Indecomposable Decompositions of Torsion-free Abelian Groups

TL;DR

This paper investigates the possible partitions of a torsion-free abelian group's rank that can result from its indecomposable decompositions, aiming to characterize which sets of partitions are realizable.

Contribution

It provides a characterization of the sets of partitions of the group's rank that can occur from indecomposable decompositions of torsion-free abelian groups.

Findings

Identifies conditions under which certain partitions can arise

Characterizes all possible sets of partitions for given ranks

Provides a framework for understanding indecomposable decompositions

Abstract

An indecomposable decomposition of a torsion-free abelian group $G$ of rank $n$ is a decomposition $G=A_1\oplus\cdots\oplus A_t$ where $A_i$ is indecomposable of rank $r_i$ so that $\sum_i r_i=n$ is a partition of $n$. The group $G$ may have decompositions that result in different partitions of $n$. We address the problem of characterising those sets of partitions of $n$ which can arise from indecomposable decompositions of a torsion-free abelian group.

Equations

$$\mathcal{S}(n,k)=\{P\in \mathcal{P}(n)\mid \overline{P}\le k\text{ and }k+|P|\le n+1\}.$$

$$X=A+\mathbb{Z}{e}^{-1}a,A=\underset{\rho \in {\mathrm{T}}_{\mathrm{cr}}(A)}{⨁}{A}_{\rho },a=\sum _{\rho \in {\mathrm{T}}_{\mathrm{cr}}(A)}{a}_{\rho },a_{\rho }\in A_{\rho }$$

$$[X:A]=e\text{ if and only if }{\gcd }^A(e,a)=1,$$

$$A=R(X)\text{ if and only if for all }\tau \in {\mathrm{T}}_{\mathrm{cr}}(A),A(\tau )=A_{\tau }=X(\tau ).\text{ (cite[cite][@@bibrefMA00, 4.5.1])}$$

$${\mu }_{\tau }(X)=\mathrm{ord}(\overline{a_{\tau }})=\mathrm{ord}(a_{\tau }+eA).$$

$$\frac{X}{R}\cong \frac{X_1}{R_1}\oplus \frac{X_2}{R_2}$$

$${\mu }_{\tau }(X)={\mu }_{\tau }(X_1)\cdot {\mu }_{\tau }(X_2).$$

$$e={\alpha }_{\rho }{\beta }_{\rho },u_{\rho }={\alpha }_{\rho }{v}_{\rho },\text{and}{\gcd }^R({\beta }_{\rho },v_{\rho })=1.$$

$$X=A+\frac{\mathbb{Z}}{e}({\alpha }_1{v}_1+\cdots +{\alpha }_r{v}_r)$$

$$X=({\tau }_1{v}_1+{\tau }_2{v}_2+{\tau }_3{v}_3+{\tau }_4{v}_4+{\tau }_5{v}_5)+\frac{\mathbb{Z}}{e}(3^2{v}_1+7v_2+3v_3+5v_4+v_5)$$

$$R=\tau u_1\oplus \cdots \oplus \tau u_k\oplus {\tau }_1{v}_1\oplus \cdots \oplus {\tau }_{n-k}{v}_{n-k},$$

$$X=R+\frac{\mathbb{Z}}{q_1}(u_1+v_1)+\frac{\mathbb{Z}}{q_2}(u_1+v_2)+\cdots +\frac{\mathbb{Z}}{q_{n-k}}(u_1+v_{n-k}).$$

$$X=R_0\oplus \left(R_1+\frac{\mathbb{Z}}{e}((e_1+\cdots +e_{n-k})u_1+e_2{v}_2+\cdots +e_{n-k}{v}_{n-k})\right)\in \mathcal{B}$$

$$X_1\oplus X_2=(R_1\oplus R_2)+\frac{\mathbb{Z}}{e_1{e}_2}(e_2{a}_1+e_1{a}_2)\text{ and }R(X_1\oplus X_2)=R_1\oplus R_2.$$

$$A_1$$

$$A_2$$

$$X_1=A_1+\frac{\mathbb{Z}}{e_1}({\alpha }_1{v}_1+{\alpha }_2{v}_2+\cdots +{\alpha }_d{v}_d+{\alpha }_{d+1}{v}_{d+1}+\cdots +{\alpha }_r{v}_r),$$

$$X_2=A_2+\frac{\mathbb{Z}}{e_2}({\beta }_1{w}_1+{\beta }_2{w}_2+\cdots +{\beta }_d{w}_d+{\beta }_{r+1}{v}_{r+1}+\cdots +{\beta }_s{v}_s)$$

$$A={\tau }_1{v}_1^{\prime }\oplus \cdots \oplus {\tau }_d{v}_d^{\prime }\oplus {\tau }_{d+1}{v}_{d+1}\oplus \cdots \oplus {\tau }_s{v}_s,$$

$$X_1\oplus X_2$$

$$Y$$

$$X_1\oplus X_2=(A_1\oplus A_2)+\mathbb{Z}\frac{1}{e_1{e}_2}\left({\sum }_{i=1}^d(e_2{v}_i+e_1{w}_i)+{\sum }_{i=d+1}^r{e}_2{\alpha }_i{v}_i+{\sum }_{i=r+1}^s{e}_1{\beta }_i{v}_i\right)$$

$$[{\tau }_1\dots {\tau }_d{\tau }_{d+1}\dots {\tau }_r]\oplus [{\tau }_1\dots {\tau }_d{\tau }_{r+1}\dots {\tau }_s]=[{\tau }_1\dots {\tau }_d{\tau }_{d+1}\dots {\tau }_s]\oplus [{\tau }_1]\oplus \cdots \oplus [{\tau }_d]$$

$$X_1=({\tau }_1{v}_1\oplus {\tau }_2{v}_2)+\mathbb{Z}\frac{1}{p_1}(v_1+v_2)\text{ and }{X}_2=({\tau }_1{w}_1\oplus {\tau }_2{w}_2)+\mathbb{Z}\frac{1}{p_2}(w_1+w_2).$$

$$\textstyle\begin{array}[]{cc}X=X_{1}\oplus X_{2}=\left(\tau_{1}(-u_{1}v_{1}+u_{2}w_{1})\oplus\tau_{2}(-u_{1}v_{2}+u_{2}w_{2})\right)\oplus\\ \oplus\left(\left(\tau_{1}(p_{2}v_{1}+p_{1}w_{1})\oplus\tau_{2}(p_{2}v_{2}+p_{1}w_{2})\right)+{\mathbb{Z}}\frac{1}{p_{1}p_{2}}\left((p_{2}v_{1}+p_{1}w1)+(p_{2}v_{2}+p_{1}w_{2})\right)\right),\end{array}$$

$${\mu }_{{\tau }_1}(X)=p_1{p}_2{p}_3,{\mu }_{{\tau }_2}(X)=p_2{p}_3,{\mu }_{{\tau }_3}(X)=p_1.$$

$${\mu }_{{\tau }_1}(Y)=p_1{p}_2{p}_3,{\mu }_{{\tau }_2}(Y)=p_2{p}_3,{\mu }_{{\tau }_3}(Y)=p_1.$$

$$\mathcal{S}(6,4)=\{(4,2),(4,1,1),(3,3),(3,2,1),(2,2,2)\}.$$

$${\mu }_{{\tau }_1}(X)=5\cdot 7\cdot 11,{\mu }_{{\tau }_2}(X)=5\cdot 11\cdot 3,{\mu }_{{\tau }_3}(X)=7,{\mu }_{{\tau }_4}(X)=3.$$

$${\mu }_{{\tau }_1}(Y)$$

$${\mu }_{{\tau }_3}(Y)$$

$${\mu }_{{\tau }_1}(Z)$$

Full text

1. Introduction

All the groups considered in this paper belong to the class $\mathcal{G}$ of finite rank torsion–free abelian groups; we shall refer to them simply as ‘groups’. Any group $G\in\mathcal{G}$ is viewed as an additive subgroup of a finite dimensional ${\mathbb{Q}}$–vector space and its rank, $\operatorname{rk}(G)$, is the dimension of the subspace ${\mathbb{Q}}G$ that $G$ generates. If a group $G$ of rank $n$ has an indecomposable decomposition $G=A_{1}\oplus\cdots\oplus A_{t}$ where $A_{i}$ is indecomposable of rank $r_{i}$, then $\sum_{i}r_{i}=n$ so $P=(r_{1},\ldots,r_{t})$ is an unordered partition of $n$ into $t$ parts, and we say that $G$ realizes $P$.

A group $G$ may realize several different partitions and we denote by $\mathcal{P}(G)$ the set of all partitions that are realized by $G$. Determining $\mathcal{P}(G)$ for a given group $G$ and characterizing realizable sets of partitions have received a great deal of attention in the literature. For example, [7, Section 90] displays groups realizing $(1,3)$ and $(2,2)$, groups realizing partitions into $2$ and $t$ parts for each $2\leq t<n$, and groups realizing every partition of $n$ into $t$ parts (Corner’s Theorem 2.1 below).

On the other hand, there are classes of groups, such as completely decomposable groups, direct sums of indecomposable fully invariant subgroups, and almost completely decomposable groups with primary regulating index (see [6, Theorem 3.5]) that realize a unique partition. Furthermore, some sets of partitions are incompatible, in the sense that no group can realize them. For example, we showed in [11, Theorem 3.5] (Theorem 2.3) that no group can realize both $(k,1^{n-k})$ and $(m,1^{n-m})$ if $k\neq m$, where $(i,1^{j})$ stands for the partition $(i,1,\ldots,1)$ with $j$ copies of $1$. In this paper, we consider three problems:

1. (1)

Characterize families of partitions that can be realized by some group. 2. (2)

For a given family $\mathcal{P}$ of partitions of $n$ that can be realized, determine the groups $G$ that realize $\mathcal{P}$. 3. (3)

For a given class $\mathcal{A}$ of groups, determine the families $\mathcal{P}(G)$ of partitions realized by $G\in\mathcal{A}$.

For the most part, we adopt the notation of [7]. The definitions and properties of almost completely decomposable, rigid and block–rigid cyclic regulator quotient groups can be found in [10]. The properties we need are described in Section 2.

This paper contains the following results. By $\mathcal{B}$ we denote the class of all block–rigid almost completely decomposable groups with cyclic regulator quotient.

1. (1)

The “Hook condition” is shown to be necessary for the sets of partitions $\mathcal{P}(X)$ of groups $X\in\mathcal{B}$ (Proposition 3.1(1)). 2. (2)

A useful “standard” description of $\mathcal{B}$–groups is established (Lemma 5.12). 3. (3)

The workhorse Lemma 6.7 describes the direct sum of two $\mathcal{B}$–groups given in standard description. 4. (4)

General obstructions to realization are established (Lemma 4.2, Lemma 4.3, Lemma 6.1). 5. (5)

The maximal sets of partitions of $n$ that satisfy the Hook condition are described and denoted by $\mathcal{S}(n,k)$ (Proposition 3.3). The (few) $\mathcal{S}(n,k)$ that can be realized by groups in $\mathcal{B}$ and the (many) that cannot so realized are determined (Theorem 7.2).

2. Background

The decomposition properties of torsion–free groups of finite rank in general are extremely complex (see [4] and [7]). For example, A. L. S. Corner [5] proved an intriguing theorem on ‘pathological’ decompositions:

Theorem 2.1**.**

(Corner’s Theorem)* Given integers $n\geq k\geq 1$, there exists a group $G$ that realizes every partition of $n$ into $k$ parts.*

Specializations and simplifications are needed in order to have any hope for results.

Our essential specialization consists in passing to the subclass of almost completely decomposable groups, first defined and systematically investigated by Lady ([8]), and further to the much more special but still interesting and rich subclass $\mathcal{B}$ of block rigid groups with cyclic regulator quotient. To our knowledge all published examples of groups with “pathological” decompositions are almost completely decomposable groups. The essential simplification occurs by passing from isomorphism to Lady’s near–isomorphism ([9], [10, Chapter 9]). A theorem by David Arnold ([2, Corollary 12.9], [10, Theorem 10.2.5]) says that passing from isomorphism to the coarser equivalence relation of Lady does not mean a loss of generality for our line of questioning: if $G$ and $G^{\prime}$ are nearly isomorphic groups, then $\mathcal{P}(G)=\mathcal{P}(G^{\prime})$.

A completely decomposable group $A$ is the direct sum of rank–$1$ groups. We always use its ‘homogeneous decomposition’ $A=\bigoplus_{\rho\in\operatorname{T_{cr}}(A)}A_{\rho}$ where $A_{\rho}$ is a non–zero direct sum of rank–$1$ groups all of type $\rho$. If so, the index set $\operatorname{T_{cr}}(A)$ is the critical typeset of $A$. An almost completely decomposable group is a group containing a completely decomposable subgroup of finite index. Let $X$ be an almost completely decomposable group. Then $X$ contains completely decomposable subgroups of least index in $X$, and these are Lady’s regulating subgroups. The intersection of all regulating subgroups is a fully invariant completely decomposable subgroup of $X$, Burkhardt’s regulator $\operatorname{R}(X)$ of $X$. Let $\operatorname{R}(X)=\bigoplus_{\rho\in\operatorname{T_{cr}}(\operatorname{R}(X))}R_{\rho}$ be the homogeneous decomposition of $R$. An almost completely decomposable group $X$ is block–rigid if $\operatorname{T_{cr}}(\operatorname{R}(X))$ is an antichain and rigid if $\operatorname{T_{cr}}(X)$ is an antichain and for all $\rho\in\operatorname{T_{cr}}(X)$, $\operatorname{rk}R_{\rho}=1$. We denote by $\mathcal{B}$ the class of all block–rigid groups for which $X/\operatorname{R}(X)$ is cyclic.

Let $G$ and $H$ be groups. We say that $G$ is nearly isomorphic to $H$, denoted $G\cong_{\text{\rm nr}}H$, if there is a group $K$ such that $G\oplus K\cong H\oplus K$. The properties of near isomorphism are described in [2, Section 12] and in [10, Chapter 10]. In our context, the most important such property is [2, Corollary 12.9], [10, Theorem 10.2.5]: for any groups $G$ and $H$, if $G\cong_{\text{\rm nr}}H$ and $G=G_{1}\oplus\cdots\oplus G_{t}$, then $H=H_{1}\oplus\cdots\oplus H_{t}$ with $H_{i}\cong_{\text{\rm nr}}G_{i}$, in particular $\mathcal{P}(G)=\mathcal{P}(H)$.

E. Blagoveshchenskaya proved an interesting obstruction to realization ([4], [10, Chapter 13]) which we quote as Theorem 2.2.

Theorem 2.2**.**

([4]) Let $(r_{1},\ldots,r_{s})$ and $(r^{\prime}_{1},\ldots,r^{\prime}_{t})$ be two partitions of $n$. Then there exists a block–rigid crq–group realizing $(r_{1},\ldots,r_{s})$ and $(r^{\prime}_{1},\ldots,r^{\prime}_{t})$ if and only if $r_{i}\leq n-t+1$ for all $i=1,\ldots,s$ and $r^{\prime}_{j}\leq n-s+1$ for all $j=1,\ldots,t$.

Another obstruction to realization is implicit in the “Main Decomposition Theorem” below. A group is clipped if it does not have a rank–one summand. Clearly any group in $\mathcal{G}$ has a completely decomposable direct summand of maximal rank (that may be [math]) and the complementary summand is necessarily clipped.

Theorem 2.3**.**

([11, Theorem 2.5])* Let $G\in\mathcal{G}$ and let $G=G_{cd}\oplus G_{cl}$ be a decomposition of $G$ with completely decomposable summand $G_{cd}$ and clipped summand $G_{cl}$. Then $G_{cd}$ is unique up to isomorphism and $G_{cl}$ is unique up to near–isomorphism, i.e., if also $G=G_{cd}^{\prime}\oplus G_{cl}^{\prime}$ is such that $G_{cd}^{\prime}$ is completely decomposable and $G_{cl}^{\prime}$ is clipped, then $G_{cd}\cong G_{cd}^{\prime}$ and $G_{cl}\cong_{\text{\rm nr}}G_{cl}^{\prime}$.*

Corollary 2.4**.**

If $G$ realizes $(r,1^{n-r})$ and $(r^{\prime},1^{n-r^{\prime}})$, then $r=r^{\prime}$.

3. Hooked Partitions

Recall that $\mathcal{P}(G)$ denotes the set of all partitions $P=(r_{1},\ldots,r_{t})$ of $\operatorname{rk}(G)$ for which there is an indecomposable decomposition $G=G_{1}\oplus\cdots\oplus G_{t}$ such that for all $i\in\{1,\dots,t\}$, $r_{i}=\operatorname{rk}(G_{i})$. There are two extreme values of evident interest: Let $r^{G}$ be the largest rank of an indecomposable summand of $G$, and let $t^{G}$ be the largest number of summands in an indecomposable decomposition of $G$. Intuitively, if $r^{G}$ is large, then $t^{G}$ should be small and conversely. For example, if $r^{G}=\operatorname{rk}(G)$, then $t^{G}=1$, $G$ is indecomposable and $\mathcal{P}(G)=\{(\operatorname{rk}(G))\}$; if $t^{G}=\operatorname{rk}(G)$, then $r^{G}=1$, $G$ is completely decomposable and $\mathcal{P}(G)=\{(1,\ldots,1)\}$. Thus if $G$ is indecomposable or completely decomposable, then $r^{G}+t^{G}=\operatorname{rk}(G)+1$.

In view of Corner’s Theorem denote by $\mathcal{C}(n,k)$ the set of all partitions of $n$ into $k$ parts. The following proposition describes restrictions on realizable partitions.

Proposition 3.1**.**

1. (1)

For any $G\in\mathcal{B}$, it is true that $r^{G}+t^{G}\leq\operatorname{rk}(G)+1$. 2. (2)

For any $G\in\mathcal{G}$ with $\mathcal{P}(G)=\mathcal{C}(n,k)$ it is true that $r^{G}+t^{G}=n+1$.

Proof.

(1) Let $G$ be a $\mathcal{B}$–group of rank $n$ and suppose that $G=C\oplus D=B_{1}\oplus\cdots\oplus B_{t}$ are decompositions of $G$ with $C$ indecomposable, $r^{G}=\operatorname{rk}C$ and all $B_{i}$ indecomposable with $t=t^{G}$. According to Theorem 2.2 $r^{G}\leq n-t^{G}+1$ or $r^{G}+t^{G}\leq n+1$.

(2) For a group $G$ with $\mathcal{P}(G)=\mathcal{C}(n,k)$ we have $t^{G}=k$ and as $(n-k+1,1^{k-1})\in\mathcal{C}(n,k)$ we have $r^{G}=n-k+1$, hence $r^{G}+t^{G}=n+1$. ∎

Let $\mathcal{P}(n)$ be the set of all partitions of $n$ and let $\mathcal{P}\subseteq\mathcal{P}(n)$, so $\mathcal{P}$ is a family of partitions of $n$. A hook in $\mathcal{P}(n)$ is a partition $(k,1^{n-k})$ where $k\in\{1,\ldots,n\}$. Let $r^{\mathcal{P}}$ be the largest entry of any partition in $\mathcal{P}$ and let $t^{\mathcal{P}}$ be the length of the longest partition in $\mathcal{P}$. We say that $\mathcal{P}$** satisfies the Hook Condition** or simply that $\mathcal{P}$** is hooked** if $r^{\mathcal{P}}+t^{\mathcal{P}}\leq n+1$.

By Proposition 3.1 the family $\mathcal{P}(G)$ is hooked if $G\in\mathcal{B}$ or if $\mathcal{P}(G)=\mathcal{C}(n,k)$ for some $k$.

Question 3.2**.**

Is $\mathcal{P}(G)$ hooked for every $G\in\mathcal{G}$?

We conjecture, but so far have no proof that the Hook Condition is indeed necessary for $\mathcal{P}(G)$. On the other hand, we show below that the Hook Condition is far from sufficient to characterize $\mathcal{P}(G)$.

Let $P\in\mathcal{P}(n)$, so $P=(r_{1},r_{2},\cdots,r_{t})$, $r_{i}\geq 1$, $\sum r_{i}=n$. Then $\overline{P}$ denotes the largest term of the partition and $|P|=t$ is its length.

For $n\geq 1$ and $1\leq k\leq n$ we define

[TABLE]

The sets $S(n,k)$ are well known in Partition Theory under the name of ‘restricted partitions’. They are studied, enumerated and tabulated for example in [1, Section 14.7].

Proposition 3.3**.**

1. (1)

The family of partitions $\mathcal{S}(n,k)$ is hooked, contains a unique hook $(k,\,1^{n-k})$ and is maximal with respect to inclusion among the hooked families of partitions of $n$. 2. (2)

If a group $G$ realizes $\mathcal{S}(n,k)$ and $\mathcal{P}(G)$ is hooked, then $\mathcal{P}(G)=\mathcal{S}(n,k)$. 3. (3)

If a group $G$ realizes $\mathcal{C}(n,k)$ and $\mathcal{P}(G)$ is hooked, then $\mathcal{P}(G)=\mathcal{C}(n,k)$.

Proof.

(1) [1, Section 14.7].

(2) By hypothesis $\mathcal{S}(n,k)\subseteq\mathcal{P}(G)$. As $\mathcal{P}(G)$ is hooked and $\mathcal{S}(n,k)$ is a maximal hooked family of partitions we must have equality.

(3) Suppose that $\mathcal{C}(n,k)\subseteq\mathcal{P}(G)$ and $\mathcal{P}(G)$ is hooked. Then $t^{G}\geq k$ and as $(n-k+1,1^{k-1})\in\mathcal{C}(n,k)\subseteq\mathcal{P}(G)$, further $r^{G}\geq n-k+1$. Now $\mathcal{P}(G)$ being hooked we get $n+1\geq r^{G}+t^{G}\geq(n-k+1)+k=n+1$ and therefore $r^{G}+t^{G}=n+1$. Furthermore $r^{G}=n-k+1$ and $t^{G}=k$ because either $r^{G}>n-k+1$ or $t^{G}>k$ contradicts the Hook condition. ∎

Question 3.4**.**

Suppose $G\in\mathcal{G}$ realizes $(k,1^{n-k})$; then is $r^{G}=k$ and/or $t^{G}=n-k+1$?

The answer is yes for a group $G$ with $\mathcal{P}(G)=\mathcal{C}(n,k)$ and any group realizing some $\mathcal{S}(n,k)$. On the other hand, it is not known if there are groups realizing, for example, both $(6,1,1)$ and $(2,2,2,2)$.

4. Fully invariant direct summands

Recall that $\mathcal{P}(X)$ denotes that set of partitions of $\operatorname{rk}(X)$ that can be realized by indecomposable decompositions of $X$. Recall that $\mathcal{S}(n,k)=\{P\in\mathcal{P}(n)\mid\overline{P}\leq k\text{ and }k+|P|\leq n+1\}$. For an indecomposable decomposition $X=\bigoplus_{\rho}X_{\rho}$ let $\operatorname{ptn}(\bigoplus_{\rho}X_{\rho})=(\ldots,\operatorname{rk}(X_{\rho}),\ldots)\in\mathcal{P}(\operatorname{rk}X)$.

Proposition 4.1**.**

Suppose that $X=Y\oplus Z$ and that both $X$ and $Y$ are fully invariant in $X$. Then the following statements are true.

1. (1)

Suppose that $X=\bigoplus_{\omega\in\Omega}X_{\omega}$ is an indecomposable decomposition of $X$. Then there is a disjoint union $\Omega=\Omega_{Y}\cup\Omega_{Z}$ such that $Y=\bigoplus_{\omega\in\Omega_{Y}}X_{\omega}$ and $Z=\bigoplus_{\omega\in\Omega_{Z}}X_{\omega}$. 2. (2)

$\mathcal{P}(X)=\mathcal{P}(Y)\times\mathcal{P}(Z)$.

Proof.

(1) As $Y$ and $Z$ are fully invariant in $X$ it follows that $Y=\bigoplus_{\omega}(Y\cap X_{\omega})$ and $Z=\bigoplus_{\omega}(Z\cap X_{\omega})$. It is easily checked that $X_{\omega}=Y\cap X_{\omega}\oplus Z\cap X_{\omega}$. As $X_{\omega}$ is indecomposable it follows that either $Y\cap X_{\omega}=0$ and $Z\cap X_{\omega}=X_{\omega}$ or $Y\cap X_{\omega}=X_{\omega}$ and $Z\cap X_{\omega}=0$. Let $\Omega_{Y}=\{\omega\in\Omega\mid Y\cap X_{\omega}=X_{\omega}\}$ and $\Omega_{Z}=\{\omega\in\Omega\mid Z\cap X_{\omega}=X_{\omega}\}$. Then $Y=\bigoplus_{\omega\in\Omega_{Y}}X_{\omega}$ and $Z=\bigoplus_{\omega\in\Omega_{Z}}X_{\omega}$ as claimed.

(2) It follows from (1) that $\operatorname{ptn}(\bigoplus_{\omega\in\Omega}X_{\omega})=\operatorname{ptn}(\bigoplus_{\omega\in\Omega_{Y}}X_{\omega})\times\operatorname{ptn}(\bigoplus_{\omega\in\Omega_{Z}}X_{\omega})$, so $\mathcal{P}(X)\subseteq\mathcal{P}(Y)\times\mathcal{P}(Z)$. Conversely, if $Y=\bigoplus_{\omega\in\Omega_{Y}}Y_{\omega}$ and $Z=\bigoplus_{\omega\in\Omega_{Z}}Z_{\omega}$ are indecomposable decompositions of $Y$ and $Z$, then $X=(\bigoplus_{\omega\in\Omega_{Y}}Y_{\omega})\oplus(\bigoplus_{\omega\in\Omega_{Z}}Z_{\omega})$ is an indecomposable decomposition of $X$ and

$\operatorname{ptn}((\bigoplus_{\omega}Y_{\omega})\oplus(\bigoplus_{\omega}Z_{\omega}))=\operatorname{ptn}(\bigoplus_{\omega}Y_{\omega})\times\operatorname{ptn}(\bigoplus_{\omega}Z_{\omega})$, so $\mathcal{P}(Y)\times\mathcal{P}(Z)\subseteq\mathcal{P}(X)$. ∎

Lemma 4.2 and Lemma 4.3 exhibit obstructions to realization. Assuming that certain decompositions are present, they say that certain other decompositions and certain critical typesets are impossible (see Lemma 6.2).

Lemma 4.2**.**

Let $G\in\mathcal{G}$ and suppose that $H$ is a non–zero indecomposable fully invariant summand of $G$. Then $H$ appears as a summand in any indecomposable decomposition of $G$.

Proof.

Let $G=G_{1}\oplus\cdots\oplus G_{t}$ be an indecomposable decomposition of $G$. As $H$ is fully invariant in $G$ we have $H=H\cap G_{1}\oplus\cdots\oplus H\cap G_{t}$. As $H$ is indecomposable, without loss of generality $H=H\cap G_{1}$, and moreover $G_{1}=H\oplus H^{\prime}$. But $G_{1}$ is also indecomposable and so it follows that $G_{1}=H$. ∎

Lemma 4.3**.**

Let $G\in\mathcal{G}$ and suppose that $H$ is a non–zero completely decomposable $\tau$–homogeneous fully invariant summand of $G$. Then every indecomposable decomposition of $G$ contains a rank–$1$ summand of type $\tau$.

Proof.

Let $G=G_{1}\oplus\cdots\oplus G_{t}$ be an indecomposable decomposition of $G$. As $H$ is fully invariant in $G$ we have $H=H\cap G_{1}\oplus\cdots\oplus H\cap G_{t}$. Without loss of generality $H\cap G_{1}\neq 0$, and moreover $G_{1}=(H\cap G_{1})\oplus H^{\prime}$. But $G_{1}$ is indecomposable and so it follows that $G_{1}=H\cap G_{1}$. This means that $G_{1}\subseteq H$ and furthermore $G_{1}$ is a direct summand of the $\tau$–homogeneous completely decomposable group $H$, hence is itself $\tau$–homogeneous completely decomposable. Being indecomposable, $G_{1}$ has rank $1$ and type $\tau$. ∎

5. $\mathcal{B}$–groups

This section lists essential details on block–rigid groups $X$ with cyclic regulator quotient, i.e., groups in $\mathcal{B}$, also referred to as ‘$\mathcal{B}$–groups ’or ‘block–rigid crq–groups’. We will use greatest common divisors $\gcd^{A}(e,a)$ where $e$ is a positive integer and $a$ is an element of the torsion-free group $A$. Greatest common divisors are explained in [10, Section 11.2] in great generality and they have the usual properties. In our special case $\gcd^{A}(e,a)=d$ when $m=\operatorname{ord}(a+eA)$ and $e=md$.

A $\mathcal{B}$–group $X$ has a representation

[TABLE]

with the following properties

[TABLE]

[TABLE]

There is a more practical version of the Regulator Criterion (5.3).

Lemma 5.4**.**

(Regulator Criterion)* Assume that $\gcd^{A}(e,a)=1$ in (5.1). Then $A=\operatorname{R}(X)$ if and only if for all $\tau\in\operatorname{T_{cr}}(A),\ \gcd^{A}(e,\sum_{\rho\neq\tau}a_{\rho})=1$.*

Proof.

Set $B:=A_{\tau}$, $C=\bigoplus_{\rho\neq\tau}A_{\rho}$ and $c=\sum_{\rho\neq\tau}a_{\rho}$. Then $A=B\oplus C$. By the Purification Lemma ([10, 3.1.2,11.4.1]), $B_{*}^{X}=B+{\mathbb{Z}}e_{\tau}^{-1}a_{\tau}$ where $e_{\tau}=\gcd^{A}(e,c)$.

So $e_{\tau}\mid e$ and $c=e_{\tau}c^{\prime}$ for some $c^{\prime}\in A$. Suppose that $A=\operatorname{R}(X)$. Then $B=X(\tau)$ is pure in $X$, hence $e_{\tau}^{-1}a_{\tau}\in A_{\tau}\subseteq A$ and $e_{\tau}(e_{\tau}^{-1}a_{\tau}+c^{\prime})=a_{\tau}+c=a$. By assumption $\gcd^{A}(e,a)=1$, so $e_{\tau}=1$. Conversely, if $\forall\,\tau\in\operatorname{T_{cr}}(A):e_{\tau}=1$, then $\forall\,\tau\in\operatorname{T_{cr}}(A):A_{\tau}=\left(A_{\tau}\right)_{*}^{X}=X(\tau)$ and $A=\operatorname{R}(X)$. ∎

We next restate well–known invariants and the classification of $\mathcal{B}$–groups.

Definition 5.5**.**

[10, Definition 12.6.2]* Let $X$ be as in (5.1). Let $e$ be any positive integer such that $eX\subset A$. Let $\,\overline{\phantom{n}}\,\>A\rightarrow A/eA=\overline{A}$ be the natural epimorphism. For any type $\tau$, set*

[TABLE]

The values $\mu_{\tau}(X)$, do not depend on the choices of $a$ and $e$ involved in their definition ([10, Lemma 12.6.3]).

The classification theorem for block–rigid crq–groups is easy to state.

Theorem 5.6**.**

[10, Theorem 12.6.5] (Near–Classification)* Let $X$ and $Y$ be block–rigid crq–groups. Then $X\cong_{\text{\rm nr}}Y$ if and only if $\operatorname{R}(X)\cong\operatorname{R}(Y)$ and $\mu_{\tau}(X)=\mu_{\tau}(Y)$ for all types $\tau$. ∎*

Theorem 5.7 says in particular that direct summands of $\mathcal{B}$–groups are again $\mathcal{B}$–groups and relates the invariants of group and summands.

Theorem 5.7**.**

[10, Theorem 13.1.1] (Product Formula)* Let $X$ be a block–rigid crq–group with regulator $R$. Suppose $X=X_{1}\oplus X_{2}$ is a direct decomposition of $X$. Then each $X_{i}$ is a block–rigid crq–group with regulator $R_{i}=X_{i}\cap R$, and*

[TABLE]

is a decomposition of the cyclic group $X/R$. The groups $X_{i}/R_{i}$ are cyclic and have relatively prime orders. Furthermore,

[TABLE]

∎

Lemma 5.8 deals with direct sums in $\mathcal{B}$ and their regulators. This lemma justifies the direct sums that we use in the later sections.

Lemma 5.8**.**

Let $X_{1},X_{2}\in\mathcal{B}$ with regulators $R_{1},R_{2}$ respectively. Assume that $\operatorname{T_{cr}}(X_{1})\cup\operatorname{T_{cr}}(X_{2})$ is an antichain and $\gcd([X_{1}:R_{1}],[X_{2}:R_{2}])=1$. Then $X=X_{1}\oplus X_{2}\in\mathcal{B}$ and $\operatorname{R}(X)=R_{1}\oplus R_{2}$.

Proof.

Set $R=R_{1}\oplus R_{2}$. Then $\frac{X}{R}\cong\frac{X_{1}}{R_{1}}\oplus\frac{X_{2}}{R_{2}}$ is a finite cyclic group of order $[X_{1}:R_{1}][X_{2}:R_{2}]$ and $\operatorname{T_{cr}}(X)=\operatorname{T_{cr}}(R)=\operatorname{T_{cr}}(R_{1})\cup\operatorname{T_{cr}}(R_{2})=\operatorname{T_{cr}}(X_{1})\cup\operatorname{T_{cr}}(X_{2})$ is an antichain. For any $\tau\in\operatorname{T_{cr}}(X)$, we have $R(\tau)=R_{1}(\tau)\oplus R_{2}(\tau)=X_{1}(\tau)\oplus X_{2}(\tau)=X(\tau)$, so $\operatorname{R}(X)=R$ by (5.3). So $X\in\mathcal{B}$. ∎

A criterion for the absence of rank–one direct summands is needed.

Proposition 5.9**.**

[10, 13.1.5]* A block–rigid crq–group $X$ is clipped if and only if $X$ is rigid and $\mu_{\tau}(X)>1$ for all $\tau\in\operatorname{T_{cr}}(X)$. ∎*

The indecomposability of a $\mathcal{B}$–group is easily recognized by associating a certain graph with the group.

Definition 5.10**.**

(Blagoveshchenskaya)* Let $X$ be a rigid $\mathcal{B}$–group. The frame of $X$ is a graph whose vertices are the elements of $\operatorname{T_{cr}}(X)$ and two vertices $\sigma$ and $\tau$ are joined by an edge if and only if $\gcd(\mu_{\sigma}(X),\mu_{\tau}(X))>1$.*

Proposition 5.11**.**

[10, Corollary 13.1.10]* A $\mathcal{B}$–group is indecomposable if and only if it is rigid and its frame is connected. ∎*

Recall that any completely decomposable group $A$ of rank $r$ can be written in the form $A=\bigoplus_{1\leq i\leq r}A_{i}=\bigoplus_{1\leq i\leq r}\tau_{i}v_{i}$ where the $A_{i}$ are rank $1$ summands of $A$, $0\neq v_{i}\in A_{i}$ and $\tau_{i}=\{x\in{\mathbb{Q}}\mid xv_{i}\in A_{i}\}$. The $\tau_{i}$ are rational groups, i.e. additive subgroups of ${\mathbb{Q}}$ containing $1$. We identify a rational group $\tau$ with its type, and so speak of the rational group $\tau$ of type $\tau$. The set $\{v_{1},\ldots,v_{r}\}$ is called a decomposition basis of $A$.

The following Lemma 5.12 is restricted to rigid groups because this simplifies the notation considerably and is all that is used later.

Lemma 5.12**.**

Let $X$ be a rigid $\mathcal{B}$–group and let $R:=\operatorname{R}(X)=\bigoplus_{\rho\in\operatorname{T_{cr}}(X)}R_{\rho}$. Then $X$ has a description $X=R+\frac{{\mathbb{Z}}}{e}\sum_{\rho\in\operatorname{T_{cr}}(X)}\alpha_{\rho}v_{\rho}$ where

1. (1)

$\{v_{\rho}\mid\rho\in\operatorname{T_{cr}}(R)\}$* is a decomposition basis of $R$, so $R=\bigoplus_{\rho\in\operatorname{T_{cr}}(X)}\rho v_{\rho}$,* 2. (2)

$\forall\,\tau\in\operatorname{T_{cr}}(X):\gcd^{R}(e,\sum_{\rho\neq\tau}\alpha_{\rho}v_{\rho})=1$, 3. (3)

$e=[X:R]$, 4. (4)

$\alpha_{\rho}$* an integral factor of $e$, and* 5. (5)

$\forall\,\tau\in\operatorname{T_{cr}}(X):\mu_{\tau}(X)=\frac{e}{\alpha_{\tau}}$.

If $\operatorname{rk}(X)=2$, $R=\operatorname{R}(X)=\tau_{1}v_{1}\oplus\tau_{2}v_{2}$ and $[X:R]=e$, then we may assume the representation $X=A+{\mathbb{Z}}e^{-1}(v_{1}+v_{2})$ with $\gcd^{R}(e,v_{i})=1$.

Proof.

Certainly, $X=R+{\mathbb{Z}}\frac{1}{e}\sum_{\rho\in\operatorname{T_{cr}}(X)}u_{\rho}$, where $u_{\rho}\in R_{\rho}$ and (Regulator Criterion) for all $\tau\in\operatorname{T_{cr}}(X)$, $\gcd^{R}(e,\sum_{\rho\neq\tau}u_{\rho})=1$. Let $\rho\in\operatorname{T_{cr}}(X)$ and $\alpha_{\rho}=\gcd^{R}(e,u_{\rho})$. Then

[TABLE]

Let $\rho=\{x\in{\mathbb{Q}}\mid xv_{\rho}\in R_{\rho}\}$. Then $\rho$ is a rational group of type $\rho$ and $R_{\rho}=\rho v_{\rho}$.

It remains to show that $\mu:=\mu_{\tau}(X)=\operatorname{ord}(\alpha_{\tau}v_{\tau}+eR)=\frac{e}{\alpha_{\tau}}$. Clearly $\frac{e}{\alpha_{\tau}}(\alpha_{\tau}v_{\tau}+eR)=0$, so $\mu\mid\frac{e}{\alpha_{\tau}}$. On the other hand $\mu=\operatorname{ord}(\alpha_{\tau}v_{\tau}+eR)=\operatorname{ord}(\alpha_{\tau}v_{\tau}+\alpha_{\tau}\beta_{\tau}R)=\operatorname{ord}(v_{\tau}+\beta_{\tau}R_{\tau})=\operatorname{ord}(v_{\tau}+\beta_{\tau}\tau v_{\tau})$. Hence $\mu v_{\tau}=\beta_{\tau}\frac{a}{b}v_{\tau}$ where $\frac{a}{b}$ is a reduced fraction in the rational group $\tau$. It follows that $b\mu=a\beta_{\tau}$ and hence $\beta_{\tau}=bb^{\prime}$. Furthermore as $\frac{a}{b}v_{\tau}\in\tau v_{\tau}=R_{\tau}$ we also have $\frac{1}{b}v_{\tau}\in R_{\tau}$ and $v_{\tau}=b\left(\frac{1}{b}v_{\tau}\right)$. Thus $b$ is a common factor of $\beta_{\tau}$ and $v_{\tau}$. But $\gcd^{R}(\beta_{\tau},v_{\tau})=1$, so $b=1$ and $\frac{e}{\alpha_{\tau}}=\beta_{\tau}\mid\mu_{\tau}$. Together with the previous observation that $\mu\mid\frac{e}{\alpha_{\beta}}$ we have the desired equality.

Consider $X=R+{\mathbb{Z}}e^{-1}(\alpha_{1}v_{1}+\alpha_{2}v_{2})$ where $R=\operatorname{R}(X)=\tau_{1}v_{1}\oplus\tau_{2}v_{2}$ and $\alpha_{i}\mid e$. The Regulator Criterion requires that $\gcd^{R}(e,\alpha_{i}v_{i})=1$ this forces $\alpha_{i}=1$. But then $\gcd^{R}(e,v_{i})=1$. ∎

Given a rigid $\mathcal{B}$–group we can assume without loss of generality that it is in the “standard form” described in Lemma 5.12. On the other hand, in constructing rigid $\mathcal{B}$–groups, in particular indecomposable $\mathcal{B}$–groups, we must make judicious choices in order to achieve the standard form and the easy way of computing invariants.

Let $e$ be a positive integer. A type $\tau$ is $e$–free if for all prime divisors $p$ of $e$, $p\tau\neq\tau$. It is easily seen that a rank–$1$ group $A$ of $e$–free type $\tau$ has a representation $A=\tau v$ such that $\gcd^{A}(e,v)=1$. Consequently, a rigid completely decomposable group $A$ all of whose critical types are $e$–free has a description $A=\bigoplus_{\rho\in\operatorname{T_{cr}}(A)}\rho v_{\rho}$ with $\gcd^{A}(e,v_{\rho})=1$. In this case $\{v_{\rho}:\rho\in\operatorname{T_{cr}}(A)\}$ is called an $e$–basis of $A$.

It is easy to construct indecomposable $\mathcal{B}$–groups with evident invariants.

Lemma 5.13**.**

Let $e$ be a positive integer, and let $\tau_{1},\ldots\tau_{r}$ be pairwise incomparable types such that \forall\,i:\tau_{i}\text{ is e--free}. Let $A=\bigoplus_{1\leq i\leq r}\tau_{i}v_{i}$ such that $\forall\,i:\gcd^{A}(e,v_{i})=1$. Then

[TABLE]

with $\alpha_{i}$ integral factors of $e$ is an indecomposable $\mathcal{B}$–group with $A=\operatorname{R}(X)$, $[X:A]=e$ and $\mu_{\tau_{i}}(X)=e/\alpha_{i}$ provided that the $\alpha_{i}$ are so chosen that

1. (1)

$\forall\,i:\gcd^{A}(e,\sum_{j\neq i}\alpha_{j}v_{j})=1$, 2. (2)

the frame of $X$ is connected.

Proof.

Evident by Lemma 5.4 and Proposition 5.11. ∎

For example, for $e=3^{2}\cdot 5\cdot 7^{3}$ and $\tau_{i}$ $e$–free, the group

[TABLE]

is an indecomposable $\mathcal{B}$–group with invariants $\mu_{1}(X)=5\cdot 7^{3},\mu_{2}(X)=3^{2}\cdot 5\cdot 7^{2},\mu_{3}(X)=3\cdot 5\cdot 7^{3},\mu_{4}(X)=3^{2}\cdot 7^{3},\mu_{5}(X)=3^{2}\cdot 5\cdot 7^{3}$.

The indecomposable decompositions of rigid $\mathcal{B}$–groups are unique. Hence different indecomposable decompositions can only occur in properly block–rigid groups.

Theorem 5.14**.**

The endomorphism ring of a rigid $\mathcal{B}$–group is commutative. Consequently, rigid $\mathcal{B}$–groups have a unique indecomposable decomposition.

Proof.

Let $X$ be a rigid $\mathcal{B}$–group. The restriction map imbeds $\operatorname{End}(X)$ in $\operatorname{End}(\operatorname{R}(X))$ and we will see shortly that $\operatorname{End}(\operatorname{R}(X))$ is commutative. Let $R:=\operatorname{R}(X)=\bigoplus_{\rho\in\operatorname{T_{cr}}(X)}R(\rho)$ where $\operatorname{rk}(R(\rho))=1$ and let $\varphi\in\operatorname{End}(R)$. Then $\varphi$ restricts to an endomorphism $\varphi_{\rho}$ of $R(\rho)$ and hence is multiplication by a rational number. Hence $\varphi=(\ldots,\varphi_{\rho},\ldots)$ and such maps commute with one another.

In a group with commutative endomorphism ring every endomorphic image is a fully invariant subgroup. In particular, any direct summand is fully invariant. Let $X=\bigoplus_{i}X_{i}=\bigoplus_{j}Y_{j}$ be indecomposable decompositions. Let $\tau\in\operatorname{T_{cr}}(X_{1})$. Then there is $j$ such that $\tau\in\operatorname{T_{cr}}(Y_{j})$. Without loss of generality $j=1$ and for convenience write $X=Y_{1}\oplus Y_{1}^{\prime}$. As $X_{1}$ is fully invariant, $X_{1}=(X_{1}\cap Y_{1})\oplus(X_{1}\cap Y_{1}^{\prime})$. As $X_{1}$ is indecomposable and $0\neq X(\tau)=R(\tau)\subset X_{1}\cap Y_{1}$ it follows that $X_{1}\subset Y_{1}$ and $X_{1}\cap Y_{1}^{\prime}=0$. Then $X_{1}$ is a summand of $Y_{1}$ and since $Y_{1}$ is indecomposable, so $X_{1}=Y_{1}$. The claim follows by induction on the number of summands $X_{i}$. ∎

Corollary 5.15**.**

If $X$ is a clipped $\mathcal{B}$–group, then $X$ has a unique indecomposable decomposition.

Proof.

Let $X$ be a clipped $\mathcal{B}$–group. By Proposition 5.9 $X$ is rigid and by Theorem 5.14 $X$ has a unique indecomposable decomposition. ∎

6. Tools and Obstructions

We start with a consequence of realization.

Lemma 6.1**.**

Let $X\in\mathcal{B}$. Let $R=\operatorname{R}(X)=\bigoplus_{\rho\in\operatorname{T_{cr}}(X)}R_{\rho}$. Suppose that $X=X_{1}\oplus\cdots\oplus X_{s}$ is an indecomposable decomposition of $X$. Then $\forall\,\tau\in\operatorname{T_{cr}}(X):\operatorname{rk}(R_{\tau})\leq s$. In particular, if $X$ realizes $(r_{1},r_{2})$, then $\forall\,\tau\in\operatorname{T_{cr}}(X):\operatorname{rk}(R_{\tau})\leq 2$.

Proof.

Indecomposable $\mathcal{B}$–groups are rigid. Let $\tau\in\operatorname{T_{cr}}(X)$. Then $R_{\tau}=X(\tau)=X_{1}(\tau)\oplus\cdots\oplus X_{s}(\tau)$ with $\operatorname{rk}X_{i}(\tau)\leq 1$ (5.3). Hence $\operatorname{rk}R_{\tau}=\operatorname{rk}X(\tau)\leq s$. ∎

We use the following convenient shorthand that displays essential features of indecomposable $\mathcal{B}$–groups but not the whole story.

**Notation. ** The expression $[\tau_{1}\,\tau_{2}\,\ldots\,\tau_{r}]$ denotes an indecomposable $\mathcal{B}$–group $X$ with critical types $\tau_{1},\ldots,\tau_{r}$ and rank $r$. It is tacitly assumed that $X$ is given in the form $X=R+\frac{{\mathbb{Z}}}{e}a$ where $R=\operatorname{R}(X)=\bigoplus_{\rho\in\operatorname{T_{cr}}(X)}\rho v_{\rho}$, $a\in R$, and $[X:R]=e$.

On occasion the regulator index $e$ is relevant to the discussion and we encode the groups $X=R+\frac{{\mathbb{Z}}}{e}a$ by $[\tau_{1}\,\tau_{2}\,\ldots\,\tau_{r};e]$. In a direct sum $[\tau_{1}\,\tau_{2}\,\ldots;e_{1}]\oplus[\sigma_{1},\sigma_{2},\ldots;e_{2}]$ it is tacitly assumed that the poset of types $\{\tau_{i},\sigma_{j}\}$ is an antichain, and the regulator indices $e_{1},e_{2}$ are relatively prime so that the direct sum is a group in $\mathcal{B}$ (Lemma 5.8). Of course, when the invariants $\mu_{\tau}(X)$ are needed, the explicit representations (5.1) must be used.

If a group realizes particular partitions, then there are consequences for the critical typeset of the group.

Lemma 6.2**.**

Suppose that $X$ realizes $(k,1^{n-k})$ and $(r_{1},r_{2})$ where $r_{1},r_{2}\geq 2$. Then $\operatorname{T_{cr}}(X)=\{\tau_{1},\ldots,\tau_{k}\}$ and $X=[\tau_{1}\,\cdots\,\tau_{k}]\oplus[\sigma_{1}]\oplus\cdots[\sigma_{n-k}]$ where $\{\sigma_{1},\ldots,\sigma_{n-k}\}\subseteq\{\tau_{1},\ldots,\tau_{k}\}$.

Proof.

Let $\operatorname{R}(X)=\bigoplus_{\rho\in\operatorname{T_{cr}}(X)}R_{\rho}$. As $X$ realizes $(r_{1},r_{2})$, by Lemma 6.1, $\forall\,\rho\in\operatorname{T_{cr}}(X):\operatorname{rk}(R_{\rho})\leq 2$. As $r_{1},r_{2}\geq 2$ the group $X$ has no fully invariant summand of rank $1$ (Lemma 4.3). As $X$ realizes $(k,1^{n-k})$, we have $X=[\tau_{1}\,\cdots\,\tau_{k}]\oplus[\sigma_{1}]\oplus\cdots\oplus[\sigma_{n-k}]$. If $\sigma_{i}\notin\{\tau_{1},\ldots,\tau_{k}\}$, then $[\sigma_{i}]\oplus\cdots$ is a fully invariant direct summand of $X$ and hence any indecomposable decomposition of $X$ contains a summand $[\sigma_{i}]$ (Lemma 4.3), contradicting our earlier observation. ∎

We now consider Corner’s groups although as a consequence of Lemma 6.1 Corner’s groups are of no help in trying to realize $\mathcal{S}(n,n-k+1)$. In particular, we show that Corner groups are block–rigid crq–groups. Theorem 6.4(2) is not new ([5], [7, Exercise 3, p.139]) but it is easily derived from our results.

Definition 6.3**.**

Given

1. (1)

integers $k,n$ with $1\leq k\leq n$, 2. (2)

pairwise relatively prime integers $>1$, $q_{1},\ldots,q_{n-k}$, 3. (3)

$e$–free pairwise incomparable types $\tau,\tau_{1},\ldots,\tau_{n-k}$,

a Corner group $X$ has base group

[TABLE]

and is defined to be

[TABLE]

Hence $\operatorname{rk}(R_{\tau})=k$ and for all $i\in[n-k],\ \operatorname{rk}(R_{\tau_{i}})=1$.

Theorem 6.4**.**

Let $X$ be a Corner group as in Definition 6.3. Then the following hold.

1. (1)

Setting $e:=q_{1}q_{2}\cdots q_{n-k}$, and $e_{i}:=e/q_{i}$, $R_{0}:=\tau u_{2}\oplus\cdots\oplus\tau u_{k}$ and $R_{1}:=\tau u_{1}\oplus\tau_{1}v_{1}\oplus\cdots\oplus\tau_{n-k}v_{n-k}$,

[TABLE]

realizing $(n-k+1,1^{k-1})\in\mathcal{P}(n)$. 2. (2)

$\mathcal{P}(X)=\mathcal{C}(n,k)$.

Proof.

(1) $\frac{1}{q_{1}}(u_{1}+v_{1})+\frac{1}{q_{2}}(u_{1}+v_{2})+\cdots+\frac{1}{q_{n-k}}(u_{1}+v_{n-k})=\frac{1}{e}((e_{1}+\cdots+e_{n-k})u_{1}+e_{2}v_{2}+\cdots+e_{n-k}v_{n-k})$. Hence $X$ contains $(\tau u_{2}\oplus\cdots\oplus\tau u_{k})\oplus$

$\left((\tau u_{1}\oplus\tau_{1}v_{1}\oplus\cdots\oplus\tau_{n-k}v_{n-k})+\frac{{\mathbb{Z}}}{e}((e_{1}+\cdots+e_{n-k})u_{1}+e_{2}v_{2}+\cdots+e_{n-k}v_{n-k})\right)$. Equality follows from Lemma 6.5.

(2) By (1) a Corner group $X$ is a $\mathcal{B}$–group hence $\mathcal{P}(X)$ is hooked. The claim follows from Proposition 3.3(3). ∎

Lemma 6.7 is basic. In its proof Lemma 6.5 will be used.

Lemma 6.5**.**

Let $R$ be a torsion-free group, $a,b\in R$, $e_{1},e_{2}$ relatively prime integers, and $1=u_{1}e_{1}+u_{2}e_{2}$. Then, in ${\mathbb{Q}}R$,

1. (1)

$\frac{1}{e_{1}}a+\frac{1}{e_{2}}b=\frac{1}{e_{1}e_{2}}(e_{2}a+e_{1}b)$, but also 2. (2)

$u_{1}e_{1}\left(\frac{1}{e_{1}e_{2}}(e_{2}a+e_{1}b)\right)=\frac{1}{e_{2}}b+u_{1}a-u_{2}b$, and 3. (3)

$u_{2}e_{2}\left(\frac{1}{e_{1}e_{2}}(e_{2}a+e_{1}b)\right)=\frac{1}{e_{1}}a-u_{1}a+u_{2}b$**

Hence $R+\frac{1}{e_{1}e_{2}}(e_{2}a+e_{1}b)=R+\frac{1}{e_{1}}a+\frac{1}{e_{2}}b$.

Proof.

Easy computation. ∎

Corollary 6.6**.**

For $i=1,2$, let $X_{i}=R_{i}+\frac{{\mathbb{Z}}}{e_{i}}a_{i}\in\mathcal{B}$ where $\operatorname{T_{cr}}(X_{1})\cup\operatorname{T_{cr}}(X_{2})$ is an antichain, $e_{1},e_{2}$ are relatively prime integers, $R_{i}=\operatorname{R}(X_{i})$, and $a_{i}\in R_{i}$. Then $X_{1}\oplus X_{2}\in\mathcal{B}$,

[TABLE]

Proof.

Lemma 6.5 says that $X_{1}\oplus X_{2}=(R_{1}\oplus R_{2})+\frac{{\mathbb{Z}}}{e_{1}e_{2}}(e_{2}a_{1}+e_{1}a_{2})$. By hypothesis $\operatorname{R}(X_{i})=R_{i}$, and, by Lemma 5.8, $\operatorname{R}(X_{1}\oplus X_{2})=R_{1}\oplus R_{2}$. ∎

We now can freely add two $\mathcal{B}$–groups by adding generators, the description of the sum and its regulator being clear. We clarify in all generality what the direct sum of two rigid $\mathcal{B}$–groups looks like. Note that the essential feature is the overlap in critical types of the two groups that are added.

Lemma 6.7**.**

Let $\tau_{1},\ldots,\tau_{d},\tau_{d+1},\ldots,\tau_{r},\tau_{r+1},\ldots\tau_{s}$ be pairwise incomparable types, $d\geq 1$, $r\geq d$, $s\geq r$

[TABLE]

and

[TABLE]

[TABLE]

where $\alpha_{i},\beta_{j}\in{\mathbb{Z}}$, $\alpha_{i}$ divides $e_{1}$, $\beta_{j}$ divides $e_{2}$, $A_{i}=\operatorname{R}(X_{i})$ and $e_{1},e_{2}$ are relatively prime integers. For $i=1,\ldots,d$, let $f_{i}=\gcd(e_{2}\alpha_{i},e_{1}\beta_{i})$. Then there exists a decomposition basis $\{v_{1}^{\prime},w_{1}^{\prime},\ldots,v_{d}^{\prime},w_{d}^{\prime},v_{d+1},\ldots,v_{r},v_{r+1},\ldots v_{s}\}$ of $A_{1}\oplus A_{2}=\operatorname{R}(X_{1}\oplus X_{2})$ such that with

[TABLE]

we have

[TABLE]

Furthermore, if $X_{1}$ and $X_{2}$ are both indecomposable, then so is $Y$. All generator coefficients of $Y$ are divisors of $e_{1}e_{2}$. In passing from the decomposition $X_{1}\oplus X_{2}$ to $Y\oplus\tau_{1}w_{1}^{\prime}\oplus\cdots\oplus\tau_{d}w_{d}^{\prime}$ the number of summands does not decrease.

In Lemma 6.7 we assume that the coefficients $\alpha_{i}$ are factors of $e_{1}$ and the coefficients $\beta_{i}$ are factors of $e_{2}$. By Lemma 5.12 this can be done without loss of generality, also, in examples, this will always be the case.

Proof.

By Corollary 6.6

[TABLE]

and $A_{1}\oplus A_{2}=\operatorname{R}(X_{1}\oplus X_{2})$.

For $i=1,\ldots,d$, we defined $f_{i}=\gcd(e_{2}\alpha_{i},e_{1}\beta_{i})$. Then $f_{i}=\alpha_{i}\beta_{i}$ and setting $e_{2}\alpha_{i}=f_{i}\alpha_{i}^{\prime}$, $e_{1}\beta_{i}=f_{i}\beta_{i}^{\prime}$, there exist integers $a_{i},b_{i}$ such that $a_{i}\alpha_{i}^{\prime}+b_{i}\beta_{i}^{\prime}=1$ and hence we have that $\tau_{i}v_{i}\oplus\tau_{i}w_{i}=\tau_{i}(\alpha_{i}^{\prime}v_{i}+\beta_{i}^{\prime}w_{i})\oplus\tau_{i}(-b_{i}v_{i}+a_{i}w_{i})$. Setting $v_{i}^{\prime}=\alpha_{i}^{\prime}v_{i}+\beta_{i}^{\prime}w_{i}$ and $w_{i}^{\prime}=-b_{i}v_{i}+a_{i}w_{i}$, we have the new decomposition basis $\{v_{1}^{\prime},w_{1}^{\prime},\ldots,v_{d}^{\prime},w_{d}^{\prime},v_{d+1},\ldots,v_{r},v_{r+1},\ldots v_{s}\}$ of $A_{1}\oplus A_{2}=\operatorname{R}(X_{1}\oplus X_{2})$. Let $Y$ be as stated. Then $X_{1}\oplus X_{2}=Y\oplus\tau_{1}w_{1}^{\prime}\oplus\cdots\oplus\tau_{d}w_{d}^{\prime}$ and by Theorem 5.7 $A=\operatorname{R}(Y)$.

For clarity note that $\operatorname{T_{cr}}(Y)=\operatorname{T_{cr}}(X_{1})\cup\operatorname{T_{cr}}(X_{2})$ and $\operatorname{T_{cr}}(X_{1})\cap\operatorname{T_{cr}}(X_{2})=\{\tau_{1},\ldots,\tau_{d}\}\neq\emptyset$ which makes it possible to merge the generators of $X_{1}$ and $X_{2}$.

Suppose now that $X_{1}$ and $X_{2}$ are indecomposable. By Proposition 5.11 the frames of $X_{1}$ and $X_{2}$ are connected. We will show that the frame of $Y$ is connected. To do so recall that $\mu_{\tau}(Y)=\mu_{\tau}(X_{1})\mu_{\tau}(X_{2})$. So for any $\tau\in\operatorname{T_{cr}}(X_{1})$, we have that $\mu_{\tau}(X_{1})$ and $\mu_{\tau}(X_{2})$ divide $\mu_{\tau}(Y)$. This shows that any edge between types of either the frame of $X_{1}$ or $X_{2}$ is also an edge in the frame of $Y$. As the frames of $X_{1}$ and $X_{2}$ are connected and $\operatorname{T_{cr}}(X_{1})\cap\operatorname{T_{cr}}(X_{2})\neq\emptyset$ it follows that the frame of $Y$ is connected and $Y$ is indecomposable. ∎

We use our previous shorthand to display the content of Lemma 6.7. The formula

[TABLE]

correctly reflects the content of Lemma 6.7. We frequently use this formula without explicit reference.

Example 6.8 explicitly describes $[\tau_{1}\,\tau_{2}]\oplus[\tau_{1}\,\tau_{2}]=[\tau_{1}\,\tau_{2}]\oplus[\tau_{1}]\oplus[\tau_{2}]$.

Example 6.8**.**

There are groups realizing $\mathcal{S}(4,2)=\{(2,2),(2,1,1)\}$.

In fact, let $p_{1},p_{2}$ be relatively prime integers $\geq 2$, let $\tau_{1},\tau_{2}$ be incomparable $p_{1}p_{2}$–free types, and let

[TABLE]

Then $\operatorname{R}(X_{1})=\tau_{1}v_{1}\oplus\tau_{2}v_{2}$ and $\operatorname{R}(X_{2})=\tau_{1}w_{1}\oplus\tau_{2}w_{2}$ and $X_{1}$ and $X_{2}$ are indecomposable. With $u_{1}p_{1}+u_{2}p_{2}=1$,

[TABLE]

so $X$ realizes $(1,1,2)$ in addition to $(2,2)$.

Proof.

The summands $X_{1}$ and $X_{2}$ are indecomposable by Proposition 5.11 and the completely decomposable base groups are the regulators in each case (Theorem 5.7. Now $\frac{1}{p_{1}}(v_{1}+v_{2})+\frac{1}{p_{2}}(w_{1}+w_{2})=\frac{1}{p_{1}p_{2}}\left((p_{2}v_{1}+p_{1}w_{1})+(p_{2}v_{2}+p_{1}w_{2})\right)$. Set $v_{1}^{\prime}=p_{2}v_{1}+p_{1}w_{1}$, $v_{2}^{\prime}=p_{2}v_{2}+p_{1}w_{2}$, $w_{1}^{\prime}=-u_{1}v_{1}+u_{2}w_{1}$, and $w_{2}^{\prime}=-u_{1}v_{2}+u_{2}w_{2}$. Then (change of basis) $\tau_{1}v_{1}\oplus\tau_{1}w_{1}=\tau_{1}v_{1}^{\prime}\oplus\tau_{1}w_{1}^{\prime}$ and $\tau_{2}v_{2}\oplus\tau_{2}w_{2}=\tau_{2}v_{2}^{\prime}\oplus\tau_{2}w_{2}^{\prime}$. Hence $X=\left(\tau_{1}w_{1}^{\prime}\oplus\tau_{2}w_{2}^{\prime}\right)\oplus\left(\left(\tau_{1}v_{1}^{\prime}\oplus\tau_{2}v_{2}^{\prime}\right)+{\mathbb{Z}}\frac{1}{p_{1}p_{2}}(v_{1}^{\prime}+v_{2}^{\prime})\right)$. ∎

A Corner group is of the form $X=[\tau_{1}\,\cdots\,\tau_{\ell}]\oplus[\tau_{1}]\oplus\cdots\oplus[\tau_{1}]$. Proposition 6.9 shows that certain Corner families of partitions may be realized by groups that are not Corner groups.

Proposition 6.9**.**

The group $X_{2n}:=[\tau_{1}\ \tau_{2}]\oplus[\tau_{2}\ \tau_{3}]\oplus\cdots\oplus[\tau_{n}\ \tau_{n+1}]$ realizes $\mathcal{C}(2n,n)$.

Proof.

Let $(r_{1},r_{2},\ldots,r_{n})\in\mathcal{C}(2n,n)$ where $\forall\,i\leq m:r_{i}\geq 2$ and $\forall\,i>m:r_{i}=1$. Note that $r_{1}+\cdots+r_{m}+(n-m)=2n$, so $r_{1}+\cdots+r_{m}=n+m$ and $(r_{1}-1)+\cdots+(r_{m}-1)=n$. The group $X_{2n}$ has $n$ summands of rank $2$. We partition the $n$ summands of $X_{2n}$ into groups of $r_{1}-1,r_{2}-1,\ldots,r_{m}-1$ summands. A typical summand is of the form $[\sigma_{1}\ \sigma_{2}]\oplus[\sigma_{2}\ \sigma_{3}]\oplus\cdots\oplus[\sigma_{r-1}\ \sigma_{r}]=[\sigma_{1}\ \sigma_{2}\cdots\sigma_{r}]\oplus[\sigma_{2}]\oplus\cdots\oplus[\sigma_{r-1}]$ where $[\sigma_{1}\ \sigma_{2}\cdots\sigma_{r}]$ is a summand of rank $r$. Note that in combining these summands the total number of summands has not changed. Doing this for each summand of the partition we get summands of rank $r_{1},r_{2},\ldots,r_{m}$ and $n-m$ summands of rank $1$, a total of $n$ summands. The partition $(r_{1},r_{2},\ldots,r_{n})\in\mathcal{C}(2n,n)$ has been realized by our particular decomposition of $X_{2n}$. ∎

7. Realizing families $S(n,k)$

The following proposition illustrates why some groups cannot realize $S(n,k)$.

Proposition 7.1**.**

Suppose that $G=Y\oplus Z$ is a decomposition with fully invariant summands neither of which is completely decomposable. Let $n_{G}=\operatorname{rk}(G)$, $n_{Y}=\operatorname{rk}(Y)$, and $n_{Z}=\operatorname{rk}(Z)$. If $Y$ realizes $\mathcal{S}(n_{Y},k_{Y})$, and $Z$ realizes $\mathcal{S}(n_{Z},k_{Z})$. then $\mathcal{P}(G)=\mathcal{S}(n_{Y},k_{Y})\times\mathcal{S}(n_{Z},k_{Z})\subsetneqq\mathcal{S}(n_{G},k_{Z})$, so $G$ does not realize $\mathcal{S}(n_{G},k_{Z})$.

Proof.

By Proposition 3.3(2) we have that $\mathcal{P}(Y)=\mathcal{S}(n_{Y},k_{Y})$ and $\mathcal{P}(Z)=\mathcal{S}(n_{Z},k_{Z})$ and by Proposition 4.1(2) $\mathcal{P}(X)=\mathcal{P}(Y)\times\mathcal{P}(Z)=\mathcal{S}(n_{Y},k_{Y})\times\mathcal{S}(n_{Z},k_{Z})$. Without loss of generality, suppose $k_{Z}\geq k_{Y}$. By way of contradiction assume that $\mathcal{S}(n_{Y},k_{Y})\times\mathcal{S}(n_{Z},k_{Z})=\mathcal{S}(n_{X},k_{Z})$. Then $X$ realizes $\mathcal{S}(n_{X},k_{Z})$ and, in particular, $(k_{Z},1^{n_{X}-k_{Z}})\in\mathcal{S}(n_{X},k_{Z})$. It follows that $(1^{n_{Y}})\in\mathcal{P}(Y)$ and $Y$ is completely decomposable, or else $(1^{n_{Z}})\in\mathcal{P}(Z)$ and $Z$ is completely decomposable, either way a contradiction. ∎

Our main result Theorem 7.2 determines completely which families $\mathcal{S}(n,k)$ can be realized in $\mathcal{B}$ and which cannot.

Theorem 7.2**.**

1. (1)

The families $\mathcal{S}(n,n),\mathcal{S}(n,1),\mathcal{S}(n,2),\mathcal{S}(n,n-1)$ are realizable in $\mathcal{B}$. 2. (2)

The special families $\mathcal{S}(5,3),\mathcal{S}(6,4),\mathcal{S}(4,2)$ can be realized in $\mathcal{B}$. 3. (3)

All other hooked families $\mathcal{S}(n,k)$ are not realizable in $\mathcal{B}$. Specifically, the families $\mathcal{S}(n,k)$ for $k\geq 3,n\geq 2k$, $\mathcal{S}(n,6)$ for $n\geq 10$, $\mathcal{S}(7,4)$, $\mathcal{S}(7,5)$, $\mathcal{S}(8,5)$, $\mathcal{S}(9,5)$, $\mathcal{S}(8,6)$, $\mathcal{S}(9,6)$, $\mathcal{S}(9,7)$ cannot be realized in $\mathcal{B}$.

Proof.

Details of the proof are contained in the following lemmas, corollaries and examples:

(1) Lemma 7.3.

(2) Example 7.5, Example 7.6, Example 6.8.

(3) Corollary 7.12, Example 7.13, Corollary 7.14, Example 7.15, Example 7.16, Example 7.17, Example 7.18. ∎

Lemma 7.3**.**

1. (1)

$S(n,1)=\{(1,\ldots,1)\}$* is realized by any completely decomposable group of rank $n$.* 2. (2)

$S(n,n)=\{(n)\}$* is realized by any indecomposable $\mathcal{B}$–group of rank $n$.* 3. (3)

The set of partitions $\mathcal{S}(n,2)$ can be realized. 4. (4)

$\mathcal{S}(n,n-1)$* can be realized by suitable Corner groups.*

Proof.

(3) First note that $P\in S(n,2)$ means that $P=(2,\ldots,2,1,\ldots,1)=(2^{i},1^{n-2i})$. So, for $n$ even, $S(n,2)=\{(2^{n/2}),\ldots,(2,1^{n-2})\}$. Using Lemma 6.7 repeatedly we have for $n$ even, $n[\tau_{1},\tau_{2}]=[\tau_{1}]\oplus[\tau_{2}]\oplus(n-1)[\tau_{1},\tau_{2}]=\cdots=[\tau_{1},\tau_{2}]\oplus(n/2-1)[\tau_{1}]+(n/2-1)[\tau_{2}]$. For odd $n$ we have an obvious variant.

(4) $\mathcal{S}(n,n-1)=\{(n-1,1),(n-2,2),\ldots\}$ can be realized by Corner’s Theorem. ∎

Remark 7.4**.**

In all examples that follow we assume that the groups are given in the form of Lemma 5.12 so that we have a handy formula for the invariants. The naming of the elements of the decomposition bases are irrelevant as only the associated types matter, i.e., in two decompositions of a group the $v_{1}$ in one decomposition and the $v_{1}$ in the other decomposition need not be the same.

Example 7.5**.**

Let $p_{1},p_{2},p_{3}$ be pairwise relatively prime integers $\geq 2$, let $\tau_{1},\tau_{2},\tau_{3}$ be pairwise incomparable $p_{1}p_{2}p_{3}$–free types, and let

1. (1)

$R=\tau_{1}v_{1}\oplus\tau_{2}v_{2}\oplus\tau_{3}v_{3}\oplus\tau_{1}w_{1}\oplus\tau_{2}w_{2}$, 2. (2)

$X_{1}=(\tau_{1}v_{1}\oplus\tau_{2}v_{2})+{\mathbb{Z}}\frac{1}{p_{2}p_{3}}(v_{1}+v_{2})$, 3. (3)

$X_{2}=(\tau_{1}w_{1}\oplus\tau_{3}v_{3})+{\mathbb{Z}}\frac{1}{p_{1}}(w_{1}+v_{3})$, and 4. (4)

$X_{3}=\tau_{2}w_{2}$.

Then $X=X_{1}\oplus X_{2}\oplus X_{3}$ realizes $S(5,3)=\{(3,1,1),(3,2),(2,2,1)\}$.

Proof.

By Lemma 6.7 $X_{1}\oplus X_{2}=[\tau_{1}\ \tau_{2}]\oplus[\tau_{1}\ \tau_{3}]\oplus[\tau_{2}]=[\tau_{1}\ \tau_{2}\ \tau_{3}]\oplus[\tau_{1}]\oplus[\tau_{2}]$ Thus $X=X_{1}\oplus X_{2}\oplus X_{3}$ realizes $(2,2,1)$ and $(3,1,1)$.

Let

1. (1)

$Y_{1}=(\tau_{1}v_{1}\oplus\tau_{2}v_{2}\oplus\tau_{3}v_{3})+{\mathbb{Z}}\frac{1}{p_{1}p_{2}}(v_{1}+p_{1}v_{2}+p_{2}v_{3})$, 2. (2)

$Y_{2}=(\tau_{1}w_{1}\oplus\tau_{2}w_{2})+{\mathbb{Z}}\frac{1}{p_{3}}(w_{1}+w_{2})$.

Then

$Y_{1}\oplus Y_{2}=[(\tau_{1}v_{1}^{\prime\prime}\oplus\tau_{2}v_{2}^{\prime\prime}\oplus\tau_{3}v_{3})+\frac{{\mathbb{Z}}}{p_{1}p_{2}p_{3}}(v_{1}^{\prime\prime}+p_{1}v_{2}^{\prime\prime}+p_{2}p_{3}v_{3})]\oplus\tau_{1}w_{1}^{\prime\prime}\oplus\tau_{2}w_{2}^{\prime\prime}$ where $\tau_{1}v_{1}\oplus\tau_{1}w_{1}=\tau_{1}v_{1}^{\prime\prime}\oplus\tau_{1}w_{1}^{\prime\prime}$, and $\tau_{2}v_{2}\oplus\tau_{2}w_{2}=\tau_{2}v_{2}^{\prime\prime}\oplus\tau_{2}w_{2}^{\prime\prime}$.

Then $Y=Y_{1}\oplus Y_{2}$ realizes $(3,2)$ by definition.

So far $X=X_{1}\oplus X_{2}\oplus X_{3}$ and $Y=Y_{1}\oplus Y_{2}$ are independently constructed groups although $R=\operatorname{R}(X)=\operatorname{R}(Y)$ and they are both contained in ${\mathbb{Q}}R$, the divisible hull of $R$. We will show that they are nearly isomorphic and therefore they share their decomposition properties.

Using the Product Formula (Theorem 5.7) we immediately see that

[TABLE]

and

[TABLE]

Hence $X\cong_{\text{\rm nr}}Y$, and we are done. ∎

Example 7.6**.**

The $\mathcal{B}$–group $X=X_{1}\oplus X_{2}\oplus X_{3}$ where

1. (1)

$X_{1}=(\tau_{1}v_{1}\oplus\tau_{2}v_{2})+{\mathbb{Z}}\frac{1}{5\cdot 11}(v_{1}+v_{2})$,

where $\operatorname{R}(X_{1})=\tau_{1}v_{1}\oplus\tau_{2}v_{2}$ and $\{v_{1},v_{2}\}$ is a $5\cdot 11$–basis of $\operatorname{R}(X_{1})$, 2. (2)

$X_{2}=(\tau_{1}w_{1}\oplus\tau_{3}v_{3})+{\mathbb{Z}}\frac{1}{7}(w_{1}+v_{3})$,

where $\operatorname{R}(X_{2})=\tau_{1}w_{1}\oplus\tau_{3}v_{3}$ and $\{w_{1},v_{3}\}$ is a $7$–basis of $\operatorname{R}(X_{2})$, 3. (3)

$X_{3}=(\tau_{2}w_{2}\oplus\tau_{4}v_{4})+{\mathbb{Z}}\frac{1}{3}(w_{2}+v_{4})$,

where $\operatorname{R}(X_{3})=\tau_{2}w_{2}\oplus\tau_{4}v_{4}$ and $\{w_{2},v_{4}\}$ is a $3$–basis of $\operatorname{R}(X_{3})$.

realizes $\mathcal{S}(6,4)$.

Proof.

For easy reference note that

[TABLE]

It is assumed that all critical types $\tau_{i}$ are $3\cdot 5\cdot 7\cdot 11$–free. By definition $X:=[\tau_{1}\,\tau_{2}]\oplus[\tau_{1}\,\tau_{3}]\oplus[\tau_{2}\,\tau_{4}]=[\tau_{1}\,\tau_{2}\,\tau_{3}]\oplus[\tau_{1}]\oplus[\tau_{2}\,\tau_{4}]=[\tau_{1}\,\tau_{2}\,\tau_{3}\,\tau_{4}]\oplus[\tau_{1}]\oplus[\tau_{2}]$ showing that any such $X$ will indeed realize $(4,1,1),(3,2,1)$ and $(2,2,2)$. We record the invariants of $X$.

[TABLE]

Next let $Y=Y_{1}\oplus Y_{2}$ where

1. (1)

$Y_{1}=(\tau_{1}v_{1}\oplus\tau_{2}v_{2}\oplus\tau_{3}v_{3}\oplus\tau_{4}v_{4})+{\mathbb{Z}}\frac{1}{7\cdot 3}(3v_{1}+7v_{2}+3v_{3}+7v_{4})$, 2. (2)

$Y_{2}=(\tau_{1}w_{1}\oplus\tau_{2}w_{2})+{\mathbb{Z}}\frac{1}{5\cdot 11}(w_{1}+w_{2})$.

It is easily checked that $\operatorname{R}(Y_{1})=\tau_{1}v_{1}\oplus\tau_{2}v_{2}\oplus\tau_{3}v_{3}\oplus\tau_{4}v_{4}$, $\operatorname{R}(Y_{2})=\tau_{1}w_{1}\oplus\tau_{2}w_{2}$ and $Y_{1},Y_{2}$ are indecomposable. The invariants of $Y$ are as follows.

[TABLE]

We see that $X$ and $Y$ have the same regulator and the same invariants, hence $X\cong_{\text{\rm nr}}Y$ and $X$ realizes $(4,2)$.

Finally, let $Z=Z_{1}\oplus Z_{2}$ where

1. (1)

$Z_{1}=(\tau_{1}v_{1}\oplus\tau_{2}v_{2}\oplus\tau_{3}v_{3})+{\mathbb{Z}}\frac{1}{7\cdot 5}(v_{1}+7v_{2}+5v_{3})$, 2. (2)

$Z_{2}=(\tau_{1}w_{1}\oplus\tau_{2}w_{2}\oplus\tau_{4}v_{4})+{\mathbb{Z}}\frac{1}{3\cdot 11}(3w_{1}+w_{2}+11v_{4})$.

It is routine to verify that $\operatorname{R}(Z_{1})=\tau_{1}v_{1}\oplus\tau_{2}v_{2}\oplus\tau_{3}v_{3}$ and $\operatorname{R}(Z_{2})=\tau_{1}w_{1}\oplus\tau_{2}w_{2}\oplus\tau_{4}v_{4}$, hence $\operatorname{R}(Z)=\operatorname{R}(X)$. We now compute invariants.

[TABLE]

Hence $Z_{1},Z_{2}$ are indecomposable, $Z\cong_{\text{\rm nr}}X$ and $X$ realizes $(3,3)$. ∎

It is intuitively clear that “larger” sets of partitions cannot be realized if “smaller” ones cannot.

Proposition 7.7**.**

Let $n^{\prime}>n\geq k$. If $\mathcal{S}(n^{\prime},k)$ can be realized, then so can $\mathcal{S}(n,k)$. Equivalently, if $\mathcal{S}(n,k)$ cannot be realized, then neither can $\mathcal{S}(n^{\prime},k)$.

Proof.

Let $(r_{1},\ldots,r_{t})\in\mathcal{S}(n,k)$ and $n^{\prime}=n+1$. Then $k+t\leq n+1$. Consider $P=(r_{1},\ldots,r_{t},1)\in\mathcal{P}(n^{\prime})$. Then $k+(t+1)\leq n+1+1=n^{\prime}+1$, so $P\in\mathcal{S}(n^{\prime},k)$. Suppose $\mathcal{S}(n^{\prime},k)$ can be realized. Then there is a group $X$ and a decomposition $X=Y_{1}\oplus\cdots\oplus Y_{t}\oplus Y$ such that $\operatorname{rk}Y_{i}=r_{i}$, the $Y_{i}$ are indecomposable and $\operatorname{rk}(Y)=1$. Then $X/Y$ realizes $(r_{1},\ldots,r_{t})$. Repeat for arbitrary $n^{\prime}>n$ until $\mathcal{S}(n,k)$ is reached. ∎

Proposition 7.8**.**

There is no $\mathcal{B}$–group realizing $\mathcal{S}(2m,m)$ provided $m\geq 3$.

Proof.

Suppose $X\in\mathcal{B}$ realizes $\mathcal{S}(2m,m)$ for $m\geq 3$. Note that $\mathcal{S}(2m,m)$ contains the partitions $(m,1^{m})$ and $(m,m)$. Then $X$ cannot have completely decomposable fully invariant summands (Lemma 4.3) and cannot be the direct sum of fully invariant subgroups (Lemma 7.1). Lemma 6.2 implies that $\operatorname{T_{cr}}(X)=\{\tau_{1},\ldots,\tau_{m}\}$, $X=[\tau_{1}\,\cdots\,\tau_{m}]\oplus[\tau_{1}]\oplus\cdots[\tau_{m}]$, and by Lemma 6.1 $\forall\,\tau\in\operatorname{T_{cr}}(X):\operatorname{rk}(X(\tau))\leq 2$. The group $X$ also realizes $(2^{m})$. We claim that, without loss of generality,

[TABLE]

where

[TABLE]

In fact, the first rank–$2$ summand of $X$ may be assumed to be $[\tau_{1}\ \tau_{2};p_{1}]$. The second rank–$1$ summand of $\operatorname{R}(X)$ of type $\tau_{2}$ appears in a summand $[\tau_{2}\ \tau;p_{2}]$. The type $\tau$ cannot be $\tau_{1}$ because this would create a decomposition with fully invariant summands. So without loss of generality $\tau=\tau_{3}$. Continuing in this fashion we obtain the claimed decomposition (7.9). The summands may be written in the form (7) by Lemma 5.12. Thus we have the following invariants $\mu_{i}=\mu_{i}(X)$.

[TABLE]

The group $X$ also realizes $(2^{m-1},1,1)$ and hence may be assumed to be of the form

[TABLE]

where without loss of generality $i<j$. Again assume the presentation of Lemma 5.12 and obtain the invariants

[TABLE]

We obtain in particular that

[TABLE]

Suppose that $\sigma_{1}=\tau_{k}$, so $k\neq i$. Assume first that $k\geq 2$. Then $\mu_{k}=p_{k}p_{k-1}=e_{1}e_{2}=p_{i}p_{i-1}e_{2}$. So $p_{i}$ divides $p_{k}p_{k-1}$ and it follows that $p_{i}=p_{k-1}$, so $i=k-1$. Hence $p_{k}p_{k-1}=p_{k-1}p_{k-2}e_{2}$, and $p_{k-2}\mid p_{k}$, a contradiction. This leaves the case $k=1$. In this case $\mu_{1}=p_{1}p_{m}=e_{1}e_{2}=p_{i}p_{i-1}e_{2}$. So $p_{i}$ divides $p_{1}p_{m}$ and it follows that $p_{i}=p_{m}$ and $p_{i-1}=p_{1}$, so $i=2$. Hence $p_{1}p_{m}=p_{2}p_{1}e_{2}$, and $p_{2}\mid p_{m}$, again a contradiction as $m\geq 3$. ∎

Corollary 7.12**.**

There is no $\mathcal{B}$–group realizing $\mathcal{S}(n,m)$ for $m\geq 3$ and $n\geq 2m$.

Proof.

Proposition 7.8 and Proposition 7.7. ∎

Example 7.13**.**

The hooked family $\mathcal{S}(10,6)$ cannot be realized by $\mathcal{B}$–groups.

Proof.

For easy reference

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Suppose that $X$ realizes $(6,1,1,1,1)$ and $(6,4)$. Then $\operatorname{T_{cr}}(X)=\{\tau_{1},\tau_{2},\tau_{3},\tau_{4},\tau_{5},\tau_{6}\}$ and $X=[\tau_{1}\,\tau_{2}\,\tau_{3}\,\tau_{4}\,\tau_{5}\,\tau_{6}]\oplus[\tau_{1}]\oplus[\tau_{2}]\oplus[\tau_{3}]\oplus[\tau_{4}]$ by Lemma 6.2. Note that the types $\tau_{1},\ldots,\tau_{4}$ are interchangeable, and $\tau_{5},\tau_{6}$ are interchangeable. Assume further that $X$ realizes $(2,2,2,2,2)$. We claim that after renaming types if necessary we may assume that $X=[\tau_{1}\,\tau_{2}]\oplus[\tau_{2}\,\tau_{3}]\oplus[\tau_{3}\,\tau_{4}]\oplus[\tau_{4}\,\tau_{5}]\oplus[\tau_{1}\,\tau_{6}]$. In fact, $\tau_{5}$ and $\tau_{6}$ cannot belong to the same summand and must be paired with different types to avoid fully invariant summands, so without loss of generality $X=[\tau_{4}\,\tau_{5}]\oplus[\tau_{1}\,\tau_{6}]\oplus\cdots$. The remaining three summands must contain $\tau_{1}$ and $\tau_{4}$ once each and $\tau_{2},\tau_{3}$ twice each. With proper labeling we get $X=[\tau_{1}\,\tau_{2}]\oplus[\tau_{2}\,\tau_{3}]\oplus[\tau_{3}\,\tau_{4}]\oplus[\tau_{4}\,\tau_{5}]\oplus[\tau_{1}\,\tau_{6}]$. In order to compute invariants we need the explicit description of $X$. By Lemma 5.12 we may assume that

[TABLE]

where $\tau_{1},\tau_{2}$ are $p_{1}$–free, $\tau_{2},\tau_{3}$ are $p_{2}$–free, $\tau_{3},\tau_{4}$ are $p_{3}$–free, $\tau_{4},\tau_{5}$ are $p_{4}$–free, $\tau_{1},\tau_{6}$ are $p_{5}$–free, and the $p_{i}$ are pairwise relatively prime integers. The invariants $\mu_{i}=\mu_{i}(X)$ are easily read off.

[TABLE]

The partition $(4,4,1,1)$ cannot be realized by $X$. By way of contradiction suppose $X$ realizes $(4,4,1,1)$. The exceptional types $\tau_{5}$ and $\tau_{6}$ cannot be the types of the rank–$1$ summands because they would be fully invariant. So $\tau_{5},\tau_{6}$ appear together in one of the rank–$4$ summands or they are distributed over the rank–$4$ summands. We have to consider two cases.

Case I. $X=[\tau_{1}\,\tau_{2}\,\tau_{3}\,\tau_{4}]\oplus[\tau_{1}\,\tau_{2}\,\tau_{5}\,\tau_{6}]\oplus[\tau_{3}]\oplus[\tau_{4}]$

In this case $X=X_{1}\oplus X_{2}\oplus[\tau_{3}]\oplus[\tau_{4}]$ where

[TABLE]

and

[TABLE]

We now see that $\mu_{4}=p_{3}p_{4}=\frac{e_{1}}{\alpha_{4}}$ and $\mu_{5}=p_{4}=\frac{e_{2}}{\beta_{3}}$. This means that $p_{4}$ is a factor both of $e_{1}$ and $e_{2}$ contradicting $\gcd(e_{1},e_{2})=1$.

Case II. $X=[\tau_{1}\,\tau_{2}\,\tau_{3}\,\tau_{5}]\oplus[\tau_{1}\,\tau_{2}\,\tau_{4}\,\tau_{6}]\oplus[\tau_{3}]\oplus[\tau_{4}]$

In this case $X=X_{1}\oplus X_{2}\oplus[\tau_{3}]\oplus[\tau_{4}]$ where

[TABLE]

and

[TABLE]

We now see that $\mu_{3}=p_{2}p_{3}=\frac{e_{1}}{\alpha_{3}}$ and $\mu_{4}=p_{3}p_{4}=\frac{e_{1}}{\beta_{4}}$. This means that $p_{3}$ is a factor both of $e_{1}$ and $e_{2}$, a contradiction. ∎

Corollary 7.14**.**

$\mathcal{S}(n,6)$* cannot be realized in $\mathcal{B}$ for $n\geq 10$.*

Example 7.15**.**

The hooked family $\mathcal{S}(7,4)$ cannot be realized in $\mathcal{B}$.

Proof.

For easy reference

[TABLE]

Suppose that $X\in\mathcal{B}$ realizes $\mathcal{S}(7,4)$. Then $X=[\tau_{1}\,\tau_{2}\,\tau_{3}\,\tau_{4}]\oplus[\tau_{1}]\oplus[\tau_{2}]\oplus[\tau_{3}]$ (Lemma 6.2). So there is the exceptional critical type $\tau_{4}$ that appears only once in decompositions. By assumption $X$ realizes $(2,2,2,1)$. The rank–$1$ summand cannot have type $\tau_{4}$ because, if so, it would be a fully invariant summand. Hence without loss of generality $X=[\tau_{4}\,\tau_{1}]\oplus[\tau_{1}\,\tau_{2}]\oplus[\tau_{2}\,\tau_{3}]\oplus[\tau_{3}]$. Explicitly we have

[TABLE]

and $\tau_{1}$ is $p_{1}p_{2}$–free, $\tau_{2}$ is $p_{2}p_{3}$–free, $\tau_{3}$ is $p_{3}$–free, and $\tau_{4}$ is $p_{1}$–free. We can now compute the invariants $\mu_{i}=\mu_{\tau_{i}}(X)$.

[TABLE]

Our group $X$ also realizes $(3,2,2)$. There are two cases depending on whether the exceptional type $\tau_{4}$ belongs to the summand of rank $3$ or not.

Case I. $X=[\tau_{4}\,\tau_{1}\,\tau_{2}]\oplus[\tau_{2}\tau_{3}]\oplus[\tau_{3}\,\tau_{1}]$ and, explicitly,

[TABLE]

where $\tau_{2}$ is $e_{2}$–free, $\tau_{3}$ is $e_{2}e_{3}$–free, and $\tau_{1}$ is $e_{3}$–free. By way of invariants we obtain that

[TABLE]

It follows that $\alpha_{2}p_{2}e_{2}e_{3}=\alpha_{2}p_{2}p_{3}=e_{1}e_{2}$ and hence $e_{3}\mid e_{1}$, a contradiction.

Case II. $X=[\tau_{1}\,\tau_{2}\,\tau_{3}]\oplus[\tau_{3}\tau_{4}]\oplus[\tau_{1}\,\tau_{2}]$ and, explicitly,

[TABLE]

where $\tau_{1},\tau_{4}$ are $e_{2}$–free, and $\tau_{1},\tau_{2}$ are $e_{3}$–free. By way of invariants we obtain that

[TABLE]

Hence $e_{2}p_{2}=\frac{e_{1}}{\alpha_{1}}e_{3}$, an immediate contradiction as $e_{1},e_{2},e_{3}$ are pairwise relatively prime. ∎

Example 7.16**.**

The hooked family $\mathcal{S}(7,5)$ cannot be realized in $\mathcal{B}$. Consequently, neither can $\mathcal{S}(8,5)$ and $\mathcal{S}(9,5)$.

Proof.

For easy reference

[TABLE]

Suppose that $X\in\mathcal{B}$ realizes $\mathcal{S}(7,5)$. Then $X=[\tau_{1}\,\tau_{2}\,\tau_{3}\,\tau_{4}\,\tau_{5}]\oplus[\tau_{1}]\oplus[\tau_{2}]$ (Lemma 6.2). So there are the exceptional critical types $\tau_{3},\tau_{4},\tau_{5}$ that appear only once in decompositions. By assumption $X$ realizes $(3,2,2)$. Each rank–$2$ summand must contain exactly one of $\tau_{1}$ and $\tau_{2}$ in order to avoid a fully invariant summand. Hence without loss of generality $X=[\tau_{1}\,\tau_{2}\,\tau_{3}]\oplus[\tau_{1}\,\tau_{4}]\oplus[\tau_{2}\,\tau_{5}]$. Explicitly we have

[TABLE]

where $\tau_{1},\tau_{4}$ are $p_{2}$–free, and $\tau_{2},\tau_{5}$ are $p_{3}$–free. We can now compute the invariants $\mu_{i}=\mu_{\tau_{i}}(X)$.

[TABLE]

It is easily seen that, in addition to $(3,2,2)$, our $X$ also realizes $(4,2,1)$ and $(5,1,1)$, but it does not realize $(3,3,1)$ as we shall see, so not all of $\mathcal{C}(7,3)$ is realized. Suppose that $X$ realizes $(3,3,1)$. So $X=Y_{1}\oplus Y_{2}\oplus Y_{3}$ with $Y_{3}=\sigma w$ having rank $1$. Then $\sigma$ cannot be one of the exceptional type $\tau_{3},\tau_{4},\tau_{5}$ to avoid a fully invariant summand of rank $1$. So without loss of generality $X=[\tau_{1}\,\tau_{2}\,\tau_{3}]\oplus[\tau_{2}\,\tau_{4}\,\tau_{5}]\oplus\tau_{1}w_{1}$. Explicitly, we have

[TABLE]

By way of invariants we obtain that

[TABLE]

From the last two statements show that $p_{2}\mid q_{2}$ and $p_{3}\mid q_{2}$, and hence $q_{2}=p_{2}p_{3}p_{1}^{\prime}$ for some $p_{1}^{\prime}$. Using that $[X:R(X)]=p_{1}p_{2}p_{3}=q_{1}q_{2}$ it follows that $p_{1}=q_{1}p_{1}^{\prime}$. From the equation for $\mu_{1}$, it now follows that $\beta_{1}q_{1}p_{1}^{\prime}p_{2}=\alpha_{1}q_{1}$, so $\beta_{1}p_{1}^{\prime}p_{2}=\alpha_{1}$. We arrive at the contradiction $p_{2}\mid\alpha_{1}\mid p_{1}$. ∎

Example 7.17**.**

The hooked families $\mathcal{S}(8,6)$ and $\mathcal{S}(9,6)$ cannot be realized in $\mathcal{B}$.

Proof.

For easy reference

[TABLE]

Suppose that $X\in\mathcal{B}$ realizes $\mathcal{S}(8,6)$. Then $X=[\tau_{1}\,\tau_{2}\,\tau_{3}\,\tau_{4}\,\tau_{5}\,\tau_{6}]\oplus[\tau_{1}]\oplus[\tau_{2}]$. The critical types $\tau_{3},\ldots,\tau_{6}$ appear only once in decompositions.

By assumption $X$ realizes $(4,2,2)$. It then also realizes $(6,1,1),(5,2,1),(4,3,1)$ Either one of $\tau_{1},\tau_{2}$ appears in the rank-$4$ summand, or both do. We have two cases:

1. (1)

Case I. $X=[\tau_{1}\,\tau_{3}\,\tau_{4}\,\tau_{5}]\oplus[\tau_{1}\,\tau_{2}]\oplus[\tau_{2}\,\tau_{6}]$ and 2. (2)

Case II. $X=[\tau_{1}\,\tau_{2}\,\tau_{3}\,\tau_{4}]\oplus[\tau_{1}\,\tau_{5}]\oplus[\tau_{2}\,\tau_{6}]$.

Case I. Explicitly we have

[TABLE]

$\tau_{1},\tau_{2}$ are $p_{2}$–free, and $\tau_{2},\tau_{6}$ are $p_{3}$–free. We can now compute the invariants $\mu_{i}=\mu_{\tau_{i}}(X)$.

[TABLE]

Case II. Explicitly we have

[TABLE]

and $\tau_{1},\tau_{5}$ are $p_{2}$–free, $\tau_{2},\tau_{6}$ are $p_{3}$–free. We can now compute the invariants $\mu_{i}=\mu_{\tau_{i}}(X)$.

[TABLE]

Our group $X$ also realizes $(3,3,2)$. There are two cases depending on whether the types $\tau_{1},\tau_{2}$ belongs to different summands of rank $3$ or not.

1. (1)

Case A. $X=[\tau_{1}\,\tau_{3}\,\tau_{4}]\oplus[\tau_{2}\,\tau_{5}\,\tau_{6}]\oplus[\tau_{1}\,\tau_{2}]$ 2. (2)

Case B. $X=[\tau_{1}\,\tau_{2}\,\tau_{3}]\oplus[\tau_{2}\,\tau_{4}\,\tau_{5}]\oplus[\tau_{1}\,\tau_{6}]$

Case A. Explicitly,

[TABLE]

where $\tau_{1},\tau_{2}$ are $q_{3}$–free. The invariants are

[TABLE]

Case B. Explicitly,

[TABLE]

where $\tau_{6},\tau_{1}$ are $q_{2}$–free. The invariants are

[TABLE]

Case I.A. The relevant invariants are

[TABLE]

The equation for $\mu_{6}$ implies $q_{2}=p_{3}\gamma_{6}$. Together with the equation for $\mu_{5}$ we get $\frac{p_{1}}{\alpha_{5}}=\frac{p_{3}\gamma_{6}}{\gamma_{5}}$, further $\frac{p_{1}}{\alpha_{5}}\gamma_{5}=p_{3}\gamma_{6}$, so $\gamma_{6}=\frac{p_{1}}{\alpha_{5}}\gamma_{6}^{\prime}$ for some $\gamma_{6}^{\prime}$. Next we get $q_{2}=p_{3}\frac{p_{1}}{\alpha_{5}}\gamma_{6}^{\prime}$. Using the equation for $\mu_{2}$ we conclude that $\gamma_{2}p_{2}=\frac{p_{1}}{\alpha_{5}}\gamma_{6}^{\prime}q_{3}$