an arbitrary normal closed subgroup which we consider acting on $G$ (on the right) by conjugation, and $H$ the centraliser of $\Gamma$ in $G$. Then $H$ is an isolated normal closed subgroup of $G$, and $G/H$ is $p$-valuable.

Proof.

Given a sequence $h_{n}\in H$ converging to $h\in G$, we must have $\gamma=h_{n}^{-1}\gamma h_{n}\to h^{-1}\gamma h$ for all $\gamma\in\Gamma$, showing that $h$ centralises $\Gamma$. Hence $h\in H$, and so as the convergent sequence $(h_{n})$ was arbitrary, $H$ must be closed in $G$. The normality of $H$ follows from the normality of $\Gamma$.

To show that $H$ is isolated using the condition given in Property 1.2(ii), we need only show that, if $h^{p}\in H$ for some $h\in G$, then in fact $h\in H$. So let $\gamma\in\Gamma$ be arbitrary: then we know that $\gamma^{-1}h^{p}\gamma=h^{p}$, i.e. $(\gamma^{-1}h\gamma)^{p}=h^{p}$, and so by Property 1.2(v) we get $\gamma^{-1}h\gamma=h$, i.e. $h$ centralises $\Gamma$. So $H$ is isolated, and $G/H$ is $p$-valuable by Property 1.2(iv). ∎

1.2 Profinite spaces, groups and rings

Our groups and rings are all profinite, so we recall the basics of inverse limits necessary for our purposes: cf. [6, §1], [23, §1].

Definition 1.4**.**

Let $(I,\leq)$ be a directed set, i.e. a nonempty partially ordered set with the property that, for all $i,j\in I$, there exists $k\in I$ such that $k\geq i$ and $k\geq j$.

Let $\mathcal{C}$ be the category of topological spaces, topological groups or topological rings. An inverse system in $\mathcal{C}$ (over $I$) is a collection of objects $A_{i}$ for all $i\in I$, together with a collection of morphisms $\pi_{ij}:A_{i}\to A_{j}$ for all $i,j\in I$ satisfying $i\geq j$, with the property that $\pi_{ik}=\pi_{jk}\circ\pi_{ij}$ for all $i\geq j\geq k$: we will usually write this inverse system simply as $(A_{i})_{i\in I}$, omitting the $\pi_{ij}$ if they are known from context.

The inverse limit of the inverse system $(A_{i})_{i\in I}$ is denoted $\underset{\longleftarrow}{\lim}_{i\in I}A_{i}$, and is defined to be the object

[TABLE]

This comes equipped with natural projection maps $\varphi_{j}:\underset{\longleftarrow}{\lim}_{i\in I}A_{i}\to A_{j}$, and is given the coarsest topology making all the projection maps $\varphi_{j}$ continuous, making it again an object of $\mathcal{C}$.

Finally, if the $A_{i}\in\mathcal{C}$ are finite and discrete, then $\underset{\longleftarrow}{\lim}_{i\in I}A_{i}\in\mathcal{C}$ is called profinite.

Properties 1.5**.**

**

1. (i)

Let $(A_{i})_{i\in I}$ be an inverse system of finite discrete objects in $\mathcal{C}$. Then $\underset{\longleftarrow}{\lim}_{i\in I}A_{i}$ is a closed subobject of $\prod_{i\in I}A_{i}$ (cf. **[6, proof of Proposition 1.3]**). 2. (ii)

Suppose that $J\subseteq I$ is a directed set. Then there is a canonical morphism $\underset{\longleftarrow}{\lim}_{i\in I}A_{i}\to\underset{\longleftarrow}{\lim}_{j\in J}A_{j}$. Moreover, if $J$ is cofinal with $I$, i.e. for all $i\in I$ there exists $j\in J$ such that $j\geq i$, then this canonical map is an isomorphism in $\mathcal{C}$. 3. (iii)

If $G$ is a profinite group, then $G\cong\underset{\longleftarrow}{\lim}_{N}(G/N)$, where $N$ ranges over the inverse system of open normal subgroups of $G$. The partial order of this directed set is reverse containment, i.e. $N_{2}\geq N_{1}$ if $N_{2}\subseteq N_{1}$, and the morphisms $G/N_{2}\to G/N_{1}$ of this inverse system are the natural projection homomorphisms.

1.3 Topological rings

Let $R$ be a ring equipped with a topology $\mathcal{T}$. We say that $\mathcal{T}$ is a ring topology on $R$ if the addition and multiplication maps $R\times R\to R$ and the negation map $R\to R$ are all continuous with respect to $\mathcal{T}$, and when this is the case, we say $R$ is a topological ring (with respect to $\mathcal{T}$). All rings of interest in this paper are profinite (see Remark 1.10 below), and hence carry the profinite topology [23, §1].

Our standard reference for general topological rings is [22].

Lemma 1.6**.**

Let $R$ be a compact, totally disconnected (topological) ring and $I$ a closed ideal. Then $R/I$ is a compact, totally disconnected ring.

Proof.

Clearly $R/I$ is a ring, and is compact as a quotient space. By [22, Theorem 5.4], the topology on $R/I$ is indeed a ring topology, so that $R/I$ is a compact ring. Now it follows from [22, Theorem 5.17(3)] that $R/I$ is totally disconnected. ∎

Lemma 1.7**.**

Profinite rings are compact and totally disconnected.

Proof.

By definition, a profinite ring $R$ is the inverse limit of an inverse system of discrete finite rings, and is hence a closed subspace of the product space. But discrete finite rings are compact and totally disconnected, and these properties are preserved under arbitrary products and closed subspaces. ∎

Lemma 1.8**.**

Let $R$ be a compact, totally disconnected ring. Suppose that $\mathcal{J}$ is a fundamental system of neighbourhoods of zero for $R$ consisting of open ideals, and let $I$ be an arbitrary closed ideal of $R$. Then the canonical ring homomorphism

[TABLE]

is an isomorphism of topological rings.

Proof.

By Lemma 1.6, $R/I$ is a compact, totally disconnected ring. As $\mathcal{J}$ is a fundamental system of neighbourhoods of zero in $R$, by [22, Theorem 5.5], we have that $\{A_{J}\;:\;J\in\mathcal{J}\}$ is a fundamental system of neighbourhoods of zero in $R/I$, where we set $A_{J}:=(J+I)/I$. Each $A_{J}$ is easily seen to be an open ideal in $R$, as it is the image of $J$ under the projection map $R\to R/I$, which is an open map (see e.g. [22, Theorem 5.2]).

Now [22, Theorem 5.23] implies that $R/I$ is topologically isomorphic to the inverse limit of the inverse system of rings $\{(R/I)/A_{J}\;:\;J\in\mathcal{J}\}$. But $(R/I)/A_{J}$ is naturally isomorphic to $R/(J+I)$ by standard isomorphism theorems. (That this topological isomorphism is canonical can be read off from [22, Corollary 5.22] and the proof of [22, Theorem 5.23].) ∎

1.4 Completed group algebras

Let $G$ be a compact $p$-adic analytic group, and let $k$ be either a finite field of characteristic $p$ or a complete discrete valuation ring whose residue field is a finite field of characteristic $p$.

Definition 1.9**.**

The completed group ring of $G$ over $k$ is defined to be

[TABLE]

where $U$ ranges over the inverse system of open normal subgroups of $G$, and $k[G/U]$ is the usual group algebra of the (finite) group $G/U$ over the commutative ring $k$. The topological rings $\mathbb{Z}_{p}G$ and $\mathbb{F}_{p}G$ are sometimes called the (respectively integral and modular) Iwasawa algebras of $G$. When it is notationally convenient, we may write $k[[G]]$ instead of $kG$; when $G$ is finite, we may write $k[G]$, as the completed group ring and the usual group ring are canonically isomorphic in this case.

The reader may also wish to consult [23, Chapter 7] for more on completed group rings of profinite groups over profinite rings in the general setting, or [13, Ch. II, §2.2] for more on integral Iwasawa algebras of $p$-valuable groups.

Remarks 1.10**.**

**

1. (i)

$kG$* is always profinite. Indeed, if $k$ is finite, then each $k[G/U]$ is a finite ring. If $k$ is a complete discrete valuation ring with uniformiser $\pi$, then $kG$ may be identified with $\underset{\longleftarrow}{\lim}(k/\pi^{n}k)[G/U]$, where the inverse limit ranges over the inverse system of open normal subgroups $U\lhd G$ and $n\geq 1$: see [6, §7.1]. As the rings $(k/\pi^{n}k)[G/U]$ are all finite, it follows that $kG$ is profinite.* 2. (ii)

It follows from Lemma 1.7 and (i) that $kG$ is compact and totally disconnected. 3. (iii)

When $G$ is $p$-valuable, $kG$ admits a complete filtration by powers of its Jacobson radical, and the associated graded ring is noetherian: see e.g. **[3, §6.2]** or **[13, III, §2.3]**. Hence by **[9, p. 85, Corollary 5]** every ideal of $kG$ is closed. Combined with (ii), this says that we may apply Lemma 1.8 to arbitrary ideals $I$ of $kG$. 4. (iv)

When $G$ is $p$-valuable, the subring $k[G]$ (the ordinary group ring of $G$ over $k$) is dense in $kG$ **[13, II, 2.2.1–3]**. 5. (v)

If $G$ is pro-$p$, then $kG$ is a local ring **[23, Proposition 7.5.3]**.

1.5 Ideals

Let $k$ be a finite field of characteristic $p$ or a complete discrete valuation ring whose residue field is a finite field of characteristic $p$. Let $G$ be a $p$-valuable group.

Due to Remark 1.10(iii), we will say simply “ideal” instead of “closed ideal” throughout the remainder of this paper.

Definition 1.11**.**

Let $H$ be a closed normal subgroup of $G$, and $I$ an ideal of $kH$. Then we will say $I$ is $G$-invariant if it is invariant under conjugation by $G$. Furthermore, we will say that $I$ is $G$-prime if it is $G$-invariant and has the property that, whenever $A$ and $B$ are $G$-invariant ideals of $kH$ with $AB\subseteq I$, we must have either $A\subseteq I$ or $B\subseteq I$.

Remark 1.12**.**

If $I$ is a $G$-invariant ideal of $kH$, and $H$ is a closed normal subgroup of $G$, then $IkG$ is an ideal of $kG$.

If $H$ is also isolated normal, then $kG/IkG\cong(kH/I)[[x_{1},\dots,x_{e}]]$ as a filtered $kH/I$-module: this may be proved along the same lines as the very similar results [2, Proposition 3.6] and [3, Lemma 8.5].

Definition 1.13**.**

If $I$ is an ideal of $kG$, we write $I^{\dagger}=\ker(G\to kG^{\times}\to(kG/I)^{\times})=(I+1)\cap G$, and we say that $I$ is faithful if $I^{\dagger}=\{1\}$.

Note that, as $G$ is assumed $p$-valuable, it follows that $I^{\dagger}$ is an isolated normal closed subgroup of $G$ by [3, Lemma 5.3(a), (c)].

Lemma 1.14**.**

Let $H$ be an isolated normal closed subgroup of $G$, and let $I$ be a faithful $G$-invariant ideal of $kH$. Then $IkG$ is a faithful ideal of $kG$.

Proof.

The proof of [3, 8.6(b)] carries over to this situation mutatis mutandis. ∎

The following is also well known:

Lemma 1.15**.**

Let $N$ be a closed normal subgroup of $G$. Then the augmentation ideal $w_{N}:=\ker(kG\to k[[G/N]])$ is generated as a right ideal by $N-1$.

Proof.

This follows from [23, Proposition 7.1.2], after taking Remark 1.10(iii) into account. ∎

Lemma 1.16**.**

Let $N$ be an open normal subgroup of $G$, and write $N_{2}=[G,N]$. Then, for all $r\in kN$ and $g\in G$, we have $grg^{-1}\in kN$ and $gr-rg\in(N_{2}-1)kG=w_{N_{2}}$.

Proof.

Let $n\in N$ and $g\in G$. Then $gng^{-1}n^{-1}\in N_{2}$ implies $gng^{-1}\in N\subseteq k[N]$ and $gn-ng=(gng^{-1}n^{-1}-1)ng\in(N_{2}-1)k[G]$. It now follows for all $r\in k[N]$ that $grg^{-1}\in k[N]$ and $gr-rg\in(N_{2}-1)k[G]$ (as $r$ is just a finite $k$-linear combination of elements $n\in N$), and the lemma follows by taking closures inside $kG$ as in Remark 1.10(iv). ∎

2 Inner automorphisms

2.1 Uniform subgroups

Before we prove Theorem A for general $p$-valuable groups $G$, we first prove the analogue for uniform groups $G$, and so we recall below the relevant definitions and facts for the convenience of the reader.

Definition 2.1**.**

A (topologically) finitely generated pro-$p$ group is called powerful if $[G,G]\leq G^{p}$ if $p$ is odd, or $[G,G]\leq G^{4}$ if $p=2$. [6, Definition 3.1, Lemma 3.4]

A uniformly powerful pro-$p$ group, or a uniform group for short, is a powerful finitely generated pro-$p$ group which is torsion-free [6, Theorem 4.5].

Properties 2.2**.**

Let $G$ be uniform. Then the following properties hold:

1. (i)

For each $i\geq 1$, set $G_{i}=G^{p^{i-1}}$: this is a characteristic open subgroup of $G$ **[6, Theorem 3.6(ii)]**. The filtration $G=G_{1}>G_{2}>\dots$ is a fundamental system of neighbourhoods of the identity for $G$ **[6, Proposition 1.16(iv)]**, and also defines an integer-valued $p$-valuation on $G$ in the sense of **[13, III, 2.1.2]**. We also have $[G,G_{i}]\leq G_{i+1}$ **[6, Theorem 3.6(i)]**. 2. (ii)

There exists a binary operation $+$ on $G$ which makes $G$ into a $\mathbb{Z}_{p}$-module; moreover, any topological generating set for $G$ as a group is also a basis for $G$ as a $\mathbb{Z}_{p}$-module **[6, Definition 4.12, Theorem 4.17]**. In additive notation, the subgroup $G_{n}$ defined above is equal to the $\mathbb{Z}_{p}$-submodule $p^{n-1}G$ **[6, Lemma 4.14(iii)]**. 3. (iii)

Let $H$ be an isolated normal closed subgroup of $G$. Then there exists a topological generating set $\{a_{1},\dots,a_{d}\}$ for $G$ such that the subset $\{a_{1},\dots,a_{e}\}$ (for some $e\leq d$) is a topological generating set for $H$: cf. **[3, proof of Lemma 8.5]** or **[25, remark after Corollary 4.4]**.

Lemma 2.3**.**

Let $G$ be a uniform group and $H$ an isolated normal closed subgroup. Then $H_{n}=G_{n}\cap H$ and $[G,H_{i}]\subseteq H_{i+1}$.

Proof.

Choose a topological generating set $\{a_{1},\dots,a_{d}\}$ for $G$ such that $\{a_{1},\dots,a_{e}\}$ is a topological generating set for $H$ as in Property 2.2(iii).

Now consider $H$ and $G_{n}$ as $\mathbb{Z}_{p}$-modules. By Property 2.2(ii), $G_{n}$ has $\mathbb{Z}_{p}$-basis $\{p^{n}a_{1},\dots,p^{n}a_{d}\}$, and so $G_{n}\cap H$ has basis $\{p^{n}a_{1},\dots,p^{n}a_{e}\}$ by linear algebra. This shows that $G_{n}\cap H=H_{n}$.

Reverting to group notation: $H$ normal in $G$ by assumption, and so

[TABLE]

using Property 2.2(i) as required. ∎

Proposition 2.4**.**

Let $H$ be a uniform group, and fix some $i\geq 1$ and $a\in kH_{i}^{\times}$. Define the inner automorphism $\sigma$ of $kH$ by $\sigma(r)=ara^{-1}$. Let $T$ be a transversal to $H_{i+1}$ in $H_{i}$, so that we may write $a=\sum_{x\in T}a_{x}x$ for coefficients $a_{x}\in kH_{i+1}$.

1. (i)

If $\sigma(h)h^{-1}\in H_{i+1}$ for some $h\in H$, then $a_{x}=\sigma(h)a_{x}xh^{-1}x^{-1}$ for all $x\in T$. 2. (ii)

If $\sigma(h)h^{-1}\in H_{i+1}$ for all $h\in H$, then there exist an element $y\in H_{i}$ and $b\in kH_{i+1}^{\times}$ with the following properties. Define the inner automorphism $\tau$ of $kH$ by $\tau(r)=brb^{-1}$: then $\sigma(h)=\tau(yhy^{-1})$ and $\tau(h)h^{-1}\in H_{i+2}$ for all $h\in H$.

Proof.

1. (i)

Suppose we are given such an $h\in H$. Rearranging the equation $\sigma(h)=aha^{-1}$ gives $a=\sigma(h)ah^{-1}$, which we may rewrite as

[TABLE]

As $kH_{i}=\bigoplus_{x\in T}(kH_{i+1})x$, we will be done if we can show that $\sigma(h)a_{x}xh^{-1}x^{-1}\in kH_{i+1}$. But

• •

$\sigma(h)a_{x}\sigma(h)^{-1}\in k{H_{i+1}}$ as $H$ normalises $k{H_{i+1}}$,

• •

$\sigma(h)h^{-1}\in H_{i+1}\subseteq k{H_{i+1}}$ by assumption, and

• •

$hxh^{-1}x^{-1}\in H_{i+1}\subseteq k{H_{i+1}}$ by Lemma 2.3,

and the product of these three elements is $\sigma(h)a_{x}xh^{-1}x^{-1}$. 2. (ii)

Since $a$ is invertible, some $a_{x}$ must also be invertible, as $kH_{i}$ is local. So fix $y\in T$ such that $a_{y}$ is invertible. Define $\tau\in\mathrm{Inn}(k{H_{i+1}})$ to be conjugation (on the left) by $a_{y}$, and define $\eta\in\mathrm{Inn}(H)$ to be conjugation (on the left) by $y$. Then eqn (2.1) for $x=y$ says that $\sigma(h)=a_{y}yhy^{-1}a_{y}^{-1}=\tau(yhy^{-1})$ for all $h\in H$.

It remains to show that $\tau(h)h^{-1}\in H_{i+2}$. But $\tau(h)h^{-1}=a_{y}ha_{y}^{-1}h^{-1}$, and so by Lemma 1.16, $\tau(h)h^{-1}-1=(a_{y}h-ha_{y})a_{y}^{-1}h^{-1}\in w_{H_{i+2}}$. This means that $\tau(h)h^{-1}\in w_{H_{i+2}}^{\dagger}=H_{i+2}$.∎

Theorem 2.5**.**

Let $H$ be a uniform group, and let $\sigma$ be an inner automorphism of $kH$ that fixes $H$. Then $\sigma$ is the unique extension of an inner automorphism of $H$.

Proof.

Write $\sigma_{1}=\sigma$. First, note as in the previous proof that $\sigma_{1}(h)h^{-1}-1\in w_{H_{2}}$ by Lemma 1.16, and so in particular $\sigma_{1}(h)h^{-1}\in w_{H_{2}}^{\dagger}=H_{2}$.

We now define $\sigma_{n}$ and $y_{n}$ inductively for all $n\geq 1$. Given an inner automorphism $\sigma_{i}$ of $kH$ such that $\sigma_{i}(h)h^{-1}\in H_{i+1}$ for all $h\in H$, Proposition 2.4(ii) will give an element $y_{i}\in H_{i}$ and an inner automorphism $\sigma_{i+1}$ of $kH$ such that $\sigma_{i}(h)=\sigma_{i+1}(y_{i}hy_{i}^{-1})$ and $\sigma_{i+1}(h)h^{-1}\in H_{i+2}$ for all $h\in H$. Now set $g_{0}=1$ and $g_{i+1}=y_{i+1}g_{i}$ for all $i\geq 0$, so that $\sigma(h)=\sigma_{i+1}(g_{i}hg_{i}^{-1})$ for all $h\in H$ and all $i\geq 0$. Then the sequence $(g_{n})_{n\geq 0}$ converges inside $H$, say to some limit $g_{\infty}$, and $\sigma(h)=g_{\infty}hg_{\infty}^{-1}$ for all $h\in H$. ∎

2.2 The general result

Now let $G$ be a $p$-valuable group. The argument below is an extension of a standard argument for finite $p$-groups.

Proof of Theorem A. $G$ contains a characteristic uniform open subgroup $N$ [6, Corollary 4.3], and so we may write $kG=kN*F$ (a crossed product in the sense of [17]), where $F$ is a finite $p$-group. Suppose that $\sigma$ is an inner automorphism of $kG$ which preserves $G$, say there exists $a\in kG^{\times}$ such that $\sigma(r)=ara^{-1}$ for all $r\in kG$.

This time, a slightly different notation will be convenient. Let $T$ be a transversal to $N$ in $G$, and for each $x\in G$, write $n_{x}$ and $t_{x}$ for the unique elements $n_{x}\in N$ and $t_{x}\in T$ satisfying $n_{x}t_{x}=x$. We will also index the coefficients of $a$ using the cosets, i.e. we will write

[TABLE]

for some coefficients $a_{xN}\in kN$. In particular, if $T^{\prime}$ is any (perhaps different) transversal, we still have

[TABLE]

Given $g,x\in G$, we will write $g\mathbb{\hashsym}x=\sigma(g)xg^{-1}$; note that $(g,xN)\mapsto(g\mathbb{\hashsym}x)N$ is a well-defined left action of $G$ on the set of cosets $G/N$. As $G$ is pro-$p$ and $G/N$ is a finite set, the orbits $G\mathbb{\hashsym}xN$ must all have cardinality a power of $p$.

Fix an arbitrary $g\in G$. On the one hand, $a=\sigma(g)ag^{-1}$, or in other words

[TABLE]

On the other hand, by making the change of variables $x\mapsto g\mathbb{\hashsym}x$, we can see that

[TABLE]

As before, this implies that

[TABLE]

for all $xN\in G/N$ and (as $g$ was arbitrary) for all $g\in G$.

Now consider the augmentation map $\varepsilon:kG\to k$ sending all $g\in G$ to $1$: applying $\varepsilon$ to equation (2.2) gives $\varepsilon(a_{xN})=\varepsilon(a_{g\mathbb{\hashsym}xN})$, and so in particular, if the orbit $G\mathbb{\hashsym}yN$ of $yN\in G/N$ has size $p^{r}$, then

[TABLE]

But we can calculate $\varepsilon(a)$ as

[TABLE]

and note that this must be a unit as $a$ is invertible. In particular, some orbit $G\mathbb{\hashsym}yN$ must have size 1, and of course we may choose this element $y$ to be an element of $T$, so that $t_{y}=t_{g\mathbb{\hashsym}y}=y$ for all $g\in G$.

For this particular $y\in G$, equation (2.2) reads $a_{yN}^{\sigma(g)^{-1}}n_{g\mathbb{\hashsym}y}=a_{yN}$ for all $g\in G$, and this can be rearranged to give that $\sigma(g)=a_{yN}ygy^{-1}a_{yN}^{-1}$. It now follows that the automorphism $g\mapsto\sigma(y^{-1}gy)$ of $kG$ is given by conjugation by an element of $kN^{\times}$.

Call this automorphism $\beta$. As $\sigma$ preserves $G$, so does $\beta$, and as $N$ is characteristic in $G$, $\beta$ must also preserve $N$. In particular, the restriction $\beta|_{kN}\in\mathrm{Aut}(kN)$ satisfies the hypotheses of Theorem 2.5: that is, there exists some $n\in N$ such that $\beta(h)=nhn^{-1}$ for all $h\in N$. But now, for every $g\in G$, there exists some $r$ such that $g^{p^{r}}\in N$, and so $\beta(g)^{p^{r}}=(ngn^{-1})^{p^{r}}$; and as $G$ is $p$-valuable, the desired result now follows from Property 1.2(v). ∎

Proof of Corollary B. As in the introduction, fix any element $g$ satisfying $G=\overline{\langle H,g\rangle}$: then setting $x=g-1$ and letting $\sigma$ be conjugation by $g$ gives an equality of filtered left $kH$-modules $kG=kH[[x;\sigma,\sigma-\mathrm{id}]]$. If $\sigma$ is inner, then by Theorem A there exists $h\in H$ such that $grg^{-1}=hrh^{-1}$ for all $r\in kH$. But this means that $h^{-1}g$ centralises $H$, and so $G=\overline{\langle H,h^{-1}g\rangle}=H\times\overline{\langle h^{-1}g\rangle}$.∎

3 The action of $\Gamma$

3.1 Centralised group elements

Lemma 3.1**.**

Let $G$ be a group, $\Gamma$ a subgroup, and $H$ a normal subgroup of $G$ centralising $\Gamma$. Suppose, for some fixed $g\in G$, we have that $x^{-1}gx\in Hg$ for all $x\in\Gamma$; that is, the function $\theta:=\theta_{g}:\Gamma\to G$ defined by $\theta(x)=x^{-1}gxg^{-1}$ has image in $H$. Then $\theta$ is in fact a group homomorphism with image contained in $Z(H)$.

Remark 3.2**.**

This is mentioned (without proof) in the context of linear groups by Roseblade in [19]; the proof is easy, but we have chosen to supply it.

Proof.

To see that $\theta$ is a group homomorphism, note that

[TABLE]

and that the two boxed factors commute, as the first is an element of $H$ and the second is an element of $\Gamma$. This expression then simplifies to $\theta(x)\theta(y)=y^{-1}x^{-1}gxyg^{-1}=\theta(xy)$ as required. To see that the image is central in $H$, we compute, for arbitrary $h\in H$,

[TABLE]

noting again that the boxed expressions commute for the same reason as above. Now, as $H$ is normal in $G$, there exists some $h^{\prime}\in H$ such that $h^{-1}g=gh^{\prime-1}$ and hence that $g^{-1}h=h^{\prime}g^{-1}$. Making these substitutions, we can see that

[TABLE]

Finally, $h^{\prime-1}xh^{\prime}=x$, and so this equation reduces to $h^{-1}\theta(x)h=\theta(x)$. ∎

For the remainder of this subsection, let $G$ be a $p$-valuable group, $H$ a closed isolated normal subgroup (so that $G/H$ is $p$-valuable by Property 1.2(iv)), and $\Gamma$ a closed normal subgroup of $G$ acting on $G$ by conjugation and centralising $H$. If $I$ is a $G$-invariant ideal of $kH$, we will always view $kG/IkG$ as a filtered $kH/I$-module as in Remark 1.12.

Proposition 3.3**.**

Let $I$ be a $G$-invariant ideal of $kH$. Suppose that $y=a_{1}g_{1}+\dots+a_{n}g_{n}\in kG/IkG$ is centralised by $\Gamma$, where each $a_{i}\in kH/I$, and the $g_{i}$ are elements of $G$ which are pairwise distinct modulo $H$. Then $a_{i}g_{i}$ is centralised by $\Gamma$ for each $1\leq i\leq n$.

Proof.

$\Gamma$ permutes the submodules $(kH/I)g_{i}$, which means that it must permute the summands $a_{i}g_{i}$ and also the cosets $Hg_{i}$ in the same way. We will prove that, if $Hg\in G/H$ has finite $\Gamma$-orbit, then it has trivial $\Gamma$-orbit: this will suffice.

Let $S$ be the stabiliser in $\Gamma$ of $Hg$. Then $S$ has finite index in $\Gamma$, so its normal core $S^{\circ}:=\bigcap_{\gamma\in\Gamma}S^{\gamma}$ has finite index in $\Gamma$ too, say index $p^{f}$, as $\Gamma$ is orbitally sound by Properties 1.2(iii), (iv). Hence, for all $x\in\Gamma$, we have $x^{p^{f}}\in S^{\circ}\subseteq S$, and so

[TABLE]

an equality inside $G/H$. But now applying Property 1.2(v) to this equation shows that $x$ fixes $Hg$, i.e. $S=\Gamma$. ∎

Proposition 3.4**.**

Let $I$ be a faithful $G$-prime ideal of $kH$. Let $g\in G$ be such that $x^{-1}gx\in Hg$ for all $x\in\Gamma$. Suppose that, for some nonzero $a\in kH/I$, the element $ag\in kG/IkG$ is centralised by $\Gamma$. Then $g$ is also centralised by $\Gamma$.

Proof.

Since $I$ is faithful, we may view $G$ as a subgroup of $(kG/IkG)^{\times}$. Define $\theta:\Gamma\to G$ by $\theta(x)=x^{-1}gxg^{-1}$, so that by Lemma 3.1 we have $\theta(x)\in Z(H)$ for all $x\in\Gamma$. We will prove that in fact $\theta(x)=1$ for all $x\in\Gamma$.

Write $I$ as the intersection of a finite $G$-orbit of prime ideals $J_{1},\dots,J_{s}$ of $kH$, which is possible due to [15, Theorem 3.18]. As $a$ is assumed nonzero inside $kH/I$, we can take $a=\tilde{a}+I$ for some $\tilde{a}\in kH\setminus I$; and $\tilde{a}\not\in I$ implies that $\tilde{a}\not\in J_{i}$ for at least one $1\leq i\leq s$. Choose such a $J_{i}$, and denote it simply by $J$.

Take $x\in\Gamma$ with $\theta(x)=z\in Z(H)$. This means that

[TABLE]

so that $(z-1)\tilde{a}\in I\subseteq J$. Now, as $z$ is central in $H$, this implies that $(z-1)\cdot kH\cdot\tilde{a}\subseteq J$; and since $J$ is prime, we must now have $z-1\in J$, i.e. $z\in J^{\dagger}$.

As the $G$-orbit of $J$ is finite, the closed subgroup $J^{\dagger}$ is orbital in $G$, and we may calculate its normal core in $G$ as $(J^{\dagger})^{\circ}=\bigcap_{g\in G}(J^{\dagger})^{g}=J_{1}^{\dagger}\cap\dots\cap J_{s}^{\dagger}=I^{\dagger}$, which is $\{1\}$ by assumption. But now Property 1.2(ii) tells us that the index $[J^{\dagger}:(J^{\dagger})^{\circ}]$ is finite, and so $J^{\dagger}$ is a finite subgroup of $G$. Finally, $G$ is torsion-free by Property 1.2(i), so $J^{\dagger}=\{1\}$. In particular, this tells us that $z=1$ as required. ∎

At this stage, we note the following for later use. Let $H$ be the centraliser in $G$ of $\Gamma$. Let $T$ be an arbitrary transversal to $H$ in $G$, and write $R=\bigoplus_{t\in T}(kH/I)t$. (Note that $R$ is a subring of $kG/IkG$, as $H$ is normal in $G$.)

Proposition 3.5**.**

If $I$ is a faithful $G$-prime ideal of $kH$, then $\mathbf{C}_{\Gamma}(R)=kH/I$.

Proof.

The inclusion “$\supseteq$” is clear.

For the converse, suppose that the element $y=a_{1}t_{1}+\dots+a_{n}t_{n}\in R$ is centralised by $\Gamma$, where each $a_{i}\in kH/I$ is nonzero, and the $t_{i}$ are distinct elements of $T$. Then, as $G/H$ is $p$-valuable by Lemma 1.3, we may apply Proposition 3.3 to see that the elements $a_{i}t_{i}\in R$ are centralised by $\Gamma$ for each $1\leq i\leq n$.

Now let $x\in\Gamma$ and $1\leq i\leq n$ be arbitrary, and write $a=a_{i}$ and $t=t_{i}$. Set $t^{x}=s\in G$. Then $at=(at)^{x}\in((kH/I)t)^{x}=(kH/I)s$ implies (by intersecting both sides with the naturally embedded copy of $G$) that $Hs=Ht$, and so in particular $x^{-1}tx=s\in Hs=Ht$. As $x$ was arbitrary, Proposition 3.4 shows that $t$ is centralised by $\Gamma$, i.e. $t\in H$. And as $i$ was also arbitrary, and the $t_{i}$ were assumed distinct mod $H$, we must have $n=1$ and $y=a_{1}t_{1}\in kH/I$. ∎

3.2 Choosing a compatible family of transversals

Let $G$ be a $p$-valuable group, and $H$ a closed, isolated, normal subgroup of $G$.

In what follows, it will be convenient to fix a descending sequence $G=G_{1}\supseteq G_{2}\supseteq G_{3}\supseteq\dots$ of open normal subgroups of $G$ such that $\bigcap_{n\geq 1}G_{n}=\{1\}$. This can be done using e.g. [6, Proposition 1.16(iii)]. This sequence (viewed as an inverse system over the directed set of positive integers) is cofinal with the inverse system of all open normal subgroups of $G$, and so $G\cong\underset{\longleftarrow}{\lim}_{i}(G/G_{i})$ by Properties 1.5(ii), (iii). For each $i\geq 1$, and each closed subgroup $K$ of $G$, write $K_{i}=G_{i}\cap K$.

In this section, we will choose a family of “compatible” transversals to $H_{i}$ in $G_{i}$ for all $i\geq 1$, and use them to construct a transversal to $H$ in $G$ compatible with the quotient maps. In particular, we will write the following:

\textstyle{G\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho}$$\textstyle{\underset{\longleftarrow}{\lim}_{i}(G/G_{i})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi_{n}}$$\textstyle{G/G_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{n}}$$\scriptstyle{\rho_{n}}$$\textstyle{G/G_{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho_{n-1}}$$\textstyle{G/H\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\underset{\longleftarrow}{\lim}_{i}(G/HG_{i})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\overline{\varphi}_{n}}$$\textstyle{G/HG_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\overline{\pi}_{n}}$$\textstyle{G/HG_{n-1}}

Here, $\rho,\rho_{n},\pi_{n},\overline{\pi}_{n}$ are the natural quotient maps for each $n$, the inverse limits run over $\pi_{n}$ and $\overline{\pi}_{n}$, and $\varphi_{n},\overline{\varphi_{n}}$ are the natural projection maps. This diagram commutes.

A transversal to $H$ in $G$ is a subset $T\subseteq G$ such that, for each $g\in G$, there exists a unique $t\in T$ satisfying $Hg=Ht$: in other words, we want a map of sets $\tau:G/H\to G$ splitting the surjective map $\rho$, so that $T=\mathrm{im}(\tau)$.

Proposition 3.6**.**

There exists a sequence of transversals $T_{n}$ to $HG_{n}/G_{n}$ in $G/G_{n}$ such that $\pi_{n}(T_{n})=T_{n-1}$ for all $n$.

Proof.

We proceed by induction. The case $n=1$ is trivial: $G_{1}=G$, so $HG_{1}/G_{1}=G/G_{1}=\{G_{1}\}$ is the trivial group, which is its own transversal, so we set $T_{1}=G/G_{1}$.

Suppose now that $n>1$, and that $T_{n-1}$ is a transversal to $HG_{n-1}/G_{n-1}$ in $G/G_{n-1}$: that is, we have a splitting $\tau_{n-1}$ of the surjective map $\rho_{n-1}$. Choose an arbitrary splitting $\sigma$ of the surjective map $\pi_{n}$, and define the map $\tau_{n}=\sigma\tau_{n-1}\overline{\pi}_{n}:G/HG_{n}\to G/G_{n}$. A diagram chase shows that

[TABLE]

and as $\tau_{n}$ is injective, it is left-cancellable, so that $\rho_{n}\tau_{n}=\mathrm{id}_{G/HG_{n}}$. This tells us that $\tau_{n}$ does in fact split $\rho_{n}$, and we may set $T_{n}=\mathrm{im}(\tau_{n})$. Now $\pi_{n}(T_{n})=\mathrm{im}(\pi_{n}\tau_{n})=\mathrm{im}(\pi_{n}\sigma\tau_{n-1}\overline{\pi}_{n})$, and as $\pi_{n}\sigma=\mathrm{id}_{G/G_{n-1}}$ and $\overline{\pi}_{n}$ is surjective, this is just equal to $\mathrm{im}(\tau_{n-1})=T_{n-1}$. ∎

Corollary 3.7**.**

The inverse limit $T:=\underset{\longleftarrow}{\lim}_{i}(T_{i})$ (where the $T_{i}$ are viewed as discrete topological spaces) is a transversal to $H$ in $G$, and $\varphi_{n}(T)=T_{n}$ for all $n$.∎

We will make crucial use of this transversal later.

4 The lifting procedure

4.1 Twisted permutation modules

The proof of Theorem C essentially follows [1, Propositions 2.1 and 2.2], but as we are no longer dealing with straightforward permutation modules, it requires some extra setup.

Definition 4.1**.**

Let $A$ be a ring, $M$ a free left $A$-module with basis $X$, and $\Gamma$ a group acting $A$-linearly on $M$. If, for every $\gamma\in\Gamma$ and $x\in X$, there exist unique $x^{\prime}\in X$ and $a\in A^{\times}$ such that $x^{\prime}=ax^{\gamma}$, we call $M$ a twisted permutation module for $\Gamma$, and denote it $A^{t}[X]$. We will say $A^{t}[X]$ has finite orbits if the $\Gamma$-orbit of every element of $A^{t}[X]$ is finite.

Notation 4.2**.**

Throughout this section, we will have a group $\Gamma$ acting on various sets $S$.

1. (i)

The action of an element $\gamma\in\Gamma$ will always be written as a right superscript, i.e. we will write the action map $S\times\Gamma\to S$ as $(s,\gamma)\mapsto s^{\gamma}$. 2. (ii)

The subgroup of $\Gamma$ stabilising the element $s\in S$ will be written $\Gamma_{s}$. The orbit of $s\in S$ will be written $s^{\Gamma}$. 3. (iii)

The subset of $S$ which is centralised by $\Gamma$ will be denoted $\mathbf{C}_{\Gamma}(S)$: this is non-standard, but we hope that it will cause no confusion in practice. 4. (iv)

If $\Gamma$ acts on an additive group $A$, and $|a^{\Gamma}|$ is finite for some $a\in A$, then we will write $\widehat{a}$ for the sum of its $\Gamma$-orbit, i.e. $\sum_{b\in a^{\Gamma}}b$. Equivalently, if $T$ is any right transversal to $\Gamma_{a}$ in $\Gamma$, then $\widehat{a}=\sum_{\gamma\in T}a^{\gamma}$.

Lemma 4.3**.**

Let $A^{t}[X]$ be a twisted permutation module for $\Gamma$ with finite orbits. Then $\mathbf{C}_{\Gamma}(A^{t}[X])=\sum_{x\in X}A\widehat{x}$.

Proof.

The inclusion “$\supseteq$” is obvious. For the reverse inclusion: let $r=a_{1}x_{1}+\dots+a_{n}x_{n}\in\mathbf{C}_{\Gamma}(A^{t}[X])$, where $a_{i}\in A$ and the $x_{i}$ are pairwise distinct elements of $X$. Then $\Gamma$ must permute the set $\{a_{1}x_{1},\dots,a_{n}x_{n}\}$, and so it falls into disjoint $\Gamma$-orbits $o_{1},\dots,o_{s}$, so that $r=\widehat{o}_{1}+\dots+\widehat{o}_{s}$. It suffices to show that each $\widehat{o}_{i}$ can be written as $a\widehat{x}$ for some $a\in A$ and $x\in X$. To this end, suppose (renumbering without loss of generality) that $o_{i}=\{a_{1}x_{1},\dots,a_{m}x_{m}\}$: but then $a_{1}x_{1}+\dots+a_{m}x_{m}=\widehat{a_{1}x_{1}}=a_{1}\widehat{x}_{1}$. ∎

Given any $x\in X$, we will denote by $M(x)$ the submodule $\sum_{\gamma\in\Gamma}Ax^{\gamma}\subseteq A^{t}[X]$.

Lemma 4.4**.**

Let $A^{t}[X]$ be a twisted permutation module for $\Gamma$ with finite orbits, and let $x_{1}$ and $x_{2}\in X$. If there exist $a\in A^{\times}$ and $h\in\Gamma$ such that $x_{1}=ax_{2}^{h}$, then $\widehat{x}_{1}=a\widehat{x}_{2}$ and $A\widehat{x}_{1}=A\widehat{x}_{2}$. If not, then $M(x_{1})\cap M(x_{2})=\{0\}$.

Proof.

Let $\mathcal{S}=\{Ax\}_{x\in X}$ be the set of $A$-submodules of $A^{t}[X]$ each generated by a single element of $X$: note that the definition of twisted permutation module ensures that $\Gamma$ acts on $\mathcal{S}$.

If $Ax_{1}$ and $Ax_{2}$ are in the same orbit, then $Ax_{1}=Ax_{2}^{h}$ for some $h\in\Gamma$, and so $x_{1}=ax_{2}^{h}$ for some $a\in A$; similarly $Ax_{2}=Ax_{1}^{h^{-1}}$ and so $x_{2}=bx_{1}^{h^{-1}}$ for some $b\in A$, implying that $x_{1}=abx_{1}$ and so $a$ is invertible. In this case, $\Gamma_{x_{2}}=h\Gamma_{x_{1}}h^{-1}$, and so if $T$ is a right transversal to $\Gamma_{x_{1}}$ in $\Gamma$, then $hT$ is a right transversal to $\Gamma_{x_{2}}$ in $\Gamma$. Hence

[TABLE]

and so $A\widehat{x}_{1}=A\widehat{x}_{2}$. Otherwise, $Ax_{1}$ and $Ax_{2}$ are in disjoint orbits: as the $x\in X$ are $A$-linearly independent by definition, this ensures that $M(x_{1})$ and $M(x_{2})$ intersect only in the zero submodule. ∎

Remark 4.5**.**

We draw an immediate conclusion from Lemmas 4.3 and 4.4. Elements of $\mathbf{C}_{\Gamma}(A^{t}[X])$ can be written as $a_{1}\widehat{x}_{1}+\dots+a_{r}\widehat{x}_{r}$, for coefficients $a_{i}\in A$ and elements $x_{i}\in X$, and the $x_{i}$ can be chosen so that the $M(x_{i})$ have pairwise zero intersection. Moreover, $\mathbf{C}_{\Gamma}(A^{t}[X])$ is the direct sum of the $M(x_{i})$ (with duplicates removed), so for each $x\in X$ there is a well-defined projection map $\mathrm{pr}_{M(x)}:\mathbf{C}_{\Gamma}(A^{t}[X])\to M(x)$.

Note that, by definition, $A\widehat{x}\subseteq M(x)$ for all $x\in X$. In particular, in this language, Lemma 4.4 says the following: given $x,z\in X$, $\mathrm{pr}_{M(x)}(\widehat{z})\neq 0$ if and only if there exist $a\in A^{\times}$ and $h\in\Gamma$ such that $z=ax^{h}$, and in this case $\mathrm{pr}_{M(x)}(\widehat{z})=\widehat{z}$.

Definition 4.6**.**

Let $c\in\mathbf{C}_{\Gamma}(A^{t}[X])$ and $x\in X$. We will write $\widehat{x}\in\mathrm{supp}(c)$ if $\mathrm{pr}_{M(x)}(c)\neq 0$.

Lemma 4.7**.**

Let $A^{t}[X]$ and $B^{t}[Y]$ be twisted permutation modules for $\Gamma$ with finite orbits, and suppose that we have a map of $\Gamma$-modules $f:B^{t}[Y]\to A^{t}[X]$. Fix $x\in X$ and $y\in Y$, and suppose that $f(y)=x$. Then $f(\widehat{y})=n\widehat{x}$, where $n=|y^{\Gamma}|/|x^{\Gamma}|$.

Proof.

As $f$ is a map of $\Gamma$-sets, we have $\Gamma_{y}\subseteq\Gamma_{x}$. Then, writing $S$ for a right transversal of $\Gamma_{x}$ in $\Gamma$ and $T$ for a right transversal of $\Gamma_{y}$ in $\Gamma$, we have

[TABLE]

and $|T|=|y^{\Gamma}|$ and $|S|=|x^{\Gamma}|$ by the orbit-stabiliser theorem. ∎

Finally, we come to the main proposition of the section:

Proposition 4.8**.**

Let $A^{t}[X]$ and $B^{t}[Y]$ be twisted permutation modules for $\Gamma$ with finite orbits, and suppose that we have a map of $\Gamma$-modules $f:B^{t}[Y]\to A^{t}[X]$. Assume also that $f(Y)\subseteq X$. Suppose that we are given $c\in\mathbf{C}_{\Gamma}(B^{t}[Y])$ and $x\in X$ such that $\widehat{x}\in\mathrm{supp}(f(c))$.

If $A$ has characteristic $p>0$ and $\Gamma$ is a pro-$p$ group, then there exists $y\in Y$ such that $f(y)=x$, $|y^{\Gamma}|=|x^{\Gamma}|$ and $\widehat{y}\in\mathrm{supp}(c)$.

Proof.

Let

[TABLE]

where $a_{i}\in A$, $b_{i}\in B$, $x_{i}\in X$ and $y_{i}\in Y$, such that the $M(x_{i})\subseteq A^{t}[X]$ have pairwise zero intersection and the $M(y_{i})\subseteq B^{t}[Y]$ have pairwise zero intersection. Suppose without loss of generality that $x=x_{i}$, and write $a=a_{i}$ so that $\mathrm{pr}_{M(x)}(f(c))=a\widehat{x}\neq 0$. But

[TABLE]

where $f(y_{j})=z_{j}$ and $n_{j}=|y_{j}^{\Gamma}|/|z_{j}^{\Gamma}|$, by Lemma 4.7. Now there must be some $1\leq j\leq s$ such that $f(b_{j})n_{j}\cdot\mathrm{pr}_{M(x)}(\widehat{z}_{j})\neq 0$. For this $j$, we must have $b_{j}\neq 0$ and $n_{j}\neq 0$, and by Lemma 4.4 and Remark 4.5, $\mathrm{pr}_{M(x)}(\widehat{z}_{j})\neq 0$ implies that there exist $h\in\Gamma$ and $u\in A^{\times}$ such that $f(y_{j})=z_{j}=ux^{h}$, so that $|z_{j}^{\Gamma}|=|x^{\Gamma}|$.

Set $y^{\prime}$ to be equal to this $y_{j}$, and let $y\in Y$ be such that $y=v(y^{\prime})^{h^{-1}}$ for some $v\in B^{\times}$. It is easy to see that $|y^{\Gamma}|=|y^{\prime\Gamma}|$ and $f(y)=x$, and by definition $\widehat{y}\in\mathrm{supp}(c)$ if and only if $\widehat{y^{\prime}}\in\mathrm{supp}(c)$.

It remains to show that $|z_{j}^{\Gamma}|=|y_{j}^{\Gamma}|$, i.e. $n_{j}=1$. However, as $\Gamma$ is a pro-$p$ group, $n_{j}$ must be a power of $p$, and as $A$ has characteristic $p$, all powers of $p$ are either [math] or $1$ in $A$. As we know $n_{j}\neq 0$, we must have $n_{j}=1$. So we may take $y=y_{j}$. ∎

4.2 Lifting orbits

Let $G$ be a $p$-valuable group, $\Gamma$ an arbitrary closed normal subgroup which we consider acting on $G$ (on the right) by conjugation, and $H$ the centraliser in $G$ of $\Gamma$. Recall that this implies that $G/H$ is $p$-valuable by Lemma 1.3. Fix also a faithful $G$-prime ideal $I\lhd kH$.

We fix further notation below for the remainder of this section.

Notation 4.9**.**

**

Fix a descending sequence $G=G_{1}\supseteq G_{2}\supseteq G_{3}\supseteq\dots$ of open normal subgroups of $G$ such that $\bigcap_{n\geq 1}G_{n}=\{1\}$. For each $i\geq 1$, write $H_{i}=G_{i}\cap H$. 2. 2.

Write $\varphi_{n}:G\to G_{n}$ and $\pi_{n}:G/G_{n}\to G/G_{n-1}$ for the natural projection maps. 3. 3.

Let $(T_{n})_{n\geq 1}$ be a sequence of transversals to $HG_{n}/G_{n}$ in $G/G_{n}$ such that $\pi_{n}(T_{n})=T_{n-1}$ for all $n\geq 2$ as in Proposition 3.6, and let $T$ be the corresponding inverse limit transversal to $H$ in $G$ from Corollary 3.7. 4. 4.

For each $i\geq 1$, denote $A_{i}=kH/(w_{H_{i}}+I)$ and $R_{i}=kG/(w_{G_{i}}+IkG)$.

Remark 4.10**.**

It follows from Lemma 1.8 and Remark 1.10(iii) that the maps $R_{n+1}\to R_{n}$ form an inverse system with inverse limit $kG/IkG$.

As before, we view $kG/IkG$ as a filtered left $kH/I$-module.

Definition 4.11**.**

We define the following left $kH/I$-submodule of $kG/IkG$:

[TABLE]

Note that, as $H$ is normal in $G$, in fact $R$ is a subring of $kG/IkG$.

Lemma 4.12**.**

$R_{n}=A_{n}^{t}[T_{n}]$* is a twisted permutation module for $\Gamma$.*

Proof.

Note first that $H/H_{n}\cong HG_{n}/G_{n}\lhd G/G_{n}$, so that $k[G/G_{n}]$ can be written as a crossed product $k[H/H_{n}]*F$, where $F$ is the finite group $G/HG_{n}$. Now, writing $\theta:kG\to k[G/G_{n}]$, we have $\theta(IkG)=\theta(I)k[G/G_{n}]=\theta(I)\cdot(k[H/H_{n}]*F)$. Now, as $\theta(I)$ is assumed $F$-stable, this is equal to $\theta(I)*F$ by [17, Lemma 1.4(ii) and its proof]. Hence, again by [17, Lemma 1.4(ii)], we have

[TABLE]

Taking $T_{n}$ as a basis for the crossed product inside $R_{n}$ yields the desired result. ∎

Now let $f_{i}:R_{i}\to R_{i-1}$ be the natural projection map for each $i\geq 2$. We have the data of an inverse system of $\Gamma$-modules:

\textstyle{\dots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R_{n+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{n+1}}$$\textstyle{R_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{n}}$$\textstyle{R_{n-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\dots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R_{1}.}

This gives a cone over the diagram of $\Gamma$-fixed rings:

\textstyle{\mathbf{C}_{\Gamma}(kG/IkG)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi_{n+1}}$$\scriptstyle{\varphi_{n}}$$\scriptstyle{\varphi_{n-1}}$$\textstyle{\dots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbf{C}_{\Gamma}(R_{n+1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{n+1}}$$\textstyle{\mathbf{C}_{\Gamma}(R_{n})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{n}}$$\textstyle{\mathbf{C}_{\Gamma}(R_{n-1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\dots}

It is now easy to check that we are in a position to apply Proposition 4.8.

Theorem 4.13**.**

$\mathbf{C}_{\Gamma}(kG/IkG)=\overline{\mathbf{C}_{\Gamma}(R)}$.

Proof.

Case 1: $\mathrm{char}(k)=p>0$.

Take some nonzero element $\alpha\in\mathbf{C}_{\Gamma}(kG/IkG)$. Set $\alpha^{(i)}=\alpha$ and proceed inductively for all $i\geq 1$. If $\alpha^{(i)}=0$, set $\xi^{(i)}=0$. Otherwise, let $r$ be the minimal positive integer such that $\varphi_{r}(\alpha^{(i)})\in\mathbf{C}_{\Gamma}(R_{r})$ is nonzero.

Now, using Proposition 4.8, we can lift each $x_{r}\in T_{r}$ such that $\widehat{x}_{r}\in\mathrm{supp}(\varphi_{r}(\alpha^{(i)}))$ to an $x_{r+1}\in T_{r+1}$ whose $\Gamma$-orbit has the same size and such that $\widehat{x}_{r+1}\in\mathrm{supp}(\varphi_{r+1}(\alpha^{(i)}))$. Repeating this procedure gives a sequence of elements $x_{r},x_{r+1},x_{r+2},\dots$ such that $f_{r+j}(x_{r+j})=x_{r+j-1}$ for all $j\geq 1$, and so this sequence determines a unique element $x\in T$ whose $\Gamma$-orbit has the same size and such that $\varphi_{r}(\widehat{x})=\widehat{x}_{r}$. Lifting every $\widehat{x}_{r}\in\mathrm{supp}(\varphi_{r}(\alpha^{(i)}))$ in this way, we see that there exists $\xi^{(i)}\in\mathbf{C}_{\Gamma}(R)$ such that $\varphi_{r}(\xi^{(i)})=\varphi_{r}(\alpha^{(i)})$. Now set $\alpha^{(i+1)}=\alpha^{(i)}-\xi^{(i)}$ so that $\varphi_{r}(\alpha^{(i+1)})=0$, and repeat.

Now the sum $\alpha=\xi^{(1)}+\xi^{(2)}+\xi^{(3)}+\dots$ converges in $kG/IkG$, and each term is in $\mathbf{C}_{\Gamma}(R)$, so $\alpha\in\overline{\mathbf{C}_{\Gamma}(R)}$.

Case 2: $\mathrm{char}(k)=0$.

Let $\pi$ be a uniformiser for $k$ (which must now be a complete discrete valuation ring). Take some nonzero element $\alpha\in\mathbf{C}_{\Gamma}(kG/IkG)$. Set $\alpha^{(0)}=\alpha$ and proceed inductively for all $i\geq 0$ as follows. By reducing $\alpha^{(i)}\in\mathbf{C}_{\Gamma}(kG/IkG)$ modulo $\pi$, we are back in Case 1, so the argument of Case 1 gives us an element $\beta^{(i)}\in\overline{\mathbf{C}_{\Gamma}(R)}$ such that $\alpha^{(i)}+\pi\mathbf{C}_{\Gamma}(kG/IkG)=\beta^{(i)}+\pi\mathbf{C}_{\Gamma}(kG/IkG)$: that is, there exists some $\alpha^{(i+1)}\in\mathbf{C}_{\Gamma}(kG/IkG)$ such that $\alpha^{(i)}=\beta^{(i)}+\pi\alpha^{(i+1)}$.

Now the sum $\alpha=\beta^{(0)}+\pi\beta^{(1)}+\pi^{2}\beta^{(2)}+\dots$ converges in $kG/IkG$, and each term is in $\overline{\mathbf{C}_{\Gamma}(R)}$, so $\alpha\in\overline{\mathbf{C}_{\Gamma}(R)}$. ∎

We can now join this with Proposition 3.5.

Theorem 4.14**.**

If $I$ is faithful and $G$-prime, then $\mathbf{C}_{\Gamma}(kG/IkG)=kH/I$.

Proof.

Let $R$ be as above, so that $\mathbf{C}_{\Gamma}(kG/IkG)$ is the closure in $kG/IkG$ of $\mathbf{C}_{\Gamma}(R)$; and by Proposition 3.5, $\mathbf{C}_{\Gamma}(R)=kH/I$, which is already closed in $kG/IkG$. ∎

This proves Theorem C.

5 Applications and open questions

5.1 Ideal separation conditions

$G$ will continue to denote a $p$-valuable group throughout.

In the usual way, we will say that:

• •

$kG$ is polycentral if, for all two-sided ideals $A\lneq B$ of $kG$, the ideal $B/A$ contains a nonzero central element of $kG/A$.

• •

$kG$ is centrally separated if, for all two-sided prime ideals $A\lneq B$ of $kG$, the ideal $B/A$ contains a nonzero central element of $kG/A$.

(For the reader’s convenience, we note that polycentrality is equivalent to hypercentrality in this context since $kG$ is right noetherian, and that both terms appear commonly in the literature.)

Proof of Corollary D. Let $P$ be a faithful prime ideal of $kG$, and set $I=P\cap kZ$ and $Q=IkG$.

It will suffice to prove that $Q$ is a prime ideal of $kG$. Indeed, by Theorem C (with $\Gamma=G$), we can calculate that $Z(kG/Q)=(Q+kZ)/Q$, and hence

[TABLE]

which is the zero subring of $kG/Q$. (The second equality follows from the modular law.) But $Q\leq P$: so, if $Q$ is prime, as we have assumed that $kG$ is centrally separated, we must have $P=Q$.

So we now prove that $Q$ is a prime ideal of $kG$. Firstly, $I$ is a prime ideal of $kZ$, from which it follows by commutative algebra (see e.g. [8, Chapter II, Exercise 4.11]) that the ring $kZ/I$ admits a discrete valuation $v$ with respect to which it is complete. Without loss of generality, let $v$ have image in $\mathbb{N}\cup\{\infty\}$.

Now we may realise $kG/Q$ as an iterated skew power series ring over $kZ/I$ using [15, 3.14(iv)], and as $G/Z$ is supersoluble by assumption, this iterated extension is triangular in the sense of [28, cf. Example 2.16]. We may extend $v$ to a valuation $\tilde{v}$ on $kG/Q$ using the same argument as [28, Theorem E]. This shows that $Q$ is a prime ideal of $kG$.∎

5.2 Some parallels

Let $H$ be a discrete group and $\mathbb{Z}[H]$ its integral group ring. It is known [18, Corollary] that $\mathbb{Z}[H]$ is polycentral if and only if $H$ is finitely generated and nilpotent, and the paper [20] contains similar results over arbitrary coefficient rings.

A similar fact holds for finite-dimensional Lie algebras $\mathfrak{g}$ over a field of characteristic [math]: in this case, $U(\mathfrak{g})$ is polycentral if and only if $\mathfrak{g}$ is nilpotent [10, Theorem A.3.3].

The above facts about group rings of polycyclic groups were crucially exploited in [19, Corollary 11] to understand their full prime ideal structure. We are currently unable to prove an analogue of these facts for Iwasawa algebras, but we can immediately mimic a small part of the argument in our case:

Proposition 5.1**.**

Suppose that the $p$-valuable group $G$ contains a nontrivial abelian normal subgroup $A$, and that $\mathbb{Z}_{p}G$ is polycentral. Then $G$ is nilpotent.

Proof.

We will show that such a $p$-valuable group must have a nontrivial centre $Z$. Then $\mathbb{Z}_{p}[[G/Z]]$ will remain polycentral, as it is a quotient of $\mathbb{Z}_{p}G$; and as $Z$ is known to be isolated in $G$ [3, Lemma 8.4(a)], it will follow that $G/Z$ is still $p$-valuable [13, IV, 3.4.2] of lower rank, so the statement follows by induction on $\mathrm{rank}(G)$.

Let $\mathfrak{a}=\ker(\mathbb{Z}_{p}G\to\mathbb{Z}_{p}[[G/A]])$ and $\mathfrak{g}=\ker(\mathbb{Z}_{p}G\to\mathbb{Z}_{p})$. Consider the function $\theta:A\to\mathfrak{a}/\mathfrak{ga}$ given by $\theta(a)=(1-a)+\mathfrak{ga}$: then a very similar proof to that in [7, §2.2, Lemma 1] shows that $\theta$ is a $\mathbb{Z}_{p}$-module isomorphism. (Recall that, if $A$ is written multiplicatively, it is a $\mathbb{Z}_{p}$-module under exponentiation, so that the $\mathbb{Z}_{p}$-linearity conditions read as follows: for $a,b\in A$ and $\lambda\in\mathbb{Z}_{p}$, we have $\theta(ab)=\theta(a)+\theta(b)$ and $\theta(a^{\lambda})=\lambda\theta(a)$.)

The proof now proceeds as in [18, §2.1(c)]. As $\mathbb{Z}_{p}G$ is polycentral, there exists an element $\eta\in\mathfrak{a}\setminus\mathfrak{ga}$ such that $\eta+\mathfrak{ga}$ is central in $\mathbb{Z}_{p}G/\mathfrak{ga}$. Choose the element $a\in A$ such that $\theta(a)=\eta+\mathfrak{ga}$: then, for all $g\in G$, we have

[TABLE]

from which we may conclude that $gag^{-1}=a$. ∎

Proof of Theorem E. This follows from Proposition 5.1, as soluble groups have nontrivial abelian normal subgroups.∎

The hypothesis that $G$ contain a nontrivial abelian normal subgroup, appearing in Proposition 5.1, does not appear to be necessary in the classical cases of discrete groups or finite-dimensional Lie algebras in characteristic 0. Encouraged by the partial evidence provided by Proposition 5.1 and the results of [3], we conjecture:

Conjecture**.**

If $G$ is $p$-valuable, then $kG$ is polycentral if and only if $G$ is nilpotent.