Neat locally compact abelian p-groups

Abstract

Inspired by a result of Kulikov about splitting of pure subgroups in discrete abelian p-groups we classify all locally compact abelian p-groups with no elements of infinite p-height in which every maximal monothetic subgroup splits. Descriptions of separable compact groups with elements of finite p-height, in which every pure monothetic subgroup splits, are obtained.

1 Introduction

Let p be a prime. By a locally compact abelian p-group we mean a locally compact abelian totally disconnected group G such that every element in G topologically generates either a finite cyclic p-group ${\mathbb {Z}}(p^n)$ for some natural number n, or else, a closed subgroup algebraically and topologically isomorphic to the additive group ${\mathbb {Z}}_p$ of p-adic integers. If the topology of a locally compact abelian p-group is discrete, it is a discrete abelian p-group. A compact p-group G is a pro-p group, i.e., it is the projective limit of an inverse system of finite p-groups. The p-height of an element g in a locally compact abelian p-group G is the maximal nonnegative integer k such that $g=p^kx$ for some $x\in G$ if it exists and $\infty$ else (see [1, p.4]). A locally compact abelian p-group G has p-rank r, and we write $\textrm{rank}_p(G)=r$, provided every closed subgroup H can be generated topologically by at most r elements from H and, in addition, there is a closed subgroup L which cannot be topologically generated by less than r elements. A locally compact abelian group G will be termed finitely generated provided there is a finite set F such that $G=\overline{\langle F\rangle }$. A locally compact abelian torsion group is bounded if it has finite exponent. With $\textrm{tor}(G)$ we denote the set of elements of finite order of G (see Definition 2.9 below). Our standard references are [2, 5, 10].

We shall start from the following well-known facts, originally published by L. Kulikov in [7]:

Theorem 1.1

(Corollary 27.2, Theorem 27.5 [1]) Let G be an abelian discrete p-group.

• (a) Every element of order p and finite p-height is contained in a cyclic direct summand of G.

• (b) Every bounded pure subgroup is a direct summand of G.

Example 5.2 below shows that neither statement (a) or (b) – even for groups of finite exponent – is necessarily true for locally compact abelian p-groups. A closed subgroup M of G topologically generated by a single element $g\in G$ is called monothetic. Thus in a discrete p-group a subgroup is monothetic iff it is a finite cyclic p-group. If G is not discrete then $M\le G$ is monothetic iff it is either a finite cyclic p-group or $M=\overline{\langle g\rangle }\cong {\mathbb {Z}}_p$, as has been said above. We shall say that a closed subgroup H of G splits, provided there is a closed subgroup K of G such that $G=H\oplus K$ is topologically and algebraically the direct sum of H and K (see Notation 2.14 below). Then K is called a complement of H in G.

We shall obtain an analogue of Theorem 1.1(a) for compact p-groups and defer its proof until the end of Sect. 2.2:

Theorem 1.2

Let G be a compact p-group and x have order p. Then there is a maximal monothetic subgroup M containing x which splits in G.

Recall that a locally compact abelian group G is first countable provided there is a countable basis of open neighborhoods of the identity in G, i.e., it satisfies the first axiom of countability. For compact p-groups which are generated by their torsion elements we shall prove an analog of Theorem 1.1(b):

Theorem 1.3

In a first countable abelian compact p-group G that is topologically generated by its torsion elements, every pure monothetic subgroup splits if and only if G has finite exponent.

See Proposition 4.7 below. We do not know whether this result is true without the restriction of G being first countable.

If G is an arbitrary locally compact abelian p-group then, as one would expect, none of the statements in Theorem 1.1 needs to be true: In Example 5.2, by choosing $n:=2$, one can exhibit a locally compact abelian p-group G having exponent $p^2$ possessing a subgroup $M^*$ of order p, pure, and maximal monothetic, which does not split.

We are concerned with the following questions:

Question 1.4

• Q.1 For which locally compact abelian p-groups is every maximal monothetic subgroup a direct summand?

• Q.2 For which locally compact abelian p-groups is every pure monothetic subgroup a direct summand?

When G does not possess maximal monothetic subgroups, e.g., if G is divisible both questions can be answered affirmatively in a trivial manner. Denote by $\textrm{Mon}(G)$ the set of all monothetic subgroups of G and by ${\mathbb {M}}(G)$ the set of all maximal monothetic subgroups of G.

Definition 1.5

A locally compact p-group G is said to be neat if $\textrm{Mon}(G)$ is inductive and every $M\in {\mathbb {M}}(G)$ splits. It is said to be weakly neat if $\textrm{Mon}(G)$ is inductive and every pure monothetic subgroup splits.

Equivalent conditions for a locally compact abelian group G to be neat will be presented below in Proposition 2.5. Theorem 1.7 below provides a complete answer to Q.1 for neat groups.

Notation 1.6

For a natural number $e\ge 1$ we call a locally compact abelian p-group G $p^{e}$-homocyclic if each of its elements is contained in a cyclic subgroup of G of order $p^e$.

Theorem 1.7

Let G be a locally compact abelian p-group. The following statements are equivalent:

• (i) G is neat.

• (ii) G satisfies exclusively one of the following statements:

  • (A) G is torsion-free and any of the subsequent equivalent conditions holds:

    • (A.1) For every $k\ge 0$, the subgroup $p^kG=\{p^kg:g\in G\}$ is closed.

    • (A.2) There is a compact open subgroup C of G and a compact torsion-free p-group $\Gamma$ and continuous embedding $j:G\rightarrow \Gamma$ such that $j(G)/j(C)=\textrm{tor}(\Gamma /j(C))$. Moreover, redefining the topology on $\Gamma$ such that j(C) is compact open, turns j into an algebraic and topological isomorphism from G onto j(G).

  • (B) G is a bounded group of exponent $p^e$ for some $e\ge 1$ and is of the form $G=A\oplus B$ for A a $p^{e}$-homocyclic and B a $p^{e-1}$-homocyclic group.

Recall that a compact abelian p-group $\Gamma$ is torsion-free if and only if there is a set X such that $\Gamma$ is topologically and algebraically isomorphic to ${\mathbb {Z}}_p^X$. See Proposition 2.15 below.

2 Preliminaries

2.1 Maximal and pure monothetic subgroups

The set $\textrm{Mon}(G)$ of monothetic subgroups of a locally compact abelian p-group G is a poset with respect to the containment relation “$\subseteq$” and ${\mathbb {M}}(G)$ denotes the subset of its maximal elements. For instance, for $G={\mathbb {Q}}_p$ the additive group of the field of p-adic numbers, ${\mathbb {M}}({\mathbb {Q}}_p) = \emptyset$. We have that ${\mathbb {M}}(G)=\{G\}$ if G itself is monothetic, i.e. is isomorphic to ${\mathbb {Z}}(p^n)$, $n\in {\mathbb {N}}_0$ or ${\mathbb {Z}}_p$. We write $H\le G$ when H is a subgroup of G.

If $M\in {\mathbb {M}}(G)$, then $p{\cdot }M$ fails to be maximal. If G is compact, then every $H\in \textrm{Mon}(G)$ is contained in some $M\in {\mathbb {M}}(G)$. If $G={\mathbb {Z}}(p)^{(I)}$ (with the discrete topology) or $G={\mathbb {Z}}(p)^I$ for some set I, then $\textrm{Mon}(G)={\mathbb {M}}(G)\cup \{0\}$.

Lemma 2.1

Let G be a locally compact abelian group and C any compact open subgroup. A subgroup $H\le G$ is closed if and only if $H\cap C$ is closed.

Proof

This is an immediate consequence of [4, (5.9) Theorem]. $\square$

Proposition 2.2

Let G be a locally compact abelian p-group. For $M\in \textrm{Mon}(G)$ the following statements are equivalent:

• (1) M is not contained in some $M^*\in {\mathbb {M}}(G)$.

• (2) There is a subgroup H of G containing M and $H\cong {\mathbb {Q}}_p$ or $H\cong {\mathbb {Z}}(p^\infty )$, is the Prüfer’s p-group.

Proof

The implication (2) $\Rightarrow$ (1) is clear. So we assume (1). Then by induction we find a properly ascending sequence of $(M_n)_{n\in {\mathbb {N}}_0}$ in $\textrm{Mon}(G)$ with $M_0=M$. We set $H=\bigcup _{n\in {\mathbb {N}}_0}M_n$ and claim that H is closed. Given $x\in H$ and $n\in {\mathbb {N}}$ there is $m\in {\mathbb {N}}$ with $x\in M_m$ and if k is chosen such that the index $[M_k:M_m]\ge p^k$ then $M_k\ge M_m$ implies the existence of $y\in M_k$ with $p^ky=x$. Since $x\in H$ and $k\in {\mathbb {N}}$ have been chosen arbitrarily it shows that H is divisible. For proving that H is closed, fix any compact open subgroup K of G. The compact p-group K is reduced. Therefore there exists $k\in {\mathbb {N}}$ with $M_k\le K$ and $M_{k+1}\not \le K$. Set $N:=M_{k+1}\cap K$ and note that $M_k\le N\le M_{k+1}$. Therefore, for every $l>k$, the intersection $M_l\cap K=N$ and, as for $l\le k$ one has $M_l\cap K=M_l\le N$ one obtains

$$\begin{aligned} H\cap K=\left( \bigcup _{l\ge 1}M_l\right) \cap K =\bigcup _{l\ge 1}M_l\cap K=N. \end{aligned}$$

Since K is a compact open subgroup and $N=H\cap K$ is closed, it follows from Lemma 2.1 that H is closed. Next observe that H has $p$-rank one, since every finite subset of H is contained in some $M_k$ which is monothetic. Therefore [2, Theorem 3.97] implies that H is topologically and algebraically isomorphic either to ${\mathbb {Q}}_p$ or ${\mathbb {Z}}(p^\infty )$. By construction M is contained in H and thus (2) holds. $\square$

Corollary 2.3

Let G be a locally compact abelian p-group. The following conditions are equivalent:

• (1) The poset $(\textrm{Mon}(G),\le )$ is inductive.

• (2) Every increasing sequence in $\textrm{Mon}(G)$ is eventually constant.

• (3) Every $M\in \textrm{Mon}(G)$ is contained in some $M^*\in {\mathbb {M}}(G)$.

• (4) No closed subgroup of G is isomorphic to ${\mathbb {Z}}(p^\infty )$ or ${\mathbb {Q}}_p$.

• (5) G is reduced.

• (6) If, in addition, G is torsion-free, then conditions (1)–(4) are equivalent to:

  $$\begin{aligned} \text {Every nontrivial element}\ g\in G \ \text {has finite}\ p\text {-height}. \end{aligned}$$

Proof

(1)–(5) are immediate consequences of Proposition 2.2. We turn to proving the equivalence of property (5).

(1) $\Rightarrow$(5). Suppose first that (1) holds and $0\ne g$ has infinite p-height. Then there is a strictly increasing sequence $(n_k)_{k\ge 1}$ of positive integers and a sequence $(g_k)_{k\ge 1}$ of elements in G with $p^{n_k}g_k=g$ for all $k\ge 1$. Therefore $p^{n_{k+1}}g_{k+1}= p^{n_k}g_k$ holds for all $k\ge 1$, and, as G is assumed to be torsion-free, we may conclude that $g_k=p^{n_{k+1}-n_k}g_{k+1}$. Setting $M_k:=\overline{\langle g_k\rangle }$, it follows that $(M_k)_{k\ge 1}$ is an infinite properly ascending chain of monothetic subgroups, by (2), contrary to the assumption $\textrm{Mon}(G)$ being inductive.

(5) $\Rightarrow$(1). By way of contradiction, assume (1) to be false. Then there is a properly ascending sequence $(M_k)_{k\ge 1}$ of monothetic subgroups of G. Let $M_1=\overline{\langle g_1\rangle }$ and note that for every $k\ge 1$ there is $g_k\in M_k$ which generates $M_k$ topologically and satisfies $p^{n_k}g_k=g_1$ for a suitable positive integer $n_k$. Since $|M_k|$ tends to $\infty$, as k tends to $\infty$ we may conclude that $n_k$ tends to $\infty$, showing that $g_1$ would have infinite p-height, a contradiction. $\square$

Remark 2.4

In every compact p-group G the ordered set $(\textrm{Mon}(G),\le )$ is inductive. Indeed otherwise by Corollary 2.3 it would contain a closed subgroup isomorphic to either ${\mathbb {Z}}(p^\infty )$ or ${\mathbb {Q}}_p$, but neither of these is compact.

Proposition 2.5

In a locally compact abelian p-group G the following conditions are equivalent:

• (1) G is neat.

• (2) For every finitely generated subgroup H of G there is a finitely generated direct summand P of G such that $H\le P$ and that $\textrm{rank}_p(H)=\textrm{rank}_p(P)$.

Proof

(2) trivially implies (1).

Now assume (1). We prove (2) by induction w.r.t. $n=\textrm{rank}_p(H)$. For $n=1$ the conclusion is immediate. Now let $n>1$. Write $H=H_1\oplus H_2\le G$ with $\textrm{rank}(H_1)=1$. By induction hypothesis $G=P_2\oplus G_2$ with $H_2\le P_2$ and $\textrm{rank}_p(H_2) =\textrm{rank}_p(P_2)$. Let e be the morphism projecting $H_1$ isomorphically onto $H_1^*\le G_2$ along $P_2$. By (1), $G_2$ contains a monothetic subgroup $P_1$ such that $G_2=P_1\oplus G_3$ and $H_1^*\le P_1$. Set $P:=P_2\oplus P_1$. Moreover, $H=H_1\oplus H_2= H_1^*\oplus H_2\le P$ and $G=P_2\oplus G_2= P_2\oplus P_1\oplus G_3$. Then $h_1\oplus h_2\mapsto e(h_1)+h_2:H\rightarrow H_1^*+H_2$ is an injection of H into $P= P_1\oplus P_2$ and $\textrm{rank}_p(H)=\textrm{rank}_p(H_1^*)+\textrm{rank}_p(H_2)=1+\textrm{rank}_p(P_2) =\textrm{rank}_p(P)$. This completes the induction. $\square$

Remark 2.6

Since every pure monothetic subgroup of a locally compact abelian p-group G is in ${\mathbb {M}}(G)$ (see Lemma 2.8 below), it follows that every neat group is weakly neat, and, because in a torsion-free locally compact abelian p-group G a monothetic subgroup is maximal iff it is pure (see again Lemma 2.8 below), such a group is neat iff it is weakly neat. On the other hand, the compact p-group G in Example 5.4 below is weakly neat but is not neat. It is also clear from Theorem 1.1(b) that every discrete reduced p-group is weakly neat.

Lemma 2.7

Let G be a locally compact abelian p-group and $M=\overline{\langle x\rangle }$ be a maximal monothetic subgroup of G. Then the following statements are equivalent:

• (a) M is pure;

• (b) Whenever for some $g\in G$ and $k\ge 0$ the element $p^kg\in M$ then there is $r\ge k$ and $\lambda \in {\mathbb {Z}}_p^\times$ with

  $$\begin{aligned} p^kg=\lambda p^rx. \end{aligned}$$

Proof

(a) $\Rightarrow$(b): Since M is pure, the relation $p^kg\in M$ implies for some $h\in M$ that $p^kg=p^kh$. Since $h\in M$ there are $s\ge 0$ and $\lambda \in {\mathbb {Z}}_p^\times$ such that

$$\begin{aligned} h=\lambda p^sx. \end{aligned}$$

Therefore (b) holds for $r:=k+s\ge k$.

(b) $\Rightarrow$(a): Suppose that $p^kg\in M$. Then $p^kg=\lambda p^rx$ for $r\ge k$ implies that $p^kg=p^kh$ for $h:=\lambda p^{r-k}x\in M$. Hence M is pure. $\square$

In the sequel for establishing purity of a monothetic subgroup $M=\overline{\langle x\rangle }$ we shall employ the following criterion:

Lemma 2.8

Let G be a locally compact abelian p-group. Then every pure monothetic subgroup is maximal in G. If, in addition, G is either torsion-free or is $p^{e}$-homocyclic, then every maximal monothetic subgroup is pure.

Proof

Let $H=\overline{\langle h\rangle }$ be a pure monothetic subgroup of G and suppose that H is not a maximal monothetic subgroup of G. Then there is $g\in G$ with $pg=h$. By the pureness of H there exists $h'\in H$ with $ph'=pg=h$. Since $h'\in H$ there are $k\ge 0$ and a unit $\lambda \in {\mathbb {Z}}_p^\times$ with $h'=p^k\lambda h$. The equality $h=p^{k+1}\lambda h$ follows, i.e., $(1-p^{k+1}\lambda )h=0$. As $1-p^{k+1}\lambda \in {\mathbb {Z}}_p^\times$ this leads to the contradiction $h=0$.

For proving the second statement assume first G to be torsion-free and let $M=\overline{\langle x\rangle }\cong {\mathbb {Z}}_p$ belong to ${\mathbb {M}}(G)$. Suppose that for some $k\ge 1$ and $g\in G$ one has $p^kg\in p^kG\cap M$. Then there are $\lambda \in {\mathbb {Z}}_p^\times$ and $r\ge 0$ such that

$$\begin{aligned} p^kg=\lambda p^r x. \end{aligned}$$

We cannot have $r<k$, else by the torsion-free ness we may conclude that

$$\begin{aligned} x=\lambda ^{-1}p^{k-r}g\in pG \end{aligned}$$

would imply that M is not maximal monothetic, a contradiction. Thus $r\ge k$ and hence by Lemma 2.7M is pure in G.

Dealing with the second assumption suppose that G is $p^{e}$-homocyclic. Let $M=\langle x\rangle$ be a maximal monothetic subgroup. For showing that it is pure, we need, for any $1\le k<e$ and $g\in G$ with $0\ne g\in p^kG\cap M$, to find $m\in M$ with $p^km=g$.

Since $g\in p^kG$ there exists $y\in G$ with $g=p^ky$. On the other hand, since $g\in M$, there are $\lambda \in {\mathbb {Z}}_p^\times$ and $r\ge 0$ such that $g=\lambda p^r x$. Replacing x by a suitable power we may assume $M=\langle x\rangle$ and $\lambda =1$. Thus we have equations

$$\begin{aligned} g=p^ky=\lambda p^rx. \end{aligned}$$

(2.1)

If $r\ge k$ then, letting $m:=\lambda p^{r-k}x$, implies that $g=p^km$.

Suppose that $r<k$. Then we obtain from Eq. (2.1) the relation

$$\begin{aligned} p^{k-r}(p^ry-x)=0. \end{aligned}$$

Hence there exists $u\in G$ of order at most $p^{k-r}$ with $x=p^ry+u$. Since G is $p^{e}$-homocyclic, there are $t\in G$ of order $p^e$ and $s\ge 0$ with $u=p^st$. Therefore $x=p^ry+p^st$ and, since x is a generator of $M\cong {\mathbb {Z}}(p^e)$ either $r=0$ or $s=0$. If $r=0$ then Eq. (2.1) implies $g=\lambda x=p^ky$, and, as $\lambda \in {\mathbb {Z}}_p^\times$, one obtains that g is a generator of M. Therefore $k=0$, contradicting the assumption $k>r$. Thus $r>0$. Suppose that $s=0$. Then $u=t$ has order $p^e$. However, we know that u has order at most $p^{k-r}$ and since $r>0$ and $k<e$, one obtains $0<k-r<e$, a final contradiction.

$\square$

In general, a maximal monothetic subgroup of a locally compact abelian p-group need not be pure, as Claim (b) in Example 5.4 will show.

We shall now aim for a parallel result that is valid, more generally, for locally compact abelian p-groups and which may serve as a reformulation of Theorem 3.34 in [2]. As a first step we generalize the concept of what is commonly known as the socle of an abelian p-group G.

Definition 2.9

Let G be a locally compact abelian p-group. For each $n\in {\mathbb {N}}$ we define

$$\begin{aligned} (1)\qquad \qquad S_n(G)=\{x\in G:p^n{\cdot }x=0\},\quad n\in {\mathbb {N}}\end{aligned}$$

and call $S_n(G)$ the n-socle of G. The 1-socle is called directly the socle. For $n=1$ we simply write $S_1(G)=S(G)$ for the socle of G.

Recall that a subgroup of a locally compact abelian group G is called fully characteristic if it is mapped into itself by all (continuous) endomorphisms of G. Clearly

$$\begin{aligned} S_1(G)\le S_2(G)\le \cdots \le S_n(G)\le \cdots \end{aligned}$$

The torsion subgroup of G is denoted by $\textrm{tor}(G)$. It is a fully characteristic subgroup of G.

Remark 2.10

Let us recall some facts:

Fact 1: For each $n\in {\mathbb {N}}$, the n-socle $S_n(G)$ is a closed fully characteristic subgroup of G.

Fact 2: For any locally compact abelian p-group G we have

$$\begin{aligned} \textrm{tor}(G)=\bigcup _{n\in {\mathbb {N}}}S_n(G). \end{aligned}$$

If $S_n(G)\ne \emptyset$ and $m\le n$ in ${\mathbb {N}}$, then $S_m(G)\ne \emptyset$.

Fact 3: For all $n\in {\mathbb {N}}$ the quotient group $S_{n+1}(G)/S_n(G)$ has exponent p, i.e. $p{\cdot }(S_{n+1}(G)/S_n(G))=\{0\}$ and if $g\in S_{n+1}(G){{\setminus }} S_n(G)$ then $p^{n+1}{\cdot }g=0$ and $p^n{\cdot }g\ne 0$. Any $M\in \textrm{Mon}(S_{n+1})$, $M\not \le S_n(G)$ is isomorphic to ${\mathbb {Z}}(p^{n+1})$ and is maximal in $S_{n+1}(G)$, i.e., is a member of ${\mathbb {M}}(S_{n+1}(G))$. If $g\in G{\setminus } \textrm{tor}(G)$, then $\overline{\langle g\rangle }\cong {\mathbb {Z}}_p$.

Definition 2.11

Let G be a locally compact abelian group. If $G=\overline{\textrm{tor}(G)}$ we say that G is torsion generated.

Lemma 2.12

Any continuous homomorphic image of a torsion generated locally compact abelian group G is torsion generated. A monothetic group $M\cong {\mathbb {Z}}_p$ cannot be a continuous homomorphic image of a torsion generated locally compact abelian group G.

We shall see that a maximal monothetic subgroup $M\cong {\mathbb {Z}}_p$ can occur as a subgroup of a torsion generated compact p-group. See Example 5.1 below.

Proposition 2.13

For a compact abelian torsion-generated p-group G the following statements are equivalent:

• (i) There exists $n\in {\mathbb {N}}$ such that $G=S_n(G)$.

• (ii) $G=\textrm{tor}(G)$.

• (iii) $\textrm{tor}(G)$ is closed.

• (iv) For any fixed $x\in G$ the map $z\mapsto z{\cdot }x: {\mathbb {Z}}_p\rightarrow {\mathbb {Z}}_p{\cdot }x$ is not injective.

• (v) If $M\in {\mathbb {M}}(G)$ then $M\not \cong {\mathbb {Z}}_p$.

• (vi) There are sets $J_1$, ..., $J_n$ such that $G \cong {\mathbb {Z}}(p)^{J_1} \times \cdots \times {\mathbb {Z}}(p^n)^{J_n}$.

Proof

(i) implies (ii): Clearly $S_n(G) \le \textrm{tor}(G)$, and so (i) implies (ii).

(ii) equivalent (iii): Trivial in view of $G=\overline{\textrm{tor}(G)}$.

(iii) implies (i): (see [2, Lemma 3.10] The proof is based on the Baire Category Theorem and thus involves topology of some nontrivial type, see [5, Corollary 8.9.(iii)].

We now know that (i), (ii) and (iii) are equivalent.

(ii) is equivalent to (iv): The nonsingleton monothetic subgroups of G are, up to isomorphism, ${\mathbb {Z}}(p^n)$, $n\in {\mathbb {N}}$, and ${\mathbb {Z}}_p$.

So (i) through (iv) are equivalent.

(iv) equivalent (v): We know that, due to the compactness of G, every monothetic subgroup M is contained in a maximal one, say $M^*$.

Thus M is cyclic iff $M^*$ is cyclic, and $M\cong {\mathbb {Z}}_p$ iff $M^*\cong {\mathbb {Z}}_p$.

(ii) equivalent (vi): See [2, Lemma 3.10]. $\square$

For locally compact abelian groups G and H we let $\textrm{Hom}(G,H)$ denote the set of all continuous homomorphisms from G to H.

Notation 2.14

We say that a subgroup H of a locally compact abelian group G splits iff it is a closed direct summand iff

$$\begin{aligned} (\exists \phi \in \textrm{Hom}(G,G))\, \phi ^2=\phi \hbox { and } \phi (G)=H \end{aligned}$$

iff for the inclusion $\iota :H\rightarrow G$ we have

$$\begin{aligned} (3)\qquad \qquad (\exists \pi \in \textrm{Hom}(G,H))= \pi \circ \iota =\textrm{id}_H. \end{aligned}$$

The subgroup $K:=(\textrm{id}_G-\phi )H$ is then a complement of H in G and $G=H\oplus K$ algebraically and topologically.

Proposition 2.15

For a compact abelian p-group G the following assertions are equivalent:

• (i) $\textrm{tor}(G)$ is closed.

• (ii) There are sets I, $J_1$,..., $J_n$ such that

  $$\begin{aligned} G \cong {\mathbb {Z}}(p)^{J_1} \times \cdots \times {\mathbb {Z}}(p^n)^{J_n}\times {\mathbb {Z}}_p^I. \end{aligned}$$

• (iii) There is an $n\in {\mathbb {N}}$ such that $S_n(G)=\textrm{tor}(G)$ splits.

Proof

For the equivalence of (i) and (ii) see [2, Proposition 3.14 and Corollary 3.15]. In view of Lemma 2.12, trivially (ii) implies (iii) and (iii) is a consequence of the fact that every compact torsion free p-group is algebraically and topologically isomorphic to ${\mathbb {Z}}_p^I$ for some set I. $\square$

In this description of the structure of G the direct product representation might be also written additively where “$\times$” is used here. We observe that in Proposition 2.15(ii), none of the direct factors is unique in any way. The Axiom of Choice is used amply in the background. By contrast, the closed subgroups $S_n(G)$ and $\textrm{tor}(G)=\bigcup _{n\ge 0}S_n(G)$ are unique to the extent of being fully characteristic. However, if one searches direct product or sum decompositions, these subgroups are in some sense “transversal” to direct product decomposition such as presented in Proposition 2.13(vi).

The class of compact p-groups G emerging in Proposition 2.15(iii) seems to be a class to which the Kulikov formalism (see Theorem 1.1) for discrete abelian p-groups may be applicable in a compact environment. See Theorem 1.2.

2.2 Elementary properties of neat locally compact abelian groups

We shall show that in a neat locally compact abelian p-group for every $k\ge 0$ the subgroup $p^kG$ is closed, see Corollary 2.23 below. In the torsion-free case this will give us part of Theorem 1.7(ii)(A.1). Another crucial fact is dealt with in Lemma 2.26, which characterizes subgroups $M\oplus N$ for M and N maximal monothetic in a neat locally compact abelian p-group. As a consequence it turns out that neat locally compact abelian p-groups must be either bounded or torsion-free. See Corollary 2.27 below.

Lemma 2.16

Let K be a compact abelian neat p-group. Then, for every set I, the cartesian product $G:=K^I$ is a neat compact p-group.

Proof

The compact p-group K has inductive $\textrm{Mon}(K)$, as follows from Remark 2.4 For proving neatness of G we fix an arbitrary maximal monothetic subgroup $M:=\overline{\langle (k_i)_{i\in I}\rangle }$ and need to find a complement R for M in $G=K^I=\prod _{i\in I}K_i$, where we denote $K_i$ the image under the i-th projection from G into K. Since $M\in {\mathbb {M}}(G)$ there exists $i_0\in I$ with $M_{i_0}:=\overline{\langle k_{i_0}\rangle }$ maximal in the $i_0$-component $K_{i_0}$. Since $K_{i_0}\cong K$ is neat, there is a closed subgroup $R_{i_0}\le K_{i_0}$ with

$$\begin{aligned} K_{i_0}=M_{i_0}\oplus R_{i_0}. \end{aligned}$$

Define $P_{i_0}:=\prod _{i\ne i_0}K_i$ and note that

$$\begin{aligned} K=P_{i_0}\oplus K_{i_0}\ \hbox { and } \ \ P_{i_0}+M_{i_0}=P_{i_0}+M. \end{aligned}$$

Therefore, letting $R:=P_{i_0}\times R_{i_0}$, we find

$$\begin{aligned} K=P_{i_0}+K_{i_0}=P_{i_0}+M_{i_0}+R_{i_0}=P_{i_0}+R_{i_0}+M=R+M. \end{aligned}$$

For showing that this sum of closed subgroups is direct, we need to prove $R\cap M=\{0\}$. Suppose that

$$\begin{aligned} x=(x_i)_{i\in I}\in R\cap M. \end{aligned}$$

Since $x\in R=P_{i_0}+R_{i_0}$ we find that $x_{i_0}\in R_{i_0}$ and since $x\in M$ deduce $x_{i_0}\in M_{i_0}$. Thus $x_{i_0}\in R_{i_0}\cap M_{i_0}=\{0\}$ shows that $x\in P_{i_0}\cap M=\{0\}$. $\square$

Corollary 2.17

Let I be a set. Each of the following compact abelian p-groups G is neat:

• (i) The cartesian product $G:=M^I$ for M a monothetic compact p-group.

• (ii) The cartesian product ${\mathbb {Z}}_p^I$.

• (iii) Every torsion-free compact p-group.

• (iv) The $p^{e}$-homocyclic compact p-group ${\mathbb {Z}}(p^e)^I$.

Proof

(i) The monothetic compact p-group M is neat. The result follows from Lemma 2.16. (ii) and (iv) are a consequence of (i), and (iii) holds since in every torsion-free compact abelian p-group the torsion subgroup is trivial and hence it is topologically and algebraically isomorphic to ${\mathbb {Z}}_p^I$ by Proposition 2.15. Now the result follows from (ii). $\square$

Lemma 2.18

Let $\Gamma$ be a compact abelian neat p-group and C a closed subgroup. Refine the topology on $\Gamma$ by declaring C open. Define G by means of the equation $G/C=\textrm{tor}(\Gamma /C)$. Then G is a neat locally compact abelian p-group.

Proof

Since for every $g\in G$, by definition, there is $k\ge 0$ such that $p^kg\in C$, the monothetic subgroup $M:=\overline{\langle g\rangle }$ has the compact open subgroup $M\cap C$ and as factor group a finite cyclic p-group. Hence M is a compact p-group. Furthermore, since C is a compact open neighborhood of 0 in G, it follows that G is a locally compact abelian p-group.

Claim: Every $M\in {\mathbb {M}}(\Gamma )$ (where $\Gamma$ is considered as a compact p-group) with $M\cap C\ne \{0\}$ belongs to ${\mathbb {M}}(G)$. Conversely, every $M\in {\mathbb {M}}(G)$ also belongs to ${\mathbb {M}}(\Gamma )$.

Indeed, if $M=\overline{\langle m\rangle }$ is maximal monothetic in $\Gamma$, and there is $k\ge 0$ with $p^km\in C$, then by construction $m\in G$. Suppose that $N\in {\mathbb {M}}(G)$ contains M. Then there is $l\ge 0$ with $p^lN=M$, and thus, for a suitable topological generator n of M one can arrange $p^ln=m$. Therefore $n\in G$ and hence $N=M$, i.e., $M\in {\mathbb {M}}(G)$.

Conversely, suppose that $M=\overline{\langle m\rangle }$ belongs to ${\mathbb {M}}(G)$ and there is $N\in {\mathbb {M}}(\Gamma )$ containing M. Then, for a generator n of N there is $k\ge 0$ with $p^kn=m$ and hence $n\in G$. Thus $N\le G$ and hence $M=N\in {\mathbb {M}}(G)$. The Claim holds.

The Claim implies that every $g\in G$ belongs to some $M\in {\mathbb {M}}(\Gamma )$ which in turn belongs to ${\mathbb {M}}(G)$. Hence ${\mathbb {M}}(G)$ is inductive.

We still need to show that every $M\in {\mathbb {M}}(G)$ has a complement in G. Since $\Gamma$ is by assumption a neat compact p-group, there is a closed subgroup $R\le \Gamma$ such that $\Gamma =M\oplus R$. Since R is closed in $\Gamma$, the subgroup $R\cap C$ is closed in C and hence it is a closed subgroup of G. Since C, after refining the topology of $\Gamma$, is open in $\Gamma$, we conclude that $R\cap C$ is open in R and hence, taking Lemma 2.1 into account, one finds that $R\cap G$ is a closed subgroup of G. Hence R is algebraically and topologically a complement of M in G. $\square$

Lemma 2.19

Let a (weakly) neat locally compact abelian p-group G decompose as $G=A\oplus B$ for closed subgroups A and B. Then A and B are both (weakly) neat.

Proof

Let M be a monothetic subgroup of A. If G is neat assume $M\in {\mathbb {M}}(A)$, else, if G is weakly neat assume M to be pure in A. In the first case $M\in {\mathbb {M}}(G)$ and in the second case M is pure in G since A is pure in G. In either case there is a closed subgroup $K\le G$ with

$$\begin{aligned} G=M\oplus K. \end{aligned}$$

Hence $K_A:=A\cap K$ is the desired complement of M in A. $\square$

On the other hand, the direct sum of neat groups is not necessarily neat, see Example 5.4.

Lemma 2.20

Let G be a locally compact abelian p-group and suppose that $\overline{pG}$ contains pG properly. Pick any $h\in \overline{pG}{{\setminus }} pG$ of minimal order if $\textrm{tor}(G)\cap (\overline{pG}{\setminus } pG) \ne \emptyset$. Then $M:=\overline{\langle h\rangle }$ enjoys all of the following properties:

• (a) M is a pure subgroup of G.

• (b) M is a maximal monothetic subgroup of G.

• (c) M cannot have a complement in G.

Proof

(a) We need to show for any fixed $k\ge 1$ that $p^kG\cap M=p^kM$, and since the right hand side is contained in the left hand side, it will be enough to show for any element $g\in G$ such that for some $r\ge 0$ and unit $\lambda \in {\mathbb {Z}}_p^\times$ with $p^kg=p^r\lambda h$ there is an element $h'\in M$ with $p^kg=p^kh'$. Making use of Lemma 2.7 we may assume $r<k$. Then there exists $u\in G$ with

$$\begin{aligned} p^{k-r}g=\lambda h+u \ \ \text{ and } \ \ p^ru=0. \end{aligned}$$

Since $h\notin pG$ and $k>r$ we must have $u\notin pG$. We discuss the cases (A) when $\textrm{tor}(G)\cap (\overline{pG}{{\setminus }} pG)=\emptyset$, and (B), when the latter intersection is not trivial.

(A) Since $p^ru=0$ and $u\in \textrm{tor}(G)\cap (\overline{pG}{{\setminus }} pG)=\emptyset$ we conclude $u=0$ and therefore $h=\lambda ^{-1}p^{k-r}g$, and as $k>r$ this is a contradiction to the assumption $h\not \in pG$.

(B) Since $h\in \textrm{tor}(G)\cap (\overline{pG}{\setminus } pG)$ has minimal possible order and $p^ru=0$ we conclude from the minimality assumption on orders of elements in $\overline{pG}{\setminus } pG$ that $p^rh=0$. But then $p^kg=0$, a contradiction.

(b) is a consequence of Lemma 2.8.

(c) Suppose that G contains a closed subgroup K such that

$$\begin{aligned} G=M\oplus K \end{aligned}$$

algebraically and topologically. By assumption there is a net $(h_\nu )$ of elements in G such that $ph_\nu$ converges to h. Then there are elements $k_\nu \in K$ and $\mu _\nu \in {\mathbb {Z}}_p$ such that

$$\begin{aligned} h_\nu =\mu _\nu h + k_\nu . \end{aligned}$$

Making use of the compactness of ${\mathbb {Z}}_p$ we can pass to a cofinal subnet – by abusing notation we use the same indices – such that $\mu _\nu$ tends to $\mu \in {\mathbb {Z}}_p$. Multiplying with p the above equation yields

$$\begin{aligned} ph_\nu =p\mu _\nu h+pk_\nu . \end{aligned}$$

Since, by assumption $ph_\nu$ tends to h also the net $(pk_\nu )$ turns out to converge (inside the closed subgroup K) and passing to the limit one finds $k\in K$ such that

$$\begin{aligned} h=p\mu h+k, \end{aligned}$$

leading to the equation $(1-p\mu )h=k\in M\cap K=\{0\}$. Hence $k=0$ and since $1-p\mu$ is a unit in ${\mathbb {Z}}_p^\times$ we conclude $h=0$, a contradiction. $\square$

Lemma 2.21

Let G be a locally compact abelian p-group and pG be closed. Suppose that M is a monothetic subgroup of G. If $N=pM$ is a pure subgroup of pG, then M is pure subgroup of G.

Proof

Suppose that for some $k\ge 1$ there is $x\in p^kG\cap M$. We need to find $h\in M$ such that $p^kg=p^kh$. Since $p^kg=p^{k-1}pg\in pG\cap M=N$ and N is pure, there exists $n\in N$ such that $p^{k-1}n=p^{k-1}pg$. Since $n\in N=pM$ there is $h\in M$ with $ph=n$. Hence $p^kh=p^kg$ shows that M is indeed a pure subgroup of G. $\square$

Lemma 2.22

Let G be a weakly neat locally compact abelian p-group. The following assertions hold:

• (i) pG is closed and weakly neat.

• (ii) If G is neat, then pG is neat.

Proof

The closedness of pG is a consequence of Lemma 2.20. If G is neat, suppose that N is a maximal monothetic subgroup of pG, and if G is merely weakly neat, suppose in addition that N is pure in pG. Then there is a maximal monothetic subgroup $M=\overline{\langle x\rangle }$ of G with $pM=N$.

If N is pure in pG, Lemma 2.21 implies the pureness of M. Thus $G=M\oplus R$ holds for some closed subgroup R of G. Therefore

$$\begin{aligned} pG=pM\oplus (pG\cap R)=N\oplus (pG\cap R) \end{aligned}$$

shows that N can be complemented. Hence pG inherits weakly neat ness or neatness from G. $\square$

Lemma 2.22 and an induction argument yield an immediate consequence:

Corollary 2.23

In a neat locally compact abelian p-group G, for every $k\ge 1$, the subgroup $p^kG$ is closed and itself neat. Moreover, for any fixed compact open subgroup C of G, setting $C_k:=p^kG\cap C$, the sequence $(C_k)_{k\ge 0}$ is descending and, for every $k\ge 0$, one has $C_k\ge C_{k+1}\ge pC_k$.

Proof

We only check the last statement:

$$\begin{aligned} C_k =p^kG\cap C\ge p^{k+1}G\cap C=C_{k+1}\ge p(p^kG\cap C)=pC_k. \end{aligned}$$

$\square$

Remark 2.24

Recall from [2, Corollary 3.26, p. 63] that every locally compact abelian p-group G of exponent p is algebraically and topologically the direct sum $G={\mathbb {Z}}(p)^{(J_1)}\oplus {\mathbb {Z}}(p)^{J_2}$ for sets $J_1$ and $J_2$.

Such groups are neat, and, even more is true:

Lemma 2.25

Let G be a locally compact abelian group with exponent a prime p and A any closed subgroup. Then there is a closed subgroup B such that

$$\begin{aligned} G=A\oplus B \end{aligned}$$

algebraically and topologically.

Proof

Fix any compact open subgroup $U\le G$ and note that $A+U$ is an open subgroup of G. Therefore

$$\begin{aligned} G/(A+U)\cong (G/U)/(A+U/U) \end{aligned}$$

is a discrete exponent p-group and thus there is a discrete subgroup $C\le G$ with $G=C\oplus (A+U)$. The compact group U and its closed subgroup $A\cap U$ via duality, see [5, Definition 7.12] for the Annihilator Mechanism, give rise to a decomposition

$$\begin{aligned} \widehat{U}=(A\cap U)^{\perp }\oplus D \end{aligned}$$

which in turn renders a closed subgroup $E\le U$ with

$$\begin{aligned} U=A\cap U\oplus E. \end{aligned}$$

Therefore $G=A\oplus B$ for $B:=C+E$. $\square$

Lemma 2.26

Let G be a neat locally compact abelian p-group and M and N maximal monothetic subgroups intersecting trivially. Then exclusively one of the following statements holds:

• (i) M and N are both torsion-free.

• (ii) M and N are both torsion and, for some $k\ge 1$, either $\langle M,N\rangle =M\oplus N\cong {\mathbb {Z}}(p^k)\oplus {\mathbb {Z}}(p^k)$ or $\langle M,N\rangle =M\oplus N\cong {\mathbb {Z}}(p^{k+1})\oplus {\mathbb {Z}}(p^{k})$.

Moreover, if $G=M\oplus N$ and either $M\cong N$ or $N\cong pM$, then G is neat.

Proof

Fix topological generators x and y of respectively M and N and set $L:=\overline{\langle M,N\rangle }$, which by our assumptions equals $M\oplus N$. Assuming that (i) does not hold we need to establish (ii). Therefore we may assume that $N=\langle y\rangle \cong {\mathbb {Z}}(p^k)$ for some $k\ge 1$ is finite and, in addition, $|M|\ge |N|$.

For proving the lemma, all that is needed is to establish that $p^{k+1}x=0$.

Set $h:=px+y$ and $H:=\overline{\langle h\rangle }$.

We first claim that H is a maximal monothetic subgroup of G. Else there is $u\in G$ such that $pu=h=px+y$ and hence $y=p(u-x)$ leads to the contradiction that N is not a maximal subgroup of G.

Since $H=\langle h\rangle =\langle px+y\rangle$, being a direct summand of L, certainly is a pure subgroup of L, and as $p^kh=p^{k+1}x\in p^{k+1}L\cap H$, there must be some $h'\in H$ with $p^{k+1}x=p^{k+1}h'$. Therefore there exists $\lambda \in {\mathbb {Z}}_p$ such that $h'=\lambda h$ and hence the latter equality turns into

$$\begin{aligned} p^{k+1}x=\lambda p^{k+2} x. \end{aligned}$$

Hence $p^{k+1}(1-p\lambda )x=0$, and since $1-p\lambda \in {\mathbb {Z}}_p^\times$, one arrives at the desired equation $p^{k+1}x=0$ and, as we have assumed $|M|\ge |N|$, indeed (ii) follows.

For proving the final part of the statement, we may apply Corollary 2.17(iv) to G if M and N are isomorphic.

If M is torsion-free and $N\cong pM$ then $M\cong N\cong {\mathbb {Z}}_p$ and the result follows from what we just have proved.

Thus assume that M is torsion. Then one may assume for some $k\ge 0$ that $M=\langle a\rangle \cong {\mathbb {Z}}(p^{k+1})$ and $N=\langle b\rangle \cong {\mathbb {Z}}(p^k)$. Let $L=\langle \lambda a +\mu b\rangle$ be a subgroup. Then L is maximal iff either $\lambda \in {\mathbb {Z}}_p^\times$ or $\lambda \in p{\mathbb {Z}}_p$ and $\mu \in {\mathbb {Z}}_p^\times$. Therefore, without loss of generality, we may discuss the cases (A), when $\lambda =1$ and (B) when p divides $\lambda$ and $\mu =1$.

(A) Certainly $G=M\oplus N=L+N$. Fix $x\in L\cap N$. Then there are $\eta$ and $\rho$ in ${\mathbb {Z}}$, and equations

$$\begin{aligned} x= \rho (a+\mu b)=\eta b, \end{aligned}$$

from which $\rho a=0$ follows. Since b has smaller order than a, one also finds $\rho b=0$ and hence $\eta b=0$, showing $x=0$. Thus $L\cap N=\{0\}$, as was needed to show.

(B) Certainly $G=M+L$. Suppose that $x\in M\cap L$. Then, for some $\xi \in {\mathbb {Z}}$ there is a relation of the form

$$\begin{aligned} \xi a=\rho (\lambda a+b), \end{aligned}$$

so that $\rho b=0$ follows. Since p divides $\lambda$ and a has order $p^{k+1}$ deduce from this that $\rho \lambda a=0$. Therefore $\xi a=0$ and hence $M\cap L=\{0\}$.

In both cases the maximal monothetic subgroup L is a direct summand of G, showing that G is neat. $\square$

Corollary 2.27

Let G be a neat locally compact abelian p-group. Then either G is torsion-free or is bounded.

Proof

Suppose that $\textrm{tor}(G)\ne \{0\}$ and M is a finite maximal monothetic subgroup of G. Then, since by assumption G is neat, there exists a closed subgroup $K\le G$ with

$$\begin{aligned} G=M\oplus K \end{aligned}$$

and we note that by Lemma 2.19 the subgroup K is neat. Pick any $k\in K$. Then k must belong to some maximal monothetic subgroup N of K. Applying Lemma 2.26 to M and N shows that $|N|\le p|M|$ and hence K is bounded. Therefore G has finite exponent. $\square$

Remark 2.28

Example 5.4 below shows that ${\mathbb {Z}}_p\oplus {\mathbb {Z}}(p)$ is weakly neat. Thus there are no analogues of Lemma 2.26 and Corollary 2.27 for weakly neat locally compact abelian p-groups.

We end this subsection by proving our analogue of Kulikov’s Theorem 1.1(a) for compact p-groups:

Proof of Theorem 1.2

Since G is compact every element in G has finite p-height. Because G is a p-group it is profinite and since x has order p, there is an open subgroup N of G intersecting $\langle x\rangle$ trivially. Then G/N is finite and we may apply Theorem 1.1(a) to $(\langle x\rangle +N)/N$ in order to exhibit a maximal monothetic subgroup $H_N/N$ of G/N containing $(x+N)/N$ and an open subgroup $K_N$ of G containing N such that

$$\begin{aligned} G/N=H_N/N\oplus K_N/N. \end{aligned}$$

Then

$$\begin{aligned} G=H_N+K_N \ \ \hbox {and} \ \ H_N\cap K_N=N. \end{aligned}$$

(2.2)

Let ${\mathcal {U}}$ be a neighborhood basis of open subgroups N and select for every $N\in {\mathcal {U}}$ open subgroups $H_N$ and $K_N$ satisfying Eq. (2.2) and $\langle x\rangle \cap N=\{0\}$. The set ${\mathcal {U}}$ is directed and by the compactness of the space of all closed subgroups of G with respect to the Vietoris topology (which agrees with the Chabauty topology, see [2, Definition 1.16]) we may pass to a cofinal subnet $(H_N,K_N)_N$ converging to a pair of closed subgroups (H, K). Abusing language we let this cofinal subset be denoted by ${\mathcal {U}}$ as well. Let us recall that convergence in a profinite group with respect to the Vietoris topology means that for every $U\in {\mathcal {U}}$ there exists $N_U\in {\mathcal {U}}$ such that for every $N\in {\mathcal {U}}$ with $N\le N_U$ the relations

$$\begin{aligned} H_N+U=H+U \ \ \hbox {and} \ \ K_N+U=K+U \end{aligned}$$

hold. Then certainly $x\in H_N$ implies $x\in H+U$ for all $U\in {\mathcal {U}}$ so that $x\in H$ results. Next observe that $H_N/N$ is finite cyclic, and so is, for any given $U\in {\mathcal {U}}$, the factor group $(H_N+U)/U=(H+U)/U$ for all $N\le N_0$, and a suitable $N_0\in {\mathcal {U}}$. Therefore $(H+U)/U$ is finite cyclic for every $U\in {\mathcal {U}}$ and hence the group H is procyclic, i.e., it is monothetic. Note that $x\in H$ implies the finiteness of H. Next observe that $H_N+K_N=G$ for every $N\in {\mathcal {U}}$. Therefore every fixed $g\in G$ can be decomposed as

$$\begin{aligned} g=h_N+k_N, \ \ h_N\in H_N, \ \ k_N\in K_N. \end{aligned}$$

Fix $U\in {\mathcal {U}}$ and observe $H_N+U=H+U$ and $K_N+U=K+U$ so that $g=h'_U+k'_U+u_U$ for elements $h'_U\in H$, $k'_U\in K$, and $u_U\in U$. Passing to a convergent subnet of $(h'_U,k'_U,u_U)_{U\in {\mathcal {U}}}$ and observing that $u_U\rightarrow 0$, one establishes the existence of $h\in H$ and $k\in K$ with $g=h+k$. Thus

$$\begin{aligned} G=H+K. \end{aligned}$$

Finally consider $y\in H\cap K$. Then $y\in (H+N)\cap (K+N)=N$ for every $N\in {\mathcal {U}}$ and, since $\bigcap {\mathcal {U}}=\{0\}$ deduce from this that $y=0$. Hence $H\cap K=\{0\}$ and thus

$$\begin{aligned} G=H\oplus K \end{aligned}$$

holds. $\square$

2.3 $p^e$-homocyclic groups

For a topological group G we let $G_d$ denote a copy of G equipped with the discrete topology. A locally compact abelian p-group is $p^{e}$-homocyclic iff every element of G belongs to a maximal monothetic subgroup, which is cyclic of order $p^e$ (see Notation 1.6). We shall obtain a structure result for $p^{e}$-homocyclic groups, Proposition 2.38, and prove that $p^{e}$-homocyclic groups are neat, see Lemma 2.40 below.

Lemma 2.29

Let G be a locally compact abelian p-group having finite exponent $p^e$. Then G is $p^{e}$-homocyclic iff $G_d$ is $p^{e}$-homocyclic.

Proof

Suppose first that G is $p^{e}$-homocyclic and let $G_d$ be a copy of G, equipped with the discrete topology. If $g\in G_d$ then, as G is $p^{e}$-homocyclic, there is a maximal monothetic subgroup M of G containing g and of order $p^e$. Since M is closed in $G_d$ it follows that $G_d$ is $p^{e}$-homocyclic.

Conversely, suppose that $G_d$ is $p^{e}$-homocyclic. Pick $g\in G$ and let M be a maximal monothetic subgroup of $G_d$ containing g. Then, as $G_d$ is $p^{e}$-homocyclic, the subgroup M has order $p^e$ and hence is finite. Therefore M is closed in G and since G has exponent $p^e$, it follows that $M\in {\mathbb {M}}(G)$. Hence G is $p^{e}$-homocyclic. $\square$

Unfortunately the Pontryagin dual (see [5, Theorem 1.37] for relevant facts from duality theory) of a locally compact abelian $p^{e}$-homocyclic group need not be $p^{e}$-homocyclic:

Example 2.30

Let $G:=\prod _{i\in {\mathbb {N}}}Z_i$, for $Z_i=\langle z_i\rangle \cong {\mathbb {Z}}(p^2)$. Since every maximal cyclic subgroup of G is isomorphic to ${\mathbb {Z}}(p^2)$, the group G is $p^{e}$-homocyclic. Thus G is algebraically isomorphic to the cartesian product ${\mathbb {Z}}(p^2)^{{\mathbb {N}}}$. Define $C:=pG$ and declare C compact open in G. There is a short-exact sequence

$$\begin{aligned} \{0\}\rightarrow C=\prod _{i\ge 1}\langle pz_i\rangle \rightarrow G\rightarrow \prod _{i\ge 1} \langle z_i+C/C\rangle \cong ({\mathbb {Z}}(p)^{\mathbb {N}})_d\rightarrow \{0\}. \end{aligned}$$

Dualization yields the short exact sequence

$$\begin{aligned} \{0\}\rightarrow A\rightarrow \widehat{G}\rightarrow B\rightarrow \{0\}, \end{aligned}$$

where A and B are respectively topologically and algebraically isomorphic to the dual of $({\mathbb {Z}}(p)^{\mathbb {N}})_d$ and ${\mathbb {Z}}(p)^{\mathbb {N}}$. Hence B is discrete and isomorphic to ${\mathbb {Z}}(p)^{({\mathbb {N}})}$, which is countable. On the other hand, A is compact and infinite, hence is uncountable. If every element in A were a p-th power of an element in $\widehat{G}$, then the cardinality of $\widehat{G}/A\cong B$ would be at least that of the continuum, contradicting the countability of B. Thus $\widehat{G}$ cannot be $p^2$-homocyclic.

Lemma 2.31

Let $G=A\oplus B$ topologically and algebraically the direct sum of locally compact abelian p-groups. Then G is $p^{e}$-homocyclic iff A and B are $p^{e}$-homocyclic.

Proof

Suppose first that G is $p^e$-homocylic. It will suffice to show that A is $p^{e}$-homocyclic. Pick $0\ne a\in A$. Then, G being $p^{e}$-homocyclic, there is $g\in G$ of order $p^e$ and $e>l\ge 1$ with $a=p^lg$. Hence a has order $p^{e-l}$. Decomposing $g=g_A+g_B$ with $g_A\in A$ and $g_B\in B$ it follows that $p^lg_A=a$ and $p^lg_B=0$. Since $1\le l<e$ we have that $e-l-1\ge 0$ and hence

$$\begin{aligned} p^{e-1}g_A=p^{e-l-1}p^lg=p^{e-l-1}a\ne 0, \end{aligned}$$

since a has order $p^{e-l}$. Thus $g_A$ has order $p^e$ and $a\in \langle g_A\rangle$. Hence A is $p^{e}$-homocyclic.

Conversely, suppose that A and B are both $p^{e}$-homocyclic. Pick $x\in G$ and decompose

$$\begin{aligned} x=x_A+x_B, \ \ x_A\in A, \ \ x_B\in B. \end{aligned}$$

Since A and B are both $p^{e}$-homocyclic there are $g_A\in A$ and $g_B\in B$ of order $p^e$ and non-negative integers $l_A$ and $l_B$ less than e such that

$$\begin{aligned} x=p^{l_A}g_A+p^{l_B}g_B. \end{aligned}$$

Let $l:=\min \{l_A,l_B\}$ and $g=p^{l_A - l} g_A - p^{l_B - l} g_B$. Then g has order $p^e$ and $p^lg=x$. Thus G is $p^{e}$-homocyclic. $\square$

Lemma 2.32

A discrete p-group G is $p^{e}$-homocyclic if, and only if, $G\cong {\mathbb {Z}}(p^e)^{(I)}$ for some set I.

Proof

Let G be discrete $p^{e}$-homocyclic. Then G is bounded and hence, by [1, Theorem 17.2], it is a direct sum of cyclic p-groups $C_i$ for i in some set I. Thus G is $p^{e}$-homocyclic only if, for every $i\in I$, the finite cyclic group $C_i$ is $p^{e}$-homocyclic, i.e. $C_i\cong {\mathbb {Z}}(p^e)$. Hence $G\cong {\mathbb {Z}}(p^e)^{(I)}$.

The converse, namely that $G={\mathbb {Z}}(p^e)^{(I)}$ is $p^{e}$-homocyclic follows by induction on e: If $e=1$, then every nontrivial element has order p and generates a maximal cyclic subgroup. Assume $e>1$ and pick $0\ne g\in G$. If g has order $p^e$, then $M=\langle g\rangle$ serves the purpose. Otherwise $g\in pG\cong {\mathbb {Z}}(p^{e-1})^{(I)}$. By the induction assumption there is $x\in pG$ of order $p^e$ with $g\in \langle x\rangle$. Since $x=pt$ for some $t\in G$, we obtain for $M=\langle t\rangle$ the desired conclusion $g\in M$. $\square$

Lemma 2.33

Let G be a locally compact abelian $p^{e}$-homocyclic group. Then, for all $1\le k< e$, the subgroup $p^kG$ is closed.

Proof

If we can show that pG is closed, the Lemma follows by induction (assuming we accept that $p^kG$ is homocyclic if G is).

Since G is $p^{e}$-homocyclic, for each element $0\ne x\in G$ there is an $\ell$, $0 \le \ell < e$ and an element g of order $p^e$ with

$$\begin{aligned} x = p^\ell g. \end{aligned}$$

(2.3)

Assume now that $x \in \overline{pG}$; in order to prove $x \in pG$ we must show that $\ell \ge 1$. Now there is a net of elements $g_\nu \in G$ such that $x = \lim _\nu pg_\nu$. Then

$$\begin{aligned} p^{e-1}x=\lim _\nu p^ex_\nu =0. \end{aligned}$$

(2.4)

From Eq. (2.3) and Eq. (2.4) we conclude that $\ell + e - 1\ge e$, that is $\ell \ge 1$ which we had to show.

We still need to show that pG is $p^{e-1}$-homocyclic. Fix $x\in pG$. Since G is $p^{e}$-homocyclic there is $M\in {\mathbb {M}}(G)$ containing x. Thus, for $M=\langle g\rangle$, there is $l\ge 0$ with $x=p^l g$. Since $x\in pG$ we cannot have $l=0$ and thus $x\in N$ for $N:=\langle pg\rangle$. Hence pG is $p^{e-1}$-homocyclic. $\square$

From Proposition 2.13 we know that a compact open subgroup K of a torsion group G of exponent at most $p^e$ is isomorphic to $\prod _{j=0}^eK_j$ with $K_j\cong {\mathbb {Z}}(p^j)^{I_j}$ for suitable sets $I_j$. As a consequence we shall obtain a decomposition of a $p^{e}$-homocyclic group G as follows.

Lemma 2.34

Let G be a locally compact abelian $p^{e}$-homocyclic p-group and K a compact open subgroup. Suppose that $K=\bigoplus _{j=1}^eK_j$ where for each j the subgroup $K_j$ is isomorphic to ${\mathbb {Z}}(p^j)^{I_j}$ for some set $I_j$. Then there are $p^{e}$-homocyclic closed subgroups $H_j$ such that $G=\bigoplus _{j=0}^eH_j$ and, for all $j\ge 1$ one has that $K_j\le H_j$.

Proof

Let $H_e:=K_e$. The socle of G has a decomposition $S(G)=K_0\oplus S(K)$ for some discrete subgroup $K_0$ of G. Fix $j<e$. The compact group $K_j$ is $p^{j}$-homocyclic and therefore, according to Lemma 2.29, it is $p^{j}$-homocyclic, even when equipped with the discrete topology. Then, by Lemma 2.32 there is an index set $I_j$ and elements $k_{i_j}\in K_j$ for every $i_j\in I_j$ such that

$$\begin{aligned} K_j=\bigoplus _{i_j\in I_j}\langle k_{i_j}\rangle , \ \ \langle k_{i_j}\rangle \cong {\mathbb {Z}}(p^j). \end{aligned}$$

Since G is $p^{e}$-homocyclic, there is, for every $i_j\in I_j$, an element $x_{i_j}\in G_e$ of order $p^e$ with $k_{i_j}=p^{e-j}x_{i_j}$, if $j\ge 1$, and $k_{i_0}=p^{e-1}x_{i_0}$ for every $i_0\in I_0$. Let $H_j:=\langle x_{i_j}:i_j\in I_j\rangle$ and note that

$$\begin{aligned} H_j=&\bigoplus _{i_j\in I_j}\langle x_{i_j}\rangle \cong {\mathbb {Z}}(p^e)^{(I_j)},\\ K_j=&p^{e-j}H_j, \ \hbox {for} \ j\ge 1,\\ K_0=&p^{e-1}H_0. \end{aligned}$$

Furthermore, by construction, for all $0\le j\le e$, the subgroup $K_j=K\cap H_j$ is compact open in $H_j$. Observing that ${ S}(H_j)={ S}(K_j)$, one infers for $j\ne j'$ from

$$\begin{aligned} { S}(H_j\cap H_{j'})\le { S}(H_j)\cap { S}(H_{j'}) ={ S}(K_j)\cap { S}(K_{j'})\le K_j\cap K_{j'}=\{0\} \end{aligned}$$

that $H_j\cap H_{j'}=\{0\}$. Therefore $H:=\langle H_j:0\le j\le e\rangle =\bigoplus _{j=0}^eH_j$ and since $K=\bigoplus _{j=0}^eK_j$ is a compact open subgroup of G the direct sum H is an open subgroup of G.

We claim that $G=H$ and shall show by induction on l that $S_l(G)=S_l(H)$. Certainly, by construction, $S_1(G)=S(G)=S(H)=S_1(H)$. Assume for some $1\le l<e$ the equality $S_l(G)=S_l(H)$ to hold. Pick $g\in S_{l+1}(G)$. By the induction assumption we have that $pg\in S_l(H)$. Therefore there are elements $h_j\in H_j$ such that

$$\begin{aligned} pg=\sum _{j=0}^eh_j. \end{aligned}$$

Since $pg\in pG$ and H is the direct sum of the $p^{e}$-homocyclic subgroups $H_j$ we may conclude that $h_j=ph_j'$ for some elements $h_j'\in H_j$. Therefore

$$\begin{aligned} g-\sum _{j=0}^eh_j'\in S(G). \end{aligned}$$

Since $S(G)\le H$ we deduce from this that $g\in H$. Hence $G=H$. $\square$

Lemma 2.35

Let $e\ge 1$ be a natural number and $G={\mathbb {Z}}(p^e)^I$, equipped with the product topology. Select a closed subgroup K and change the topology on G by declaring K compact open. Then G is a locally compact abelian $p^{e}$-homocyclic group.

Proof

The group G, with the product topology, is certainly $p^e$-homocyclic. The only monothetic subgroups of G are finite cyclic subgroups of order at most $p^e$. After retopologizing G, every cyclic subgroup is still a discrete subgroup and hence closed. Furthermore, it is contained in a maximal monothetic subgroup of order $p^e$. Thus, by the very definition, G is $p^{e}$-homocyclic. $\square$

Lemma 2.36

Let G be a locally compact abelian $p^{e}$-homocyclic group, for some $e\ge 2$, and suppose that $pG\ne \{0\}$ is compact open. Then one can turn G into a compact p-group inducing the topology on pG.

Proof

Let $S:=S(G)$ and note that $S\le pG$. The map $\mu _p$ which sends $x\in G$ to px induces a continuous isomorphism $\mu :G/S\rightarrow pG$ (not necessarily a homeomorphism). Equip G/S with the compact topology with the initial topology induced by $\mu$.

As pG is compact by assumption, there is a set I such that $pG=\prod _{i\in I}\langle k_i\rangle$, where $\langle k_i\rangle \cong {\mathbb {Z}}(p^{e-1})$ holds for every $i\in I$. For any finite subset F of I let $S_F$ denote the closed subgroup of S topologically generated by the set $\{p^{e-2}k_i:i\in I{\setminus } F\}$.

The factor group $G/S_F$ has subgroup $S_{e-1}(G/S_F)$ and we claim that $S_{e-1}(G/S_F)$ is $p^{e-1}$-homocyclic. Indeed, let $\bar{g}\in S_{e-1}(G/S_F)$. Then there is $g\in G$ such that $g+S_F/S_F=\bar{g}$. Since G is $p^{e}$-homocyclic, there is $t\in G$ of order $p^e$ with $p^lt=g$ for some $0\le l<e$. Let $r\ge 0$ such that $\bar{g}$ has order $p^r$ and we note $r\le e-1$.

If $\langle g\rangle \cap S_F=\{0\}$ then g also has order $p^r$ and, as $r\le e-1$ conclude that g is a multiple of pt. Hence $\bar{g}\in \langle p\bar{t}\rangle$ for $\bar{t}=(t+S_F)/S_F$ an element of order $p^{e-1}$. Suppose next that $\langle g\rangle \cap S_F\ne 0$. Then $\langle t\rangle \cap S_F\ne \{0\}$ and thus $\bar{g}\in \langle \bar{t}\rangle$ where $\bar{t}$ has order $p^{e-1}$. Thus $S_{e-1}(G/S_F)$ is $p^{e-1}$-homocyclic.

Then there is an exact sequence of locally compact abelian p-groups

$$\begin{aligned} 0\rightarrow S/S_F\rightarrow G/S_F\rightarrow G/S\rightarrow 0. \end{aligned}$$

Since G/S is equipped with a compact topology and $S/S_F$ by construction is finite and hence compact, conclude that $G/S_F$ is compact.

Let ${\mathcal {F}}$ denote the set of all finite subsets of the index set I. When $F\subseteq F'$ for elements in ${\mathcal {F}}$ then $S_{F'}\le S_F$ by construction and there is a canonical epimorphism $\phi _{F'F}:G/S_{F'}\rightarrow G/S_F$ of compact p-groups. Let $\widehat{G}$ denote the inverse limit of the inverse system $(G/S_F,\phi _{F'F})$ and note that $\widehat{G}$ is compact being the inverse limit of compact groups.

By the universal property of the inverse limit, see [5, Lemma 1.26], there is a canonical map $\phi :G\rightarrow \widehat{G}$. Since $S_F\le S\le pG$ holds for every $F\in {\mathcal {F}}$ the induced inverse system $(pG/S_F,\phi _{F'F}\vert _{pG/S_{F'}\rightarrow pG/S_F})$ gives rise to an inverse limit $\widehat{H}:=\lim _{F}pG/S_F$. Since pG is compact by assumption and because of the universality of the inverse limit, one obtains a map $\psi :pG\rightarrow \widehat{H}$. By the exactness of the inverse limit in the category of profinite groups, see [10, Proposition 2.2.4], it follows that $\psi (pG)=\widehat{H}$. Since the inverse limit functor is left exact it follows that $\widehat{H}$ can be identified with the compact subgroup $p\widehat{G}$. Thus, abusing notation, we may assume that $\phi$ induces a topological isomorphism from pG onto $p\widehat{G}$.

We claim that $\phi$ itself is onto. Indeed, pick $\widehat{g}\in \widehat{G}$. Since $p\widehat{g}\in \phi (pG)$ there is $x\in G$ with $\phi (px)=p\phi (x)=p\widehat{g}$. Therefore there is $\widehat{s}\in S(\widehat{G})$ with $\widehat{g}=\phi (x)+\widehat{s}$. Since $S(\widehat{G})\le p\widehat{G}$ we have that $\widehat{s}\in p\widehat{G}$ and hence there is $s\in pG$ with $\phi (s)=\widehat{s}$. Hence

$$\begin{aligned} \phi (x+s)=\widehat{g} \end{aligned}$$

shows that $\phi :G\rightarrow \widehat{G}$ is onto.

Next we claim that $\phi$ is injective. Indeed, pick $x\in \ker (\phi )$. Since the restriction of $\phi$ to pG is a topological isomorphism of compact groups, we have that $\ker (\phi )\cap pG=\{0\}$. On the other hand, since $S(G)\le pG$, certainly $S(\langle x\rangle )\le \ker (\phi )\cap pG$ and hence $x=0$ follows. $\square$

Let G be $p^{e}$-homocyclic. Recall that by Corollary 2.23$p^kG$ is a closed subgroup of G for $0\le k\le e$. Suppose that for some $1\le l<e$ the subgroup $p^lG$ is compact and open in G. Then, applying Lemma 2.36 to $p^{l-1}G$ and its compact subgroup $K:=p^lG$, allows to modify the topology on $p^{l-1}G$ such that it becomes a compact p-group and the induced topology on $p^lG$ agrees with the given topology. Finally we change the topology on G by letting $p^{l-1}G$ a compact open subgroup.

Repeating this step one arrives at the following consequence:

Corollary 2.37

Let G be a locally compact abelian $p^{e}$-homocyclic group and suppose that for some $1\le l\le e$ the subgroup $p^lG$ is compact. Then one can turn G into a compact p-group such that the topology on $p^lG$ agrees with the induced topology.

Proposition 2.38

Let G be a locally compact abelian $p^{e}$-homocyclic p-group. Then there is a discrete subgroup $G_e$, a closed subgroup $\Gamma$ of G, and a compact open subgroup K of $\Gamma$ such that all of the following statements hold:

• (i) $G_e\cong {\mathbb {Z}}(p^e)^{(J_e)}$ is a discrete $p^{e}$-homocyclic subgroup of G;

• (ii) $\Gamma /K$ has exponent not exceeding $p^{e-1}$;

• (iii) $G=G_e\oplus \Gamma$ algebraically and topologically;

• (iv) $\Gamma =\prod _{i\in I}C_i$ for I a suitable set and $C_i\cong {\mathbb {Z}}(p^e)$ for every $i\in I$, considering $\Gamma$ with the product topology, K is a compact open subgroup of $\Gamma$ and hence of G.

Conversely, any locally compact abelian p-group of exponent $p^e$, satisfying (i)–(iv), is $p^{e}$-homocyclic.

Proof

Suppose first that G is a $p^{e}$-homocyclic locally compact abelian p-group. Fix a compact open subgroup K of G. Proposition 2.13 implies that $K=\bigoplus _{j=1}^eK_i$ where $K_i\cong {\mathbb {Z}}(p^i)^{J_i}$ is, according to Lemma 2.35, a $p^i$-homocyclic group. Lemma 2.34 yields a decomposition $G=\bigoplus _{j=0}^eH_j$ into $p^{e}$-homocyclic constituents such that $H_j\cap K=K_j$. Let $G_e:=H_0$ and note that $H_0\cap K=\{0\}$ so that $G_e$ is a discrete $p^{e}$-homocyclic group. Hence $G_e={\mathbb {Z}}(p^e)^{(J_e)}$ for some set $J_e$. Thus (i) holds. Let $\Gamma :=\bigoplus _{j\ge 1}H_j$, then $G=G_e\oplus \Gamma$, so that (iii) is established. Moreover, $\Gamma /K\cong \bigoplus _{j\ge 1}H_j/K_j$ has exponent not exceeding $p^{e-1}$. Thus (ii) holds. Next, fixing any $j\ge 1$, we observe that $K_j=p^{e-j}H_j$ is a compact open subgroup of $H_j$. Therefore, making use of Corollary 2.37, we may modify the topology on $H_j$ such that it becomes a compact p-group and inducing the topology on $K_j$. As $H_j$ is $p^{e}$-homocyclic, [10, Theorem 4.3.8] implies that $H_j=\prod _{i\in J_i}C_{i_j}$ with $C_{i_j}\cong {\mathbb {Z}}(p^e)$. Hence, letting I be the disjoint union of the sets $I_j$ for $1\le j\le e$, one obtains $\Gamma =\prod _{i\in I}C_i$ for all factors $C_i\cong {\mathbb {Z}}(p^e)$. Hence (iv) holds true.

Conversely, let a locally compact abelian p-group G satisfy items (i)–(iv). Lemma 2.35 shows that $\Gamma$ is $p^{e}$-homocyclic. Furthermore, $G_e$ is $p^{e}$-homocyclic in light of Lemma 2.32. Finally, making use of Lemma 2.31 with $A:=G_e$ and $B:=\Gamma$, implies that $G=G_e\oplus \Gamma$ is $p^{e}$-homocyclic. The subgroup K has been chosen to be compact open in G and hence is compact open in $\Gamma$ as well. $\square$

Lemma 2.39

Let a locally compact abelian p-group G be the direct sum $G=A\oplus B$ of a $p^{e}$-homocyclic group A and a $p^{f}$-homocyclic group B and suppose that $e-1\le f\le e$ holds. Then G is neat iff A and B are both neat.

Proof

Suppose that G is neat and $M\in {\mathbb {M}}(A)$. Then there exists a complement R for M in G and therefore $A\cap R$ is a complement for M in A. Hence A is neat. The same argument shows that B is neat.

Conversely, assume that A and B are both neat and assume that $M=\langle x\rangle \in {\mathbb {M}}(G)$. Then there are $a\in A$ and $b\in B$ with $x=a+b$. Since A and B are both neat, there are maximal monothetic subgroups $M_A$ and $M_B$ of respectively A and B with $a\in M_A$ and $b\in M_B$. Moreover, there are decompositions

$$\begin{aligned} A=M_A\oplus R_A \ \ \hbox {and } \ \ B=M_B\oplus R_B. \end{aligned}$$

Note that $M_A+M_B=M_A\oplus M_B\cong {\mathbb {Z}}(p)^e\oplus {\mathbb {Z}}(p)^f$ is neat, as a consequence of Lemma 2.26(ii). Therefore

$$\begin{aligned} M_A\oplus M_B=M\oplus R_{AB} \end{aligned}$$

for some closed subgroup $R_{AB}$. Letting $R:=R_A+R_B+R_{AB}$ one arrives at

$$\begin{aligned} G&=A\oplus B=(M_A\oplus R_A)\oplus (M_B\oplus R_B)\\&=(M_A\oplus M_B)\oplus (R_A\oplus R_B) \\&=(M\oplus R_{AB})\oplus (R_A\oplus R_B)\\&=M\oplus R. \end{aligned}$$

Hence G is neat. $\square$

Lemma 2.40

Every locally compact abelian $p^{e}$-homocyclic group G is neat.

Proof

Suppose first that G is discrete. Then, by Lemma 2.8, every maximal monothetic subgroup M of G is pure. Theorem 1.1 shows that M can be complemented.

Assume next that G is compact. Then $G\cong {\mathbb {Z}}(p^e)^I$ for some set I is neat as a consequence of Lemma 2.16.

Assume that G arises by retopologizing a compact $p^{e}$-homocyclic group by selecting a closed subgroup K and declaring it compact open in G. Then, letting G play the role of $\Gamma$ in Lemma 2.18, it follows that G is neat.

For finishing the proof we employ Proposition 2.38. Accordingly $G=G_e\oplus \Gamma$ for $G_e$ discrete and hence neat, and, $\Gamma$ the result of changing the topology on a compact $p^{e}$-homocyclic group by declaring a closed subgroup K compact open. It follows from the previous step of our proof that $\Gamma$ is neat. Finally, applying Lemma 2.39 to the direct sum $G=G_e\oplus \Gamma$, yields the neatness of G. $\square$

2.4 Compact torsion-free p-groups—C-chains

We introduce in Definition 2.41 for any compact abelian torsion-free p-groups C a so-called C-chain, a type of filtration $(C_k)_{k\ge 0}$ of C. Proposition 2.55 below provides a description of all compact groups possessing such a filtration – and will serve as the main tool in proving Proposition 3.2, where torsion-free neat locally compact abelian p-groups will be characterized.

Definition 2.41

For a compact abelian p-group G a C-chain is a sequence $(C_k)_{k\ge 0}$ of closed subgroups such that $C_0=C$, and for all $k\ge 0$ the containments

$$\begin{aligned} C_k\ge C_{k+1}\ge pC_k \end{aligned}$$

hold. We say that the locally compact abelian p-group G has a C-chain provided some compact open subgroup C of G possesses a $C$-chain.

Remark 2.42

The constant sequence $(C_k)_{k\ge 0}$ with $C_k=C=C_0$ for all $k\ge 0$ is a $C$-chain. We will have to use certain $C$-chains which converge to $\{0\}$. Recall from Corollary 2.23 that in a neat locally compact abelian p-group G and compact open subgroup C the sequence $(C_k)_{k\ge 0}$ where $C_k:=p^kG\cap C$ is a $C$-chain. In a moment we shall prove that this $C$-chain converges to the trivial group with respect to the Vietoris topology. See Lemma 2.45 below.

In a moment we shall characterize convergence to $\{0\}$ of a $C$-chain $(C_k)_{k\ge 0}$ in several ways, as a consequence from a general fact from set topology:

Lemma 2.43

Let $(C_n)_{n\in {\mathbb {N}}}$ be a descending sequence of compact subsets of some Hausdorff space G. Let D denote the nonempty intersection of the filter base $\{C_n:n\in {\mathbb {N}}\}$. Then the following statements are equivalent:

• (1) $D=\{c\}$ for some c.

• (2) For each open neighborhood U of the point c there is an $n\in {\mathbb {N}}$ such that $C_n\subseteq U$.

• (3) There is a point c such that for each sequence $(c_n)_{n\in {\mathbb {N}}}$ with $c_n\in C_n$ for all n we have $c=\lim _n c_n$.

Proof

Trivially (3) implies (1) since $d\in D$ allows us to take $c_n=d$ for all n and so $d=\lim _{n\in {\mathbb {N}}}c_n=c$.

(1) implies (2): Assume that (2) is false. Then $\{C_n{{\setminus }} U:n\in {\mathbb {N}}\}$ is a filter base of closed sets in the compact space $C_1$. Then it has a nonempty intersection. So there is a $d\in \bigcap _{n=1}^\infty C_n{{\setminus }} U$, contradicting (1).

(2) implies (3): Let U be an open neighborhood of c. Then there is an $n\in {\mathbb {N}}$ such that $C_n\subseteq U$. So for all $m\ge n$ in ${\mathbb {N}}$ we have $c_m\in C_m\subseteq C_n\subseteq U$. Thus $\lim _nc_n=c$.

$\square$

Letting G be a compact p-group and $(C_k)_{k\ge 0}$ a descending sequence of closed subgroups, Lemma 2.43 yields the following consequence:

Corollary 2.44

Let C be a compact abelian p-group and $(C_k)_{k\ge 0}$ be a sequence of closed subgroups such that $C_k\ge C_{k+1}$ holds for all $k\ge 0$. The following statements are equivalent:

• (a) The intersection $\bigcap _{k\ge 0}C_k=\{0\}$.

• (b) The sequence $(C_k)_{k\ge 0}$ tends to the trivial subgroup $\{0\}$ with respect to the Vietoris topology.

• (c) Every sequence $(c_k)_{k\ge 0}$ with $c_k\in C_k$ tends to 0.

Let G be a torsion-free neat locally compact abelian p-group. Recall from Corollary 2.23 that $p^kG$ is closed for every $k\ge 0$ and, letting C be a fixed compact open subgroup of G, we note that $C_k:=p^kG\cap C$ form a decreasing sequence of compact subgroups of G with trivial intersection, since by Corollary 2.44 this sequence converges to $\{0\}$.

Lemma 2.45

Let C be a compact open subgroup of a neat torsion-free locally compact abelian p-group G. The $C$-chain $(C_k)_{k\ge 0}$ with $C_k:=p^kG\cap C$ converges to 0.

Proof

In light of Corollary 2.44 it is enough to prove that $\bigcap _{k\ge 0}p^kG\cap C=\{0\}$. Pick any $c\in \bigcap _{k\ge 0}p^kG\cap C$ and, by way of contradiction, assume $c\ne 0$. Then, for every $k\ge 0$ there is $g_k\in G$ such that $c=p^kg_k$. Hence c has infinite p-height. Taking the definition of neatness, Definition 1.5, the torsion-freeness of G and hence Corollary 2.3(6) into account, this contradicts $\textrm{Mon}(G)$ being inductive. Thus $(C_k)_{k\ge 1}$ must converge to $\{0\}$. $\square$

Lemma 2.46

Let C be a compact abelian p-group and T a not necessarily closed subgroup. Let $f:C\rightarrow C/pC$ denote the canonical epimorphism. Then the closure $\overline{T}$ of T equals C if and only if $f(T)=f(C)$.

Proof

Suppose first that $\overline{T}=C$. Since T is dense in C and f is an epimorphism, it follows that f(T) is dense in f(C), i.e., $\overline{f(T)}=f(C)$.

For proving the converse, suppose $\overline{f(T)}=f(C)$. Then $f(T)\le f(\overline{T})$ implies

$$\begin{aligned} f(C)=\overline{f(T)}\le f(\overline{T}), \end{aligned}$$

and taking the definition of the quotient map f into account, one thus finds

$$\begin{aligned} C=\overline{T}+pC. \end{aligned}$$

By [10, Lemma 2.8.7] and as C is abelian, the subgroup pC agrees with the Frattini subgroup of C. Now Corollary 2.8.5 ibidem implies that $\overline{T}=\overline{T}+pC$. Hence $C=\overline{T}$. $\square$

Lemma 2.47

Let C be a compact abelian p-group and $C'$ and B be closed subgroups of C such that $C=B\oplus C'$. Suppose that $\Delta$ is a closed subgroup of C containing B. Then there is a closed subgroup $\Theta$ of $C'$ such that $\Delta =\Theta \oplus B$.

Proof

Let $\phi :C\rightarrow C$ be the canonical endomorphism with kernel B and projecting C onto $C'$. Then $\Theta :=\phi (\Delta )$ serves the purpose.

Lemma 2.48

Let P and C be compact abelian torsion-free p-groups and $f:P\rightarrow C$ an epimorphism inducing an isomorphism $f^*:P/pP\rightarrow C/pC$. Then f is an isomorphism.

Proof

Pick $x\in \ker (f)$. We need to show that $x=0$. Suppose by contrast that $x\ne 0$. Since P is a compact p-group, there is $m\in P$ and a maximal monothetic subgroup $M=\overline{\langle m\rangle }$ and $k\ge 0$ with $x=p^km$. Then

$$\begin{aligned} 0=f(x)=f(p^km)=p^kf(m) \end{aligned}$$

and by the torsion-free ness of P we obtain that $m\in \ker (f){\setminus } pP$. Therefore on the one hand $f^*((m+pP)/pP)=0$ and on the other hand, since by assumption $f^*$ is an isomorphism we arrive at the contradiction $f^*((m+pP)/pP)\ne 0$. $\square$

Lemma 2.49

Let C be a compact abelian torsion-free p-group containing a closed subgroup $C'$ with $pC\le C'$. Then there exists a closed subgroup $\Delta \le C$ such that all of the following statements hold:

• (a) $C'+\Delta =C$.

• (b) $C'\cap \Delta =p\Delta$.

• (c) There is an algebraic and topological isomorphism

  $$\begin{aligned} C/\Delta =(C'+\Delta )/\Delta \cong C'/C'\cap \Delta = C'/p\Delta \end{aligned}$$

  of torsion-free compact p-groups.

Proof

Consider the canonical epimorphism $\phi :C\rightarrow C/C'$. By [10, Lemma 2.8.15, p. 57] there is a minimal closed subgroup $\Delta \le C$ with $\phi (\Delta )=C/C'$. Thus $C=C'+\Delta$ and hence (a) holds.

Next we deduce from the just cited Lemma that the kernel of the restriction, say $\psi$, of $\phi$ to $\Delta$ is contained in the Frattini subgroup of $\Delta$, i.e., $\ker \psi \le p\Delta$. On the other hand, since $C/C'\cong (C/pC)/(C'/pC)$ has exponent p we note that $p\Delta \le \ker (\psi )$ and, since $\ker \psi =C'\cap \Delta$, we obtain (b).

The isomorphisms result from the isomorphism theorem and (b). For establishing torsion-free ness of $C'/pC$, suppose that $pc\in \Delta$ holds for some $c\in C$. Since $pC\le C'$ we have that $pc\in \Delta \cap C'$ and by (b) the latter subgroup equals $p\Delta$ by (b). Hence there is $\delta \in \Delta$ with $pc=p\delta$ and, as C is torsion-free, find that $c=\delta \in \Delta$. Thus $C/\Delta$ is torsion-free. $\square$

Lemma 2.50

Given a compact abelian torsion-free p-group C and a $C$-chain (see Definition 2.41). Set $\Delta _{-1}:=\{0\}$ and $T_{-1}:=\{0\}$. Then there are sequences $(\Delta _k)_{k\ge 0}$, $(T_k)_{k\ge 0}$, and $(\Theta _k)_{k\ge 0}$ of closed subgroups of C, satisfying for all $k\ge 0$ the following properties:

• (i) $T_k=\Theta _k\oplus T_{k-1}$;

• (ii) $\Theta _k\le C_k$ and $\Delta _k=\Theta _k\oplus pT_{k-1}$;

• (iii) $C_k+pT_{k-1}=\Delta _k+(C_{k+1}+pT_{k-1})=\Delta _k+C_{k+1}$.

• (iv) $\Delta _k\cap (C_{k+1}+pT_{k-1})=pT_k$.

• (v) $(C_k+pT_{k-1})/\Delta _k=((C_{k+1}+pT_{k-1}) +\Delta _k)/\Delta _k \cong (C_{k+1}+\Delta _{k})/pT_{k}$ is torsion-free.

Proof

We shall employ induction on k and start by letting $k:=0$. By assumption we have that $C_0\ge C_1\ge pC_0$ and hence we can apply Lemma 2.49 to $C:=C_0$ and $C':=C_1$. Thus there is $\Delta _0\le C_0$ such that

• (a) $C_0=\Delta _0+C_1$;

• (b) $C_1\cap \Delta _0=p\Delta _0$; and

• (c) $C_0/\Delta _0\cong C_1/p\Delta _0$ is torsion-free.

Setting $T_0:=\Delta _0$, $\Theta _0:=\Delta _0$, and, observing our agreement $\Delta _{-1}=T_{-1}=\{0\}$, shows that (i) and (ii) hold for $k=0$. From items (a), (b), and (c) we infer the validity of respectively (iii), (iv) and (v), thereby making use of the isomorphism

$$\begin{aligned} C_0/\Delta _0=(C_1+\Delta _0)/\Delta _0\cong C_1/C_1\cap \Delta _0=C_1/p \Delta _0. \end{aligned}$$

Performing the induction step, suppose that (i)–(v) hold for some $k\ge 0$. We need to establish these items for $k+1$.

Let tilde denote passing to the factor group modulo $pT_k$ and we keep $pT_k=p\Theta _k\oplus pT_{k-1}$ in mind, as follows from (i). Then, by (v), we observe that

$$\begin{aligned} \widetilde{C_{k+1}+\Delta _k}=(C_{k+1}+\Delta _k)/pT_k \end{aligned}$$

is torsion-free and so is its subgroup $\widetilde{C_{k+1}}=(C_{k+1}+pT_k)/pT_k$. Since every torsion-free abelian compact p-group is a projective ${\mathbb {Z}}_p$-module, there exists $C'_{k+1}\le C_{k+1}$ with

$$\begin{aligned} C_{k+1}=C'_{k+1}\oplus pT_k. \end{aligned}$$

(2.5)

From the assumption $C_{k+1}\ge C_{k+2}\ge pC_{k+1}$ we can deduce that

$$\begin{aligned} \widetilde{C_{k+1}}\ge \widetilde{C_{k+2}}\ge p\widetilde{C_{k+1}}. \end{aligned}$$

Hence we may apply Lemma 2.49 to $C:=\widetilde{C_{k+1}}$ and $C':=\widetilde{C_{k+2}}$, in order to find a subgroup $\widetilde{\Delta _{k+1}}$ of $\widetilde{C_{k+1}}$ such that

• (a’) $\widetilde{C_{k+1}}=\widetilde{\Delta _{k+1}} +\widetilde{C_{k+2}}$;

• (b’) $\widetilde{\Delta _{k+1}}\cap \widetilde{C_{k+2}} =p\widetilde{\Delta _{k+1}}$;

• (c’) $\widetilde{C_{k+1}}/\widetilde{\Delta _{k+1}} \cong \widetilde{C_{k+2}}/p\widetilde{\Delta _{k+1}}$ is torsion-free.

Letting $\Delta _{k+1}$ denote the preimage of $\widetilde{\Delta _{k+1}}$ in $C_{k+1}$ and observing that $\Delta _{k+1}\ge pT_k$, we may apply Lemma 2.47 to the decomposition in Eq. (2.5), thereby letting $C_{k+1}'$, $C_{k+1}$, $\Delta _{k+1}$, and, $pT_k$ play respectively the roles of $C'$, C, $\Delta$, and, B, in order to find a closed subgroup $\Theta _{k+1}\le C'_{k+1}$ with

$$\begin{aligned} \Delta _{k+1}=\Theta _{k+1}\oplus pT_k. \end{aligned}$$

Thus (ii) holds, since we arranged $\Delta _{k+1}\le C'_{k+1}\le C_{k+1}$.

Setting

$$\begin{aligned} T_{k+1}:=\Theta _{k+1}+T_k, \end{aligned}$$

(2.6)

one observes from $\Theta _{k+1}\cap pT_k=\{0\}$ that

$$\begin{aligned} \Theta _{k+1}\cap T_k=\{0\} \end{aligned}$$

and hence (i) holds for $T_{k+1}$.

Recalling that tilde denotes factoring $pT_k$, one rereads (a’) as

$$\begin{aligned} C_{k+1}+pT_k=\Delta _{k+1}+(C_{k+2}+pT_k)=\Delta _{k+1}+C_{k+2}, \end{aligned}$$

hence (iii) holds.

In the same vein, observing Eq. (2.6) item (b’) reads

$$\begin{aligned} \Delta _{k+1}\cap (C_{k+1}+pT_k)=p\Delta _{k+1}+pT_k=pT_{k+1}, \end{aligned}$$

establishing (iv).

Finally, for proving in (v) the torsion-freeness reread (c’):

$$\begin{aligned} (C_{k+1}+pT_k)/\Delta _{k+1}\cong (C_{k+2}+pT_k)/pT_k, \end{aligned}$$

which is thus torsion-free. $\square$

Corollary 2.51

For a subset $A\subseteq C=C_0$ let $\widetilde{A}$ denote the image under the canonical map $C_0\rightarrow C_0/pC_0$. The following statements hold for all $k\ge 0$:

• (a) $\widetilde{T}_k= \bigoplus _{j=0}^k\widetilde{\Theta }_j$.

• (b) $\widetilde{C}_0= \widetilde{T}_k\oplus \widetilde{C}_{k+1}$.

Proof

We shall make use of Lemma 2.50.

(a). Let $k\ge 0$. Observe first that (ii) implies

$$\begin{aligned} \widetilde{\Theta }_j=\widetilde{\Delta }_j, \ \ j\ge 0, \end{aligned}$$

and (iv) yields

$$\begin{aligned} \widetilde{\Theta }_j\cap \widetilde{C}_{j+1}=\{0\}, \ \ j\ge 0. \end{aligned}$$

(2.7)

Therefore, since $\widetilde{\Theta }_i\le \widetilde{C}_{j+1}$ for all $i\ge j+1$ we conclude from Eq. (2.7) that

$$\begin{aligned} \widetilde{\Theta }_j\cap \overline{\langle \widetilde{\Theta }_i:i \ge j+1\rangle }=\{0\}, \ 0\le j\le k. \end{aligned}$$

Therefore $\widetilde{T}_k=\overline{\langle \widetilde{\Theta }_j:0\le j\le k\rangle } =\bigoplus _{j=0}^k\widetilde{\Theta }_j$, as claimed.

(b) Item (iii), observing $\widetilde{T}_0=\widetilde{\Theta }_0$, implies for $k=0$ the basis of induction on k:

$$\begin{aligned} \widetilde{C}_0=\widetilde{T}_0\oplus \widetilde{C}_1. \end{aligned}$$

Suppose that for some $k\ge 0$ we found a decomposition $\widetilde{C}_0=\widetilde{T}_k\oplus \widetilde{C}_{k+1}$. Then, making use of Eq. (2.7) with j replaced by $k+1$, and taking Lemma 2.50(ii) and (iii) into account, we obtain

$$\begin{aligned} \widetilde{C}_0&=\widetilde{T}_k\oplus \widetilde{C}_{k+1}\\&=\widetilde{T}_k\oplus (\widetilde{\Theta }_{k+1} \oplus \widetilde{C}_{k+2})\\&=\widetilde{T}_{k+1}\oplus \widetilde{C}_{k+2}. \end{aligned}$$

Therefore (b) holds. $\square$

We recall from [8, 9] the following concept:

Definition 2.52

A family $(H_j)_{j\in J}$ of closed subgroups of a profinite group H is convergent, if for every open normal subgroup N of H the set $J_N:=\{j\in J:H_j\not \subseteq N\}$ is finite. A family $(\tau _j)_{j\in J}$ of continuous homomorphisms from profinite groups $H_j$ to a profinite group H is convergent, provided the family $(\tau _j(H_j))_{j\in J}$ is convergent.

For $P:=\prod _{j\in J}H_j$ there are natural embeddings $\tau _j:H_j\rightarrow P$ and, by the definition of the cartesian product, the family $(\tau _j)_{j\in J}$ convergent. When the factors $H_j$ are abelian, it has been noted in [8] that P together with this canonical embeddings $(\tau _j)_{j\in J}$ is a free profinite product of the family $(H_j)_{j\in J}$ in the category of profinite abelian groups. This gives rise to the following characterization of P:

Lemma 2.53

Let P be a profinite abelian group together with a family $(H_j)_{j\in J}$ and convergent family $\tau _j:H_j\rightarrow P$, where $j\in J$. Suppose further that for any profinite abelian group H and convergent family $\psi _j:H_j\rightarrow H$, for $j\in J$, there is a continuous homomorphism $\omega :P\rightarrow H$ such that $\psi _j=\omega \circ \tau _j$ holds for all $j\in J$.

Then there is a topological and algebraic isomorphism $\lambda : \prod _{j\in J} H_j\rightarrow P$ that maps each factor $H_j$ in P isomorpically onto $\tau _j(H_j)$.

Proof

We note that $P=\prod _{j\in J}H_j$ agrees with the free profinite product of the convergent family $(H_j)_{j\in J}$ of abelian profinite groups, as explained in [8, Section 5], see also [10, Ch.9]. Therefore the universal property of the free product guarantees the unique continuous homomorphism $\omega$ with $\psi _j=\omega \circ \tau _j$ for all $j\in J$. In particular, the isomorphism $\lambda$ results from the universal property of the free profinite product. $\square$

A generalization of this result to abelian pro-Lie groups can be found in a separate note, see [3].

Proposition 2.54

Let G be a torsion-free locally compact abelian p-group and C a compact open subgroup of G. Assume $p^kG$ to be closed for all $k\ge 0$ and set $C_k:=p^kG\cap C$ (note that by Remark 2.42 the sequence $(C_k)_{k\ge 0}$ with $C_0=C$ is a $C$-chain). Let the subgroups $T_k$, $\Delta _k$, and, $\Delta _k$ be formed according to Lemma 2.50. Set $T:=\bigoplus _{i\ge 0}\Theta _i$. Then

$$\begin{aligned} C_0=\prod _{i\ge 0}\Theta _i=\overline{T}. \end{aligned}$$

Proof

Let $\widetilde{A}$ denote the image of any subset A of C in C/pC. We first prove that $C_0=\overline{T}$ and observe that by Lemma 2.46 it suffices to show that $\widetilde{C}_0=\overline{\widetilde{T}}$.

Corollary 2.44 implies that the sequence $(C_k)_{k\ge 0}$ converges to $\{0\}$ and therefore so does the sequence $(\widetilde{C}_k)_{k\ge 0}$. Fix an arbitrary open subgroup $\widetilde{N}$ of $\widetilde{C}$. Then there exists $k_0\ge 0$ such that $\widetilde{\Theta }_j\le \widetilde{C}_j\le \widetilde{N}$ whenever $j\ge k_0$. Corollary 2.51(b) implies, for all such j,

$$\begin{aligned} \widetilde{C}_0=\widetilde{T}_j+\widetilde{C}_{j+1}=\widetilde{T} +\widetilde{C}_{j+1}=\overline{\widetilde{T}}+\widetilde{N}. \end{aligned}$$

Therefore, letting ${\mathcal {U}}$ denote the filter base of open subgroups of $\widetilde{C}_0$, one obtains

$$\begin{aligned} \widetilde{C}_0\le \bigcap _{\widetilde{N}\in {\mathcal {U}}} (\overline{\widetilde{T}}+N) =\overline{\widetilde{T}}, \end{aligned}$$

where the last equality is a consequence of [10, Proposition 2.1.4 (a)].

Since $\Theta _k\le C_k$ and the sequence $(C_k)_{k\ge 0}$ tends to $\{0\}$, Lemma 2.45 implies that the sequence $(\Theta _k)_{k\ge 0}$ also converges to $\{0\}$. Therefore the universal property (see Lemma 2.53) of the cartesian product $P:=\prod _{i\ge 0}\Theta _i$ in the category of profinite groups implies the existence of a unique epimorphism $\eta :P\rightarrow C_0$ sending each factor $\Theta _i$ of P identically to the subgroup $\Theta _i$ in $C_0$. For proving that $\eta$ is a topological isomorphism, in light of Lemma 2.48, we only need to prove that the induced map $\eta ^*:P/pP\rightarrow C/pC$ is an isomorphism.

From Corollary 2.51 we deduce the existence of horizontal topological isomorphisms fitting into a commutative diagram:

By the exactness of the $\varprojlim$-functor in the category of profinite groups, see [10, Proposition 2.2.4], one verifies that $\eta ^*=\varprojlim \eta _j$ agrees with the projective limit of the horizontal topological isomorphisms $\eta _j$ and is therefore itself a topological isomorphism.

In turn, $\eta$ is a topological isomorphism, and yields the desired direct product decomposition $C_0=\prod _{j\ge 0}\Theta _j$. $\square$

Recall from Remark 2.42 that for any fixed compact open subgroup C of a locally compact abelian p-group G the sequence $(C_k)_{k\ge 0}$, where $C_k:=p^kG\cap C$, is a $C$-chain. We summarize our findings:

Proposition 2.55

Let G be a torsion-free neat locally compact abelian p-group and C a compact open subgroup of G. Letting $C_0:=C$ form $C_k:=p^kG\cap C$. Then, for all $k\ge 0$, there are subgroups $\Theta _k\le C_k$ such that for $T_k:=\overline{\langle \Theta _j:0\le j\le k\rangle }$ and $C_k':=\overline{\langle \Theta _j:j\ge k\rangle }$, all of the following statements hold for every $k\ge 0$:

• (a) $T_k=\bigoplus _{j=0}^k\Theta _j$;

• (b) $C_k'=\prod _{j\ge k}\Theta _j\le C_k$;

• (c) $C_0=T_k\oplus C_{k+1}'$.

Proof

(a) is identical to statement (i) in Lemma 2.50(i).

(b) follows from the fact that $C_k'\le C_0=\prod _{j\ge 0}\Theta _j$ so that $C_k'$, being generated by factors of a direct product, turns out to be itself a direct product:

$$\begin{aligned} C_k'=\overline{\langle \Theta _j:j\ge k\rangle }=\prod _{j\ge k}\Theta _j. \end{aligned}$$

(c) follows from Proposition 2.54 and

$$\begin{aligned} C_0=C_0'=\prod _{j\ge 0}\Theta _j=\prod _{j=0}^k\Theta _j \oplus \prod _{j\ge k+1}\Theta _j=T_k\oplus C_{k+1}'. \end{aligned}$$

$\square$

3 Classifying neat locally compact p-groups

We come to prove our main result about neat locally compact abelian p-groups, namely Theorem 1.7 and first recall from Corollary 2.27 that any such group is either torsion-free or bounded. Accordingly we prove Proposition 3.2 in Sect. 3.1 and Proposition 3.3 in Sect. 3.2. Sect. 3.3 is devoted to the proof of Theorem 1.7.

3.1 Neat torsion-free locally compact abelian p-groups

Lemma 3.1

Let G be a locally compact abelian torsion-free p-group and C a compact open subgroup of G. For $k\ge 0$ set $C_k:=p^kG\cap C$. The following statements are equivalent:

• (i) For every $k\ge 0$ the subgroup $p^kG$ is closed.

• (ii) For every $k\ge 0$ one has

  $$\begin{aligned} C_k=p^k\mu _p^{-k}C, \end{aligned}$$

  and the sequence $(C_k)_{k\ge 0}$ is a $C$-chain for G.

Proof

Suppose that (i) holds. Then certainly, for every $k\ge 0$, the set $C_k=p^kG\cap C$ is closed. Moreover, since G is torsion-free,

$$\begin{aligned} C_k=p^kG\cap C=p^k(G\cap \mu _p^{-k}C)=p^k(\mu _p^{-k}C). \end{aligned}$$

From $p^{k}G\ge p^{k+1}G$ we deduce

$$\begin{aligned} C_k=p^kG\cap C\ge p^{k+1}G\cap C=C_{k+1}. \end{aligned}$$

Certainly

$$\begin{aligned} C_{k+1}=p^{k+1}G\cap C\ge p^{k+1}G\cap pC, \end{aligned}$$

and, since G is torsion-free, the last term equals $p(p^kG\cap C)$ which equals $pC_k$. Hence $(C_k)_{k\ge 0}$ is a $C$-chain for G.

Suppose that (ii) holds. We need to show that $p^kG$ is closed for every $k\ge 0$. Since C is open and, by assumption, $C_k=p^kG\cap C$ is closed, Lemma 2.1 applies, i.e., $p^kG$ is indeed closed. $\square$

Proposition 3.2

Let G be a torsion-free locally compact abelian p-group and C a compact open subgroup. The following statements are equivalent:

• (i) G is neat.

• (ii) For all $k\ge 0$ the subgroups $p^kG$ are closed.

• (iii) There is a compact torsion-free p-group $\Gamma$ and a continuous embedding $j:G\rightarrow \Gamma$ so that $j(G)/j(C)=\textrm{tor}(\Gamma /j(C))$. Moreover, redefining the topology on $\Gamma$ so that j(C) is compact open, turns j into an algebraic and topological isomorphism from G onto j(G).

Proof

(i) $\Rightarrow$(ii). This follows from Corollary 2.23.

(ii) $\Rightarrow$(iii).

Let $\widetilde{C}$ be a copy of C and $j_C:C\rightarrow \widetilde{C}$ be an algebraic and topological isomorphism. We shall construct $\Gamma$ and extend j to all of G. It will be convenient, for any subset $A\subseteq G$, to let $\widetilde{A}$ denote the image of A under j. By (ii) for every $k\ge 0$ the subgroup $p^kG$ is closed and hence, as C is compact, $C_k:=p^kG\cap C$ is closed, giving rise to a $C$-chain $(C_k)_{k\ge 0}$. Proposition 2.55 provides us, for every $k\ge 0$, closed subgroups $\Theta _k\le C_k$ and $C_k'=\prod _{j\ge k}\Theta _j$ of G. Since $\Theta _k\le C_k=p^kG\cap C$, there are closed subgroups $\Gamma _k\le G$ with $p^k\Gamma _k=\Theta _k$. Consider a copy $\widetilde{\Gamma }_k\ge \widetilde{\Theta }_k=p^k\widetilde{\Gamma }_k$ and, let $j_k:\Gamma _k\rightarrow \widetilde{\Gamma }_k$ denote any isomorphism which agrees with $j_C$ on $\Theta _k$. Our candidate for $\Gamma$ is $\prod _{k\ge 0}\widetilde{\Gamma }_k$ equipped with the product topology. Note that $\Gamma$ is a compact torsion-free p-group. We want to define $j:G\rightarrow \Gamma$ such that j coincides on C with $j_C$ and, for every $k\ge 0$ with $j_k$ on $\Gamma _k$. Pick any element $x\in G$. If $x\in C$ then $j(x):=j_C(x)$. Suppose that k is minimal such that $p^kx\in C$ and we may assume $k\ge 1$. Then $p^kx\in C$ and thus, letting $p^kx=(\theta _l)_{l\ge 0}$ for $\theta _l\in \Theta _l$, we define

$$\begin{aligned} j(x):=(\mu _p^{-k}\widetilde{\theta }_l)_{l\ge 0}\in \Gamma . \end{aligned}$$

Certainly, since $p^kx\in C$, it follows that $p^kj(x)\in \widetilde{C}$ and hence

$$\begin{aligned} j(G)/\widetilde{C}\le \textrm{tor}(\Gamma /\widetilde{C}). \end{aligned}$$

For proving the converse, namely $j(G)/\widetilde{C}\ge \textrm{tor}(\Gamma /\widetilde{C})$ we may fix $\widetilde{x}\in \Gamma$ with $p^k\widetilde{x}\in \widetilde{C}$. Decomposing

$$\begin{aligned} \widetilde{x}=(\widetilde{x}_i)_{i\ge 0}, \ \ \widetilde{x}_i \in \widetilde{\Gamma }_i \end{aligned}$$

one deduces that

$$\begin{aligned} p^k\widetilde{x}=(p^k\widetilde{x}_i)_{i\ge 0} \in \prod _{i\ge 0}\widetilde{\Theta }_i. \end{aligned}$$

Note that $\widetilde{y}$ with components $\widetilde{y}_i=0$ for $0\le i<k$ and $\widetilde{y}_i=\widetilde{x}_i$ for $i\ge k$ belongs to $\widetilde{C}_k$ and hence there is $y\in C_k'$ with $j(y)=\widetilde{y}$. Furthermore, for every $0\le i\le k-1$ there is $y_i\in G$ with $j(y_i)=\widetilde{x}_i$ since $x_i\in \widetilde{\Gamma }_i=j(\Gamma _i)$. Altogether we found that for $x:=\sum _{i=0}^{k-1}y_i+y$ the image $j(x)=\widetilde{x}$, as was needed to prove.

(iii) $\Rightarrow$(i). Since the restriction of j to G yields an algebraic and topological isomorphism onto j(G) it suffices to prove that j(G) is neat. Since, before refining the topology on $\Gamma$, the latter is a torsion-free compact p-group it is neat by Corollary 2.17(iii). We may apply Lemma 2.18 to $\Gamma$ and its compact subgroup j(C) in order to find that j(G) is indeed neat. $\square$

3.2 Neat torsion locally compact abelian p-groups

We turn to the classification of the neat locally compact abelian torsion p-groups, namely establishing the description of any neat locally compact abelian p-group as given in Theorem 1.7(B). Recall from Definition 2.9 that for a locally compact abelian p-group one denotes by $S_n(G):=\{x\in G:p^nx=0\}$ and keep in mind that $S_n(G)$ is a closed subgroup of G of exponent $\le p^n$. Let us also keep in mind the notion of G being $p^{e}$-homocyclic from Notation 1.6.

Proposition 3.3

Let G be a locally compact abelian neat torsion p-group. Then there is $e\ge 1$ and G has homocyclic closed subgroups A and B of exponent respectively $p^e$ and $p^{e-1}$ such that $G=A\oplus B$.

Proof

If $e=1$ then G has exponent p then G is $p^1$-homocyclic and thus the result follows from Lemma 2.25.

Suppose that $e>1$ and for all exponents less than $p^e$ the assertions of the proposition have been established. Then, as by Lemma 2.22, the subgroup pG is closed and is itself neat, the induction hypothesis applies; thus there are homocyclic subgroups C and D of exponents respectively $p^{e-1}$ and $p^{e-2}$ and $pG=C\oplus D$. Let $\mu _p:G\rightarrow G$ denote the endomorphism sending $x\in G$ to px. Set $\widetilde{C}:=\mu _p^{-1}(L)$ and $\widetilde{D}:=\mu _p^{-1}(D)$ and note the commutative diagrams for $\mu _p^*$ a map induced from $\mu _p$:

Making use of Lemma 2.25 one can find complements of $\ker (\mu _p^*)$ in respectively $\widetilde{C}/p\widetilde{C}$ and $\widetilde{D}/p\widetilde{D}$ and thus, by lifting, find subgroups $W\le C$ and $V\le D$ such that $\mu _p(W)=C$ and $\mu _p(V)=D$. Observe that W and V are homocyclic of exponent respectively $p^e$ and $p^{e-1}$. Therefore $S(W)=S(C)$ and $S(V)=S(D)$ and hence $W\cap V=\{0\}$. Since $\widetilde{C}=W+\ker (\mu _p)=W+S(C)+S(D)=W\oplus S(D)$ and $\widetilde{D}=V+S(C)+S(D)=V\oplus S(C)$ one obtains

$$\begin{aligned} G=\mu _p^{-1}(C)+\mu _p^{-1}(D)=\widetilde{C} +\widetilde{D}=W+S(D)+V+S(C)=V+W. \end{aligned}$$

Hence $G=W\oplus V$, as claimed. $\square$

3.3 Proving Theorem 1.7

We may compile results from the preceding subsections.

Proof of Theorem 1.7

Suppose first that G is neat. Then Corollary 2.27 implies that G is either torsion-free or bounded.

If G is torsion-free we can deduce the equivalence of the description of G in (i) with G being neat from Proposition 3.2.

Assume now that G is torsion and neat. Then Proposition 3.3 shows that G is as described in (ii)(B).

Conversely, let $G=A\oplus B$ for homocyclic groups A and B having exponents respectively $p^e$ and $p^{e-1}$ as in (ii)(B). Then Lemma 2.39 shows G is neat. $\square$

4 Weakly neat (compact) p-groups

Recall from Definition 1.5 that a locally compact abelian p-group is weakly neat provided every pure monothetic subgroup splits. As has been said in Remark 2.6, it is an immediate consequence of Theorem 1.1(b) that every discrete p-group is weakly neat. Furthermore, every neat locally compact abelian p-group is weakly neat. The latter class is considerably larger than the one of neat groups. In order to see this, we shall provide Proposition 4.2 below; this is a simple recipe for constructing a plethora of examples of weakly neat non-neat locally compact abelian p-groups. In contrast to Theorem 1.7, where among the locally compact abelian p-groups all neat ones are classified, a classification of the weakly neat ones has been obtained here only under the additional restriction of being separable and compact, see Corollary 4.8.

Lemma 4.1

Every finitely generated abelian compact p-group G is weakly neat.

Proof

By [10, Proposition 4.3.4], G admits a decomposition

$$\begin{aligned} G=X\oplus F \end{aligned}$$

where, for some $n\in {\mathbb {N}}$ the subgroup $X\cong {\mathbb {Z}}_p^n$ is torsion-free and $F=\textrm{tor}(G)$ is finite. Let M be a pure subgroup of G.

Suppose first that M is finite and contained in the direct summand $F=\textrm{tor}(G)$ and hence pure in F. Then, as a consequence of Theorem 1.1, there is a subgroup $R_F$ of F and a decomposition

$$\begin{aligned} F=M\oplus R_F. \end{aligned}$$

Therefore $G=M\oplus R$ for $R:=X\oplus R_F$.

Suppose from now on that M is infinite, i.e., $M\cong {\mathbb {Z}}_p$. Assume that $M+F/F\in {\mathbb {M}}(G/F)$. Then, since G/F is torsion-free, Theorem 1.7(A.1), implies that G/F is neat. Hence

$$\begin{aligned} G/F=(M+F)/F\oplus R/F \end{aligned}$$

for a suitable subgroup R of G. Hence $G=M+R$ and $M\cap R\le M\cap F=\{0\}$ implies that $G=M\oplus R$.

We finish the proof by showing that pureness of any infinite monothetic subgroup M of G implies that $(M+F)/F\in {\mathbb {M}}(G/F)$. Indeed, let $M=\overline{\langle m\rangle }$ and suppose that $(M+F)/F$ is properly contained in a maximal subgroup $(N+F)/F$ of G/F. Then there are $f\in F$, $n\in G$, and $k\ge 1$ such that

$$\begin{aligned} N/F=\overline{\langle (n+F)/F\rangle }, \ \ p^kn=m+f. \end{aligned}$$

Since $M\in {\mathbb {M}}(G)$, we must have $f\ne 0$. Hence, letting $p^l$ be the order of f, one observes

$$\begin{aligned} p^{k+l}n=p^lm\in M\cap \overline{\langle n\rangle }. \end{aligned}$$

Since, by assumption, M is pure in G there exists $h\in M$ with

$$\begin{aligned} p^{k+l}n=p^{k+l}h. \end{aligned}$$

As $h=\lambda p^sm$ for some $\lambda \in {\mathbb {Z}}_p^\times$ and $s\ge 0$ the equality

$$\begin{aligned} p^l m = p^{k+l}h=\lambda p^{k+l+s}m \end{aligned}$$

would follow, leading to

$$\begin{aligned} p^l(1-\lambda p^{k+s})m=0, \end{aligned}$$

a contradiction to m not being torsion. $\square$

Proposition 4.2

Let $G_i$ be a neat locally compact abelian p-group, for i in a finite index set I. Then $G:=\bigoplus _{i\in I}G_i$ is weakly neat.

Proof

We first claim that $G_i$ monothetic can be assumed. Let $M=\overline{\langle (x_i)_{i\in I}\rangle }$ be a pure monothetic subgroup of G. Then, for every $i\in I$, there is a maximal monothetic subgroup $M_i$ of $G_i$ containing $x_i$. It follows that $M\le \bigoplus _{i\in I}M_i$ is pure. Since the summands are monothetic, there exists $S\le G$ with

$$\begin{aligned} \bigoplus _{i\in I}{M_i}=M\oplus S. \end{aligned}$$

Since each $G_i$ is neat, there is $R_i\le G$ with $G_i=M_i\oplus R_i$, so that

$$\begin{aligned} G=\left( \bigoplus _{i\in I}M_i\right) \oplus \left( \bigoplus _{i\in I} R_i\right) . \end{aligned}$$

Then $R:=S\oplus \bigoplus _{i\in I}R_i$ serves the purpose.

By what we just claimed, we can assume that G is a finitely generated compact p-group. Hence the result follows from Lemma 4.1. $\square$

Corollary 4.3

Every compact p-group with closed torsion subgroup is weakly neat.

Proof

Since $G/\textrm{tor}(G)$ is a torsion-free projective ${\mathbb {Z}}_p$-module there is a section, i.e., for some torsion-free subgroup L one has $G=L\oplus \textrm{tor}(G)$. Taking this decomposition into account and the assumption that $\textrm{tor}(G)$ is closed, an application of Proposition 2.15, renders the direct decomposition

$$\begin{aligned} G=L\oplus \left( \bigoplus _{i\in I}{\mathbb {Z}}(p_i)^{J_i}\right) \end{aligned}$$

for some finite set I. Theorem 1.7 implies that G is a finite direct sum of neat compact p-groups. Therefore the result follows from Proposition 4.2. $\square$

Question 4.4

Is the direct sum $G_1\oplus G_2$ of weakly neat locally compact abelian p-groups $G_i$ weakly neat?

Lemma 4.5

(Proposition 2.3 in [6]) Let a first countable abelian pro-p group G be topologically generated by its torsion elements. Then there is an index set I such that

$$\begin{aligned} G=\prod _{i\in I}G_i \end{aligned}$$

for finite cyclic p-groups $G_i$.

Lemma 4.6

Let $G=\prod _{i\in I}G_i$ be the cartesian product of finite cyclic p-groups $G_i=\langle x_i\rangle$. Suppose that $\sup _{i\in I}|G_i|=\infty$. Then the closed subgroup $H:=\overline{\langle (x_i)_{i\in I}\rangle }$ is a pure procyclic subgroup of G.

Proof

Suppose that $p^ky\in H$ holds for some $y\in G$ and $k\ge 1$. Then there is $r\ge 1$ and a unit $\lambda \in {\mathbb {Z}}_p^\times$ such that $p^ky=p^r\lambda x$. We need to show that $p^ky=p^kh$ for some $h\in H$. In light of Lemma 2.7 it suffices to show that $r<k$ cannot happen. Suppose, by way of contradiction, that $r<k$ and $p^ky=p^kh$ holds. Then

$$\begin{aligned} p^ky_i=p^r\lambda x_i,\ \ \text {for all}\ \ i\in I. \end{aligned}$$

Now observe that $y_i=\mu _ix_i$ for $0\le \mu _i<|G_i|$. Thus the above system of equations reads

$$\begin{aligned} p^r(p^{k-r}\mu _i-\lambda )x_i=0, \ \ i\in I, \end{aligned}$$

and, as $p^{k-r}\mu _i-\lambda \in {\mathbb {Z}}_p^\times$ we may conclude that $p^rx_i=0$ for all $i\in I$. Since the orders $|G_i|$ are not bounded there is $i\in I$ with $|G_i|=|\langle x_i\rangle |>p^r$, showing that $p^rx_i\ne 0$, a contradiction. Hence H is a pure subgroup of G. $\square$

We next prove Theorem 1.3.

Proposition 4.7

Let G be a first countable abelian pro-p group generated by its torsion elements. The following statements are equivalent:

• (a) G is weakly neat.

• (b) G has finite exponent.

Proof

(b) implies (a) by Corollary 4.3.

For proving the converse, assume (a). By Lemma 4.5 the group G is a direct product $G=\prod _{i\in I}G_i$ of finite cyclic p-groups $G_i$ and I some index set. Letting $x_i$ denote generators of respectively $G_i$ the closed subgroup $H:=\overline{\langle (x_i)_{i\in I}\rangle }$ is pure by Lemma 4.6. On the other hand, H cannot be complemented, for if

$$\begin{aligned} G=H\oplus K \end{aligned}$$

for some closed subgroup K of G then, as $\textrm{tor}(G)\le K$ we would arrive at the contradiction $G=K$. $\square$

Corollary 4.8

For a compact p-group G with $\overline{\textrm{tor}(G)}$ separable the following statements are equivalent:

• (i) G is weakly neat.

• (ii) $\textrm{tor}(G)$ is closed.

Proof

Since G is a projective ${\mathbb {Z}}_p$-module and $G/\overline{\textrm{tor}(G)}$ is torsion-free, there is a torsion-free complement R, i.e.,

$$\begin{aligned} G=\overline{\textrm{tor}(G)}\oplus R. \end{aligned}$$

(i) $\Rightarrow$(ii). Lemma 2.19 implies that $\overline{\textrm{tor}(G)}$ must be weakly neat. Therefore Proposition 4.7 implies that $\overline{\textrm{tor}(G)}$ has finite exponent and hence $\textrm{tor}(G)$ is closed.

(ii) $\Rightarrow$(i). This is an immediate consequence of Corollary 4.3. $\square$

Remark 4.9

Example 5.2 below shows that for a non-compact locally compact abelian p-group G neither of the two conditions a) to be bounded (when $M\cong {\mathbb {Z}}(p^2)$) and b) to be torsion-free (when $M\cong {\mathbb {Z}}_p$) suffices for G to be weakly neat.

5 Examples

Most of the examples in the present section appeared in [2] in one way or another but in the book [2] they were not fully recognized in their function as posing limits to the extension of the Kulikov theory of torsion groups (in particular to torsion p-groups).

We first exhibit a compact p-group ${\mathbb {P}}$ containing a pure monothetic, and therefore, in light of Lemma 2.8, even maximal monothetic subgroup C not possessing a complement in ${\mathbb {P}}$.

Example 5.1

Define

$$\begin{aligned} (1)\qquad \qquad {\mathbb {P}}:={\mathbb {Z}}(p)\times {\mathbb {Z}}(p^2)\times {\mathbb {Z}}(p^3) \times \cdots =\prod _{n\in {\mathbb {N}}}{\mathbb {Z}}(p^n). \end{aligned}$$

Observe that ${\mathbb {P}}$ is a compact torsion generated p-group. Therefore Proposition 4.7 implies that ${\mathbb {P}}$ cannot be weakly neat and therefore Remark 2.6 implies that ${\mathbb {P}}$ is not neat. Letting $x_n$ be a topological generator of the n-th factor ${\mathbb {Z}}(p^n)$ one obtains that $(x_n)_{n\in {\mathbb {N}}}$ topologically generates a closed subgroup $M^*\in {\mathbb {M}}({\mathbb {P}})$ which is pure and has no complement (see the proof of Lemma 4.6).

If one realises ${\mathbb {Z}}(p^n)$ as $p^{-n}{\cdot }{\mathbb {Z}}/{\mathbb {Z}}$ for $n\in {\mathbb {N}}$, then we have

$$\begin{aligned} {\mathbb {Z}}(p)\le {\mathbb {Z}}(p^2)\le {\mathbb {Z}}(p^3)\le \cdots \le {\mathbb {Z}}(p^\infty ), \end{aligned}$$

and we notice $S_1({\mathbb {P}})={\mathbb {Z}}(p)^{\mathbb {N}}$ and

$$\begin{aligned} S_n({\mathbb {P}}) = {\mathbb {Z}}(p)\times \cdots \times {\mathbb {Z}}(p^{n-1}) \times {\mathbb {Z}}(p^n)\times {\mathbb {Z}}(p^n)\times \cdots , \ \ 1<n\in {\mathbb {N}}. \end{aligned}$$

The group ${\mathbb {P}}$ contains the dense subgroup

$$\begin{aligned} \textrm{tor}({\mathbb {P}})=\bigcup _{n\in {\mathbb {N}}} S_n(G) \end{aligned}$$

which properly contains the subgroup of all sequences $(z_n)_{n\in {\mathbb {N}}}$ with all but finitely many components vanishing. This subgroup is isomorphic to $\widehat{{\mathbb {P}}}$ and is dense in ${\mathbb {P}}$. Thus

$$\begin{aligned} (2)\qquad \qquad {\mathbb {P}}=\overline{\textrm{tor}({\mathbb {P}})}, \end{aligned}$$

and so ${\mathbb {P}}$ is torsion generated.

The group ${\mathbb {P}}$ contains the closed subgroup

$$\begin{aligned} (3)\qquad \qquad C:=\{(z_n)_{n{\in }{\mathbb {N}}}\in {\mathbb {P}}:z_n{\in }p^{-n}{\cdot } {\mathbb {Z}}/{\mathbb {Z}}\ \text { such that } \ (\forall n{\in }{\mathbb {N}})\ \ pz_{n+1}{=}z_n\}. \end{aligned}$$

We claim that $C\cong {\mathbb {Z}}_p$ algebraically and topologically. Indeed, as we present ${\mathbb {Z}}(p^n)={\mathbb {Z}}[\frac{1}{p^n}]/{\mathbb {Z}}$, it has generator $\frac{1}{p^n}+{\mathbb {Z}}$, and, for every $n\ge 1$, the canonical epimorphism $\pi _n:{\mathbb {Z}}(p^{n+1})\rightarrow {\mathbb {Z}}(p^n)$ sends $\frac{1}{p^{n+1}}+{\mathbb {Z}}$ to $\frac{1}{p^n}+{\mathbb {Z}}$. The p-adic integers ${\mathbb {Z}}_p$ are the inverse limit of the inverse system $({\mathbb {Z}}(p^{n+1}),\pi _n)_{n\ge 1}$ and hence, by the very definition of the inverse limit, we have the following chain of equalities:

$$\begin{aligned} {\mathbb {Z}}_p&=\varprojlim _n{\mathbb {Z}}(p^n)=\{(z_n)_{n\ge 1} \in {\mathbb {P}}:\pi _{n+1}(z_{n+1})=z_n\}\\&=\{(z_n)_{n\ge 1}\in {\mathbb {P}}:pz_{n+1}=z_n\}\\&=C. \end{aligned}$$

We claim that C is topologically generated by the element

$$\begin{aligned} x:=(p^{-n} +{\mathbb {Z}})_{n\in {\mathbb {N}}}\in {\mathbb {P}}. \end{aligned}$$

Certainly the components $z_n=\frac{1}{p^n}+{\mathbb {Z}}$ satisfy the equations $z_n=pz_{n+1}$ and hence $x\in C$. The canonical projection of ${\mathbb {P}}$ to ${\mathbb {Z}}(p^n)={\mathbb {Z}}\left[ \frac{1}{p^n}\right]$ sends x to $\frac{1}{p^n}+{\mathbb {Z}}$, a generator of ${\mathbb {Z}}(p^n)$, and so does its restriction to C. The claim follows, by letting $G:=C$ and $X:=\{x\}$ in [10, Lemma 2.4.1(a)]. We note that ${\mathbb {P}}$ is reduced and therefore cannot contain an isomorphic copy of the Prüfer group ${\mathbb {Z}}(p^\infty )$, see [5, Proposition 8.5], which is the dual of ${\mathbb {Z}}_p$, see [5, Example 1.38 and Table 1.1 in 1.39].

For showing the pureness of C in ${\mathbb {P}}$ suppose there are $g\in {\mathbb {P}}$, $k\ge 1$, $r\ge 0$ and a unit $\lambda \in {\mathbb {Z}}_p^\times$ with

$$\begin{aligned} p^kg=p^r\lambda x. \end{aligned}$$

We need to find $h'\in \overline{\langle x\rangle }$ with $p^kh'=p^kg$ and, taking Lemma 2.7 into account, we may assume that $k<r$. Then, letting $g=(a_n/p^n +{\mathbb {Z}})_{n\ge 1}$ the above equation reads

$$\begin{aligned} p^k\frac{a_n}{p^n}\equiv p^r\frac{1}{p^n}\pmod 1, \ \ n\ge 1 \end{aligned}$$

or equivalently

$$\begin{aligned} p^{k-r}(p^ka_n-\lambda )\equiv 0\pmod {p^n}, \ \ n\ge 1. \end{aligned}$$

For all $n\ge k-r$ one deduces from this that

$$\begin{aligned} p^ka_k\equiv \lambda \pmod {p^{n-k+r}} \end{aligned}$$

so that the contradiction $k=0$ arises. Hence C is pure in ${\mathbb {P}}$ and hence, by Lemma 2.8, it is a maximal monothetic subgroup of ${\mathbb {P}}$. By (2), ${\mathbb {P}}$ has a dense torsion subgroup, while $C\cong {\mathbb {Z}}_p$ is torsion-free. Therefore, by Lemma 2.12C cannot be a homomorphic retract of ${\mathbb {P}}$.

In fact, more is known about ${\mathbb {P}}$. For instance ${\mathbb {P}}$ contains a dense copy of the free ${\mathbb {Z}}_p$-module of rank $\aleph _0$, isomorphic to ${\mathbb {Z}}_p^{({\mathbb {N}})}$: see [2, Corollary 3.20]. Once more we see that the torsion group $\textrm{tor}({\mathbb {P}})$ fails to be closed.

Next we present an example of a locally compact abelian reduced p-group ${\mathbb {B}}$ possessing a maximal monothetic subgroup $M^*$ which is pure in ${\mathbb {P}}$ without a complement. By choosing below the monothetic p-group M finite, one can achieve ${\mathbb {B}}$ to have finite exponent. For the definition of a local product see e.g. [2, Definition 2.49].

Example 5.2

Let M be a monothetic p-group satisfying $p{\cdot }M\ne \{0\}$. Then either $M\cong {\mathbb {Z}}_p$ or $M={\mathbb {Z}}(p^n)$, $n\in {\mathbb {N}}$, $n\ge 2$. Note that $M/p{\cdot }M\cong {\mathbb {Z}}(p)$. Define

$$\begin{aligned} (4)\qquad \qquad {\mathbb {B}}:=(M,p{\cdot }M)^\mathrm{loc,{\mathbb {N}}}, \end{aligned}$$

the Braconnier local power of the pair $(M,p{\cdot }M)$.

Let ${\mathbb {B}}_c:=(p{\cdot }M)^{\mathbb {N}}$. Then ${\mathbb {B}}_c$ is a nonzero compact open subgroup of ${\mathbb {B}}$ which is isomorphic to ${\mathbb {Z}}_p^{\mathbb {N}}$, respectively, ${\mathbb {Z}}(p^{n-1})^{\mathbb {N}}$, $n\in {\mathbb {N}}$. The factor group ${\mathbb {B}}/{\mathbb {B}}_c\cong {\mathbb {Z}}(p)^{({\mathbb {N}})}$ is a discrete vector space over $\textrm{GF}(p)$ of dimension $\aleph _0$.

From this it follows that ${\mathbb {B}}_c=\overline{p{\mathbb {B}}}$ properly contains pG. Now let $\textbf{1}$ denote the generator of M, i.e. the identity 1 if $M={\mathbb {Z}}_p$, respectively, the element $p^{-n}{+}{\mathbb {Z}}_p$ if $M\cong {\mathbb {Z}}_p/p^n{\mathbb {Z}}_p$. We define $g\in {\mathbb {B}}_c$ by

$$\begin{aligned} (5)\qquad \qquad g=(e_n)_{n\in {\mathbb {N}}}\in {\mathbb {B}},\ \ e_n=p{\cdot }\textbf{1}. \end{aligned}$$

It is immediate that $g\in \overline{p{\mathbb {B}}}{\setminus } p{\mathbb {B}}$. By Lemma 2.20 the subgroup $M^*:=\overline{\langle g\rangle }$ turns out to be a pure maximal monothetic subgroup of ${\mathbb {B}}$ not possessing a complement in ${\mathbb {B}}$.

Remark 5.3

The Example 3.28 of [2] is reminiscent of the Example 5.2, as is Example 3.29 of [2].

Let us observe that in the line of Example 5.2 the special case

$$\begin{aligned} (6)\qquad \qquad {\mathbb {B}}= ({\mathbb {Z}}(4),2{\mathbb {Z}}(4))^\mathrm{loc, {\mathbb {N}}}, \quad 2{\mathbb {Z}}(4) \cong {\mathbb {Z}}(2); \quad {\mathbb {B}}/{\mathbb {B}}_c\cong {\mathbb {Z}}(2)^{({\mathbb {N}})}, \end{aligned}$$

a torsion group of exponent 4, is in some sense “the smallest” representative of the Example 5.2, and that

$$\begin{aligned} (7)\qquad \qquad {\mathbb {B}}= ({\mathbb {Z}}_p,p{\mathbb {Z}}_p)^\mathrm{loc, {\mathbb {N}}}, \quad p{\mathbb {Z}}_p \cong {\mathbb {Z}}_p; \quad {\mathbb {B}}/{\mathbb {B}}_c\cong {\mathbb {Z}}(p)^{({\mathbb {N}})}, \end{aligned}$$

a torsion-free locally compact p-group, is “the largest” one, given a fixed prime p.

Our next example will be weakly neat but not neat. Moreover, if M is chosen finite then it also shows that the direct sum of neat groups need not be neat (i.e., the converse of Lemma 2.19 does not hold for neat groups). One realises from Theorem 1.7 that for $M\notin \{{\mathbb {Z}}(p),{\mathbb {Z}}(p^2)\}$ the group cannot be neat and Lemma 4.1 implies that it is weakly neat.

Example 5.4

Let M be a monothetic compact p-group. The lead example is ${\mathbb {Z}}_p$; all the others are isomorphic to quotients of M. In particular, there is a quotient morphism $f:M\rightarrow {\mathbb {Z}}(p)={\mathbb {Z}}_p/p{\mathbb {Z}}_p$. Now we form

$$\begin{aligned} (1)\qquad \qquad {\mathbb {W}}:={\mathbb {Z}}(p)\times M\text { and } M^*:=\{(f(x),p{\cdot }x)): x\in M\}\cong p{\cdot }M. \end{aligned}$$

See [2, page 68, lines 5–9]. Thus an element $(t,m)\in {\mathbb {Z}}(p)\times M$ is in $p^n{\cdot }M^*$ for $n\in {\mathbb {N}}$ iff there is an $x\in M$ such that $(t,m)=p^n{\cdot }(f(x),p{\cdot }x)=(p^n{\cdot } f(x),p^{n+1}{\cdot }x)=(0,p^{n+1}{\cdot }x)$ that is,

$$\begin{aligned} (2)\qquad \qquad (\forall n\in {\mathbb {N}})\, p^n{\cdot }M^* =\{0\}\times p^{n+1}{\cdot }M=p^{n+1}{\cdot }{\mathbb {W}}. \end{aligned}$$

Since ${\mathbb {W}}$ is neither bounded nor torsion-free, as a consequence of Theorem 1.7, ${\mathbb {W}}$ cannot be neat. Indeed, one can show that $M^*$ is a maximal monothetic subgroup of ${\mathbb {W}}$ without complement. On the other hand, since ${\mathbb {W}}$ is finitely generated, by Lemma 4.1G is weakly neat.

But

$$\begin{aligned} (3)\qquad \qquad (\forall n\in {\mathbb {N}})\, p^n{\cdot }{\mathbb {W}}=\{0\}\times p^n{\cdot }M. \end{aligned}$$

Accordingly, an element $(f(x),p{\cdot }x)\in M^*$ is in $p^n{\cdot }{\mathbb {W}}$ iff $(f(x),p{\cdot }x)=p^n{\cdot }(t,m)=(0,p^n.m)$ for an element $(t,m)\in {\mathbb {Z}}(p)\times M$ iff $p{\cdot }x=p^n{\cdot }m$, and so

$$\begin{aligned} (4)\qquad \qquad (\forall n\in {\mathbb {N}})\, p^n{\cdot }{\mathbb {W}}\cap M^*= \{0\} \times p^n{\cdot }M. \end{aligned}$$

Claim (a) $M^*$ is maximal.

Let x be the generator of M and $y=p^{-1} + {\mathbb {Z}}\in {\mathbb {Z}}(p)=p^{-1}{\mathbb {Z}}/{\mathbb {Z}}$. Then $(y,p{\cdot }x)$ is a generator of M. If the Claim were false, there would be some $(v,u)\in {\mathbb {W}}$ such that $p{\cdot }(v,u)=(y,p{\cdot }x)$, and in particular $0=p.v=y$ which is impossible.

Claim (b) $M^*$ pure in ${\mathbb {W}}$ iff $M={\mathbb {Z}}(p)$ or ${\mathbb {Z}}(p^2)$. For all other monothetic groups M, the subgroup $M^*$ of ${\mathbb {W}}$ fails to be pure.

$M^*$ is a pure subgroup iff

$$\begin{aligned} (5)\qquad \qquad p^n{\cdot }{\mathbb {W}}\cap M^* \le p^n{\cdot } M^*. \end{aligned}$$

holds for all $n\in {\mathbb {N}}$ such that $p^n{\cdot }M\ne \{0\}$. By (2) and (4), however, (5) holds iff

$$\begin{aligned} (6)\qquad \qquad p^n{\cdot }M \le p^{n+1}{\cdot }M=p{\cdot }(p^n{\cdot }M). \end{aligned}$$

Thus whenever $p^n{\cdot }M\ne \{0\}$, (6) fails. If $M={\mathbb {Z}}(p)$, then (6) holds, since $p^n{\cdot }{\mathbb {Z}}(p)=\{0\}$ for all n. If $M={\mathbb {Z}}(p^2)$, then $p^n{\cdot }M=\{0\}$ for $n>1$.

Claim (c) $M^*$ does not split in ${\mathbb {W}}$ with the exception of the cases $M={\mathbb {Z}}(p)$ and $M={\mathbb {Z}}(p^2)$, in which it does.

If $M={\mathbb {Z}}(p)$, then ${\mathbb {W}}$ is a vector space over $\textrm{GF}(p)$ and every monothetic subgroup is a one-dimensional vector subspace which splits. If $M={\mathbb {Z}}(p^2)$, then ${\mathbb {W}}=(\{0\}\times M)+W^*$ and $M^*\cong p{\cdot }{\mathbb {Z}}(p)$.

Since homomorphic retracts are pure, the remainder follows from Claim (b). The smallest example would be $G:={\mathbb {Z}}(2)\oplus {\mathbb {Z}}(8)=\langle x\rangle \oplus \langle y\rangle$ and $M:=\langle x+2y\rangle$.

All of these examples illustrate that maximal monothetic subgroups need not be pure and thus may not split. The set of examples displayed in Example 5.4 shows that this holds even for finite p-groups of rank 2.

Example 5.1 shows that even torsion generated compact p-groups exhibit in the simplest case certain properties which needed to be avoided if monothetic subgroups are to act as a central tool, and Example 5.2 shows that this remains true even for torsion groups with finite exponent if we abandon the hypothesis of compactness.