2 Torsion and cotorsion groups
We begin with a couple of easy observations.
Lemma 2.1.
If G has the PEIS-property, then any summand of G also has the PEIS-property.
Proof.
Suppose G=A⊕B.
If γ∈EA and Aγ is pure in A, then extend γ to σ∈EG by setting Bσ=0.
It follows that Gσ=Aγ is pure in A, and hence in G.
Since G has the PEIS-property, G=Gσ⊕X=Aγ⊕X for some X⊆G so that A=Aγ⊕(X∩A).
∎
Lemma 2.2.
Suppose G is a group with a decomposition ⊕i∈IAi where each Ai is fully invariant in G.
Then G has the PEIS-property if and only if each Ai has the PEIS-property.
Proof.
One direction following immediately from Lemma 2.1, assume that each Ai has the PEIS-property.
If σ∈EG, it follows that Gσ=⊕i∈I(Ai)σ.
So if each Ai=(Ai)σ⊕Bi, then G=Gσ⊕(⊕i∈IBi).
∎
We recall a classical result due to Szele ([2, 3], Theorem 36.1): a basic subgroup of a p-group A is an epimorphic image of A.
Proposition 2.3.
Suppose T is a p-group.
Then T has the PEIS-property if and only if it is isomorphic to B⊕D, where B is bounded and D is divisible.
Proof.
Suppose first that T≅B⊕D as above.
If pkB=0 and σ∈ET, it follows that pk(Tσ)=pk(Dσ)=Dσ is divisible.
In particular, Tσ≅B′⊕Dσ, where pkB′=0.
So Tσ will be (torsion) algebraically compact.
So if Tσ is pure in T, then it must be a summand.
Conversely, suppose T has the PEIS-property, but it is not of the specified form; we will derive a contradiction.
It follows that T=G⊕D, where G is unbounded and D is divisible.
By Lemma 2.1, we can conclude that G will also have the PEIS-property.
Let B be a basic subgroup of G properly contained in G; then B must be pure in G and G/B≠0 is divisible.
By Szele’s result, there is an endomorphism σ∈EG such that B=Gσ.
However, since G/B≠0 is divisible, whereas G itself is reduced, B is clearly not a summand of G.
This contradicts the fact that G has the PEIS-property and completes the argument.
∎
Let 𝒫 be the collection of primes.
If p∈𝒫, denote the maximal p-torsion subgroup of a group A by Ap.
Putting together the last two properties, we have the following result.
Theorem 2.4.
If T≅⊕Tp is torsion, then T has the PEIS-property if and only if, for each p∈P, Tp≅Bp⊕Dp, where Bp is bounded and Dp is divisible.
The next observation will allow us to construct pure subgroups of non-reduced groups.
If G is a group and p∈𝒫, we will denote the p-height of x∈G by |x|G.
Lemma 2.5.
Suppose X is a p-local group, D is a divisible group and ϕ:X→D is injective on pωX.
Let π:X→U:=X/pωX be the canonical epimorphism.
Then μ:X→Y:=U⊕D given by xμ=(xπ,xϕ) defines a pure-exact sequence
0→X→Y→Z→0.
Proof.
By considering ϕ on pωX, we can conclude that μ is injective on pωX.
And if x∈X and |x|X<ω is finite, then |xμ|Y=|xπ|U=|x|X shows that μ preserves all finite heights computed in X and Y.
Therefore, μ must be injective, and the sequence must be pure.
∎
We review a few facts from the theory of cotorsion groups (see [2, 3], Chapter IX).
If T is a reduced p-group, then T∙:=Ext(ℤp∞,T) is the cotorsion hull of T.
The exact sequence 0→ℤ(p)→ℚ→ℤp∞→0 can be used to embed T as the torsion subgroup of T∙; as such, it will necessarily be isotype in T∙.
If T is separable and B is a basic subgroup of T with torsion-completion B¯, then the Ulm subgroup pωT∙ can be identified with the reduced torsion-free algebraically compact group Hom(ℤp∞,B¯/T) (see [2, 3], Proposition 56.5).
Furthermore, the Ulm factor T∙/pωT∙ can be identified with B¯∙, which is naturally isomorphic to the p-adic completion of B (see [2, 3], Lemma 56.6).
Lemma 2.6.
Suppose T is a reduced p-group and A is a pure subgroup of T.
Then there is a surjection T∙→A∙.
Proof.
Let X be any basic subgroup of A, which we extend to a basic subgroup B=X⊕Y of T.
There is a composite epimorphism T→B→X using Szele’s theorem.
So by the standard exact sequence for Ext, there is an epimorphism
T∙=Ext(ℤp∞,T)→Ext(ℤp∞,X)=X∙.
Next, since A/X is divisible, we have an exact sequence
→X∙→A∙→(A/X)∙(=0)→0.
Composing these surjections, T∙→X∙→A∙ gives the result.
∎
Recall that a group is cotorsion if and only if it is the epimorphic image of an algebraically compact group ([2, 3], Proposition 54.1); in particular, any algebraically compact group is necessarily cotorsion.
In addition, any reduced cotorsion group can be naturally expressed as the direct product of p-local cotorsion groups, where p ranges over 𝒫 ([2, 3], Section 54 (I)).
The reduced case of the next result was mentioned in [4].
Theorem 2.7.
Suppose we have an algebraically compact group
G≅∏p∈𝒫Ap⊕(⊕p∈𝒫Dp)⊕D,
where D is torsion-free divisible, Dp is a divisible p-group and Ap is p-local and reduced with torsion Tp.
Then G has the PEIS-property if and only if, for each p∈P, if Tp is unbounded, then Dp=0 and |D|<c (where c denotes the continuum).
Proof.
Suppose first that Dp=0 and |D|<c whenever Tp is unbounded.
Let σ∈EG have a pure image.
We need to show that Gσ is a summand of G.
Note that since G is cotorsion, so is Gσ.
Since Gσ is assumed to be pure in G, it follows that Gσ is a summand of G if and only if Gσ is algebraically compact.
And this holds precisely when its torsion subgroup agrees with the torsion subgroup of an algebraically compact group.
In other words, for every p∈𝒫, we need to show (Gσ)p=Gσ∩(Tp⊕Dp) is the direct sum of a torsion-complete group and a divisible group.
Since Gσ is pure in G, it easily follows that (Gσ)p is pure in Tp⊕Dp.
Suppose first that Tp is bounded.
Arguing as in Proposition 2.3, it is, in fact, easy to see that any pure subgroup of a group that is the direct sum of a bounded and a divisible group also has this property.
Suppose now that Tp is unbounded so that Dp=0 and |D|<c; clearly, Tp must be torsion-complete, and hence separable.
We need to show (Gσ)p is torsion-complete.
Otherwise, since (Gσ)p⊆Tp is separable,
L:=pω[(Gσ)p]∙⊆(Tp)∙⊆G
will be a non-zero torsion-free reduced cotorsion local group, in particular, a torsion-free module over the p-adic integers, Jp.
So we can conclude |L|≥c.
In addition, for all x∈L and all primes p∈𝒫, we will have |x|G≥ω.
However, the only elements of G that have infinite height at all primes are in the maximal divisible subgroup of G, which we denote by M:=(⊕p∈𝒫Dp)⊕D.
Since L is torsion-free, the projection M→D is injective on L.
However, this contradicts the fact that |D|<c.
To prove the converse, we assume p∈𝒫 is chosen such that Tp is unbounded.
We prove that if either |D|≥c or Dp≠0, then G fails to have the PEIS-property.
By showing a summand of G fails to have the PEIS-property, we may assume that G=Ap⊕M=B¯∙⊕M, where B is an unbounded Σ-cyclic group and either M=⊕cℚ or M=ℤp∞.
Let K be a pure dense subgroup of B¯ containing B such that B¯/K≅ℤp∞, so pωK∙≅Hom(ℤp∞,B¯/K)≅Jp.
In addition, we can identify K∙/pωK∙ with B¯∙=Ap.
And by Lemma 2.6, there is an epimorphism Ap=B¯∙→K∙.
Suppose first that M=⊕cℚ.
There is clearly an embedding pωK∙≅Jp→M that we can extend to a homomorphism ϕ:K∙→M.
It follows from Lemma 2.5 that there is a pure-exact sequence
0→K∙→Ap⊕M(=G)→Z→0.
So we can view K∙ as a pure subgroup of G.
And since K∙ is an epimorphic image of Ap, it is also an epimorphic image of G.
However, K∙ is not a summand of G since pωG≅M is divisible and pωK∙≅Jp≠0 is reduced.
So G will not have the PEIS-property.
Similarly, if M=ℤp∞, then we let GK:=K∙/pω+1K∙, which will also be an epimorphic image of K∙, and hence of G=Ap⊕M.
There is clearly an embedding
pωGK=pωK∙/pω+1K∙≅Jp/pJp≅ℤp→M
that we can extend to a homomorphism ϕ:GK→M.
And there are isomorphisms
GK/pωGK=[K∙/pω+1K∙]/[pωK∙/pω+1K∙]≅K∙/pωK∙≅B¯∙=Ap.
It follows from Lemma 2.5 that there is a pure-exact sequence
0→GK→Ap⊕M(=G)→Z→0,
so GK will be a pure endomorphic image of G.
However, GK is not a summand of G since pωG≅M is divisible and pωGK≅ℤp≠0 is reduced.
So again G will not have the PEIS-property.
∎
The following particular case was mentioned in [4].
Corollary 2.8.
If G is reduced and algebraically compact, then G has the PEIS-property.
It is natural to ask if Theorem 2.7, or at least Corollary 2.8, can be extended from groups that are algebraically compact to groups that are cotorsion.
The situation for cotorsion groups is considerably more complicated, however.
Recall that a cotorsion group is adjusted if it has no (non-zero) torsion-free summands.
In particular, if T is a p-group, then T∙ will be adjusted ([2, 3], Lemma 55.4).
The following will be a useful tool in this discussion.
It states that the reduced p-groups whose cotorsion hulls have the PEIS-property are precisely those with a certain projective property.
Lemma 2.9.
If T is a reduced p-group, then T∙ has the PEIS-property if and only if every pure exact sequence
(†)0→A→T→R→0,
where R is reduced, necessarily splits.
Proof.
First suppose T∙ has the PEIS-property; we want to show that the given sequence must split.
By Lemma 2.6, A∙ is a homomorphic image of T∙.
In addition, by the standard properties of Ext, since ℤp∞ is divisible and R is reduced, there is an exact sequence
(†•)0→A∙→T∙→R∙→0.
Since (†•) is pure on its torsion subgroups, it is necessarily pure, as well (or see [2, 3], Lemma 53.5).
So, since T∙ has the PEIS-property, (†∙) must split.
Therefore, restricting to its torsion subgroups, so does (†).
Conversely, suppose each such sequence (†) splits.
If σ is any endomorphism of T∙ and Kσ⊆T∙ is its kernel, then, by [2, 3], Section 54 (C), Kσ and T∙σ will be reduced and cotorsion.
Since T∙ is adjusted, it has no torsion-free summands.
Considering the short exact sequence of cotorsion groups
0→Kσ→T∙→T∙σ→0,
any torsion-free summand of T∙σ would “pull-back” to a torsion-free summand of T∙.
In other words, T∙σ will also be reduced adjusted cotorsion.
Let A=(T∙σ)p so that A∙=T∙σ.
Suppose now that A∙=T∙σ is also pure in T∙, and we consider the pure-exact sequence
0→A∙→T∙→T∙/A∙→0.
Since A∙ is cotorsion, by [2, 3], Section 54 (A), (B), T∙/A∙ is also reduced and cotorsion.
Again, any torsion-free summand of T∙/A∙ would necessarily pull-back to a summand of T∙.
Therefore, T∙/A∙ is also adjusted.
So, if R=(T∙/A∙)p, then it follows that R is reduced and R∙=T∙/A∙.
Therefore, our pure sequence can be written as 0→A∙→T∙→R∙→0.
Its sequence of torsion-subgroups 0→A→T→R→0 will also be pure, so, by hypothesis, it must necessarily split.
Therefore, T∙σ=A∙ is a summand of T∙, as required.
∎
The following consequence of Lemma 2.9 gives an interesting characterization of torsion-complete groups using the PEIS-property.
Proposition 2.10.
If T is a separable p-group, then T is torsion-complete if and only if T∙ has the PEIS-property.
Proof.
If T is torsion-complete, it follows that T∙ is reduced, algebraically compact, and so by Corollary 2.8 has the PEIS-property.
Conversely, suppose T∙ has the PEIS-property.
Let X be any pure subgroup of T.
If we define A⊇X by the property that A/X is the divisible part of T/X, then it follows that A is also pure in T and R:=T/A is reduced.
By Lemma 2.9, we have a splitting T≅A⊕R.
Since T is separable, so is R, and this implies that A is actually the closure of X in T (in the p-adic topology).
So it follows from a result of Koyama (see [2, 3], Proposition 74.9) that T must be torsion-complete.
∎
We now review a bit of terminology.
If T is a p-group, then a subgroup H⊆T is said to be high if it is maximal with respect to the condition H∩pωT=0.
Though a given group may have many high subgroups, it is a classical result that if one high subgroup of T is Σ-cyclic, then all high subgroups are Σ-cyclic (see [6, Theorem 7]).
In addition, if H is a high subgroup of T, then T/H will be divisible and H will be pω+1-pure in T (i.e., 0→H→T→T/H→0 represents an element of pω+1Ext(T/H,H); see [5, Theorem 92]).
If we start with a separable p-group H and, as in the proof of Theorem 2.7, we define GH:=H∙/pω+1H∙, then pωGH=pωH∙/pω+1H∙.
And if TH is the torsion subgroup of GH, then we can view H as a high subgroup of TH.
So again, while a p-local pω-bounded cotorsion-group must have the PEIS-property, the following shows that this statement may or may not be true for p-local pω+1-bounded cotorsion-groups.
Proposition 2.11.
If B is a Σ-cyclic p-group, then GB=B∙/pω+1B∙ has the PEIS-property if and only if B has countable final rank.
Proof.
Suppose first that B has countable final rank; choose k<ω so that pkB is countable.
Note that
pω+1Ext(ℤp∞,TB)≅pω+1GB=0 implies pω+1Ext(D,TG)=0
for any divisible p-group D.
Suppose, as in Lemma 2.9, we have a pure-exact sequence
0→A→TB→𝜋R→0,
where R is reduced; we need to show the sequence splits.
Since B can be viewed as a high subgroup of TB, TB[p]=B[p]⊕(pωTB)[p].
Therefore,
(pkTB)[p]=(pkB)[p]⊕(pωTB)[p],
(pkTB)[p]/(pωTB)[p]≅(pkB)[p]
will be countable.
By purity, we know that ((pkTB)[p])π=(pkR)[p], and clearly,
((pωTB)[p])π⊆(pωR)[p].
It follows that (pkR)[p]/(pωR)[p] will also be countable.
If H is a high subgroup of R, then (pkH)[p]≅(pkR)[p]/(pωR)[p], i.e., pkH is separable and countable.
Therefore, pkH must be Σ-cyclic, and H will be, as well.
So, since Σ-cyclic groups are pure-projective, using [2, 3], Theorem 53.7, and
→Hom(H,TB)→Hom(H,R)→0(=pωExt(H,A)),
there is a homomorphism ϕ:H→TB such that ϕπ is the inclusion H⊆R, i.e., ϕ is a splitting to π on H (remember, we are writing our morphisms on the right, so this composition goes from left to right).
Again, by [5, Theorem 92], H is pω+1-pure in R.
And since D:=R/H is divisible, by [5, Theorem 89 (b)], we have a short-exact sequence
(Hom(D,TB)=) 0→Hom(R,TB)→Hom(H,TB)→0(=pω+1Ext(D,TB)).
It follows that we can (uniquely) extend ϕ:H→TB to ϕ:R→TB.
Now, since ϕπ is the inclusion on H, R is reduced and R/H is divisible, it easily follows that ϕπ is the identity on R.
Therefore, 0→A→TB→R→0 must split, completing this direction.
Suppose now that B has uncountable final rank.
We can clearly construct a pure-exact sequence
0→K→B→C→0,
where C is a separable group that is not Σ-cyclic.
This leads to a pure-exact sequence
0→K∙→B∙→C∙→0,
which again by [2, 3], Theorem 53.7, restricts to the exact sequence
0→pωK∙→pωB∙→pωC∙→0.
Since pωC∙ is torsion-free and pωK∙ is cotorsion, this sequence actually splits.
So, modding out by the elements of height at least ω+1, we have pure sequences
(†)0→GK→GB→GC→0 and 0→TK→TB→TC→0.
We claim that (†) does not split, which implies that GB does not have the PEIS-property.
If, on the contrary, there was such a splitting, then we could conclude that TB≅TK⊕TC.
This cannot be since TB has a high subgroup (≅B) that is Σ-cyclic, whereas TK⊕TC has a high subgroup (≅K⊕C) that is not Σ-cyclic.∎
3 Torsion-free groups
We now consider the case where G is torsion-free.
We first note that, unlike the algebraically compact case, for torsion-free groups, we may ignore the divisible part of the group.
Lemma 3.1.
Suppose G≅R⊕D is torsion-free, where R is reduced and D is divisible.
Then G has the PEIS-property if and only if R has the PEIS-property.
Proof.
One direction following immediately from Lemma 2.1, suppose that R has the PEIS-property.
We want to show that G has the PEIS-property.
So assume σ∈EG and Gσ is pure in G; we need to verify that Gσ is also a summand of G.
Since G is torsion-free, E=Gσ∩D is pure in G.
And since D is divisible, so is E.
It follows that if G′:=G/E, then σ induces an endomorphism σ′ on G′ and that G′σ′ will be pure in G′.
We claim that it suffices to show G′ splits as G′σ′⊕X. If this holds, then define the subgroup Y⊆G by the formula Y/E=X.
Since E is divisible, there is a splitting Y=E⊕Z.
It is easily seen that G=Gσ⊕Z, as required.
So, replacing G by G′ and σ by σ′, without loss of generality, we may assume that D∩Gσ=0.
Since D is fully invariant, it follows that Dσ=0 and Gσ=Rσ.
Note that, since D is divisible, there is a decomposition G=R′⊕D, where Gσ⊆R′; in particular, R≅G/D≅R′.
Replacing R by R′, without loss of generality, we may assume that Gσ=Rσ⊆R is pure.
Since we are assuming that R has the PEIS-property, it follows that Gσ is a summand of R, and hence a summand of G, as required.
∎
If τ is a type, we let ℚτ⊆ℚ be a rank-one torsion-free group with that type.
For simplicity, we will always assume 1∈ℚτ, i.e., ℤ⊆ℚτ.
So, for each prime p, τp can be thought of as the supremum of the k<ω such that 1/pk∈ℚτ.
Lemma 3.2.
If τ is a fixed type and I is a set, then the following are equivalent.
The completely decomposable group G=⊕i∈Iℚτ𝐞i has the PEIS-property;
The vector group H=∏i∈Iℚτ𝐞i has the PEIS-property;
ℚτ is divisible or I is finite.
Proof.
Certainly, the result is clear if ℚτ is divisible since then G and H are also divisible.
So we may assume ℚτ is reduced.
Suppose first that I is finite.
So G=H, and it follows from a result of Baer ([2, 3], Proposition 86.8) that any pure subgroup of a homogeneous finite rank torsion-free group is a summand.
Therefore, G=H has the PEIS-property.
Suppose next that I is infinite, so we need to show that G and H do not have the PEIS-property.
By restricting to a countably infinite subset of I and using Lemma 2.1, we may assume that I=ω is the finite ordinals.
Construct a projective resolution of ℚ,
0→⊕n<ωℤ𝐞n→⊕n<ωℤ𝐞n→ℚ→0.
If we tensor this with the (flat ℤ-module) ℚτ, then, since ℚτ⊗ℤ≅ℚτ and
ℚτ⊗ℚ≅ℚ, we obtain a sequence
0→G→G→ℚ→0.
Now, since ℚ is torsion-free, G contains an isomorphic copy, and so an endomorphic image, as a pure subgroup.
However, since G is reduced, this sequence cannot split so that G does not satisfy the PEIS-property.
A dual argument shows that H also fails to satisfy the PEIS-property.
Since ℚτ is reduced, we have Hom(ℚ,ℚτ)=0, and since ⊕n<ωℤ𝐞n is projective, we have Ext(⊕n<ωℤ𝐞n,ℚτ)=0.
So if we apply the functor Hom(-,ℚτ) to our projective resolution of ℚ, we arrive at
0→H→H→Ext(ℚ,ℚτ)→0.
Since ℚ is torsion-free divisible, so is Ext(ℚ,ℚτ).
And since ℚτ is certainly not cotorsion, we can conclude that Ext(ℚ,ℚτ)≠0.
Therefore, again, H can be viewed as a pure endomorphic image of itself, which cannot be a summand, since H is reduced and Ext(ℚ,ℚτ)≠0 is divisible.
∎
The following innocent-looking result is actually the key to generalizing the last result to the non-homogeneous case.
Recall that if A is a torsion-free group and τ is a type, then A(τ) is the collection of elements whose type is at least τ, and A*(τ) is the subgroup generated by those elements whose type strictly exceeds τ.
Lemma 3.3.
Suppose τ is a type and G=Qτe⊕A is torsion-free.
If G has the PEIS-property, then A*(τ) is divisible.
Proof.
The result is obvious if ℚτ=ℚ, so assume ℚτ is reduced.
We will think of the following proof as taking place in the injective hull ℚG=ℚ𝐞⊕ℚA.
So if x∈G and β∈ℚ, then βx∈ℚG, though possibly βx∉G.
Let p be a prime.
If ℚτ is p-divisible, then A*(τ) is also clearly p-divisible.
So suppose ℚτ is not p-divisible; replacing 𝐞 by (1/pk)𝐞 for some k<ω, there is no loss of generality in assuming that 1∈ℚτ and 1/p∉ℚτ.
We still need to show that A*(τ) remains p-divisible.
If this fails, then we can find an a∈A such that the type of a is strictly greater than τ and (1/p)a∉A; i.e., (1/p)a∉G .
Since a∈A(τ), after replacing a by na for some n relatively prime to p, we may assume that ℚτa⊆A.
Define σ∈EG as follows: if α∈ℚτ and x∈A, then set
(α𝐞+x)σ=α(p𝐞+a):=αy;
in other words, 𝐞σ=p𝐞+a=y and Aσ=0.
To verify Gσ=ℚτ(p𝐞+a)=ℚτy is pure in G, suppose q is a prime, and (1/q)y∈G.
We need to show (1/q)y∈Gσ=ℚτy, i.e., 1/q∈ℚτ.
We can restate our assumption as
(p/q)𝐞+(1/q)a=(1/q)y∈G,
that is, p/q∈ℚτ and (1/q)a∈A.
Since (1/p)a∉A, the second statement implies that q≠p.
Therefore, since q/q=1∈ℚτ, p/q∈ℚτ and p,q are coprime, we can conclude that 1/q∈ℚτ, as required.
Since Gσ is a pure endomorphic image and we are assuming that G has the PEIS-property, there is a decomposition G=Gσ⊕C; let π:G→Gσ=ℚτy be the corresponding projection onto the first summand.
Since the type of a is strictly greater than τ, we can conclude that aπ=0, i.e., a∈C.
Therefore,
(1/p)y-(1/p)a=𝐞∈G=ℚτy⊕C
implies that 1/p∈ℚτ.
This contradiction completes the proof.
∎
The following consequence of Lemma 3.3 is an interesting application of the PEIS-property.
Proposition 3.4.
A finite rank torsion-free group A is the direct sum of a free group and a divisible group if and only if Ze⊕A has the PEIS-property.
Proof.
If A=R⊕D, where R is reduced and D is maximal, then, by Lemma 3.1, A (respectively, ℤ𝐞⊕A) has the PEIS-property if and only if R (respectively, ℤ𝐞⊕R) does.
So we may assume A=R is reduced.
Suppose first that A is free.
Therefore, the same holds for A^:=ℤ𝐞⊕A, and by Lemma 3.2, it has the PEIS-property.
Conversely, suppose A^ has the PEIS-property.
If 0¯ is the type of ℤ, it follows from Lemma 3.3 that A*(0¯)=0.
But this clearly means that A is homogeneous of type 0¯.
If 〈x〉 is any rank-1 pure subgroup of A, then there is clearly an endomorphism σ∈EA^ such that A^σ=〈x〉.
Therefore, every such rank-1 pure subgroup will be a summand of A.
This clearly implies that A must be free.
∎
This brings us to our description of when completely decomposable groups and vector groups have the PEIS-property.
Our argument will utilize a few basic facts about slender groups that can be found, for example, in [2, 3], Chapter XII.
Theorem 3.5.
Suppose I is a set and, for each i∈I, Qi⊆Q has type τi.
Then the following are equivalent.
The completely decomposable group G=⊕i∈Iℚi𝐞i has the PEIS-property.
The vector group H=∏i∈Iℚi𝐞i has the PEIS-property.
The following two conditions are satisfied:
for all i,j∈I, if τi<τj, then ℚj=ℚ;
for each i∈I, if ℚi≠ℚ, then [i]={j∈I:τj=τi} is finite.
Proof.
Suppose first that (b) holds; we will verify that (c) must hold as well (the argument that (a) also implies (c) is virtually identical and left to the reader).
Regarding (1), suppose i,j∈I and τi<τj; in particular, we must have i≠j.
Therefore, ℚi𝐞i⊕ℚj𝐞j will be a summand of H, so, by Lemma 2.1, it also has the PEIS-property.
It follows from Lemma 3.3 that ℚj𝐞j=(ℚj𝐞j)*(τi) is divisible, as required.
Turning to (2), let [i] be as in the statement.
Note that ∏i∈[i]ℚi𝐞i will be a summand of G, so, again by Lemma 2.1, it also has the PEIS-property.
It follows from Lemma 3.2 that [i] is finite, as required.
Suppose now that (c) holds; we will verify (a) and (b).
If J={i∈I:ℚi≠ℚ}, then it follows from Lemma 3.1 that, in either case, there is no loss of generality in assuming I=J so that G and H are reduced.
If we let i∼j mean that τi=τj, then ∼ is trivially an equivalence relation.
Let ℐ=I/∼ be the collection of equivalence classes; so each [i]∈ℐ must be finite.
If [i]∈ℐ, let Q[i]=∏j∈[i]ℚj𝐞j=⊕j∈[i]ℚj𝐞j.
Note that if i≁i′, then, since τi′≰τi, we can conclude that Hom(Q[i′],Q[i])=0.
First, we observe that (a) follows immediately from Lemma 3.2 and Lemma 2.2 since each Q[i] will have the PEIS-property and be totally invariant in G.
Regarding (b), (2) implies that |I|≤c (where c is the continuum); in particular, it is non-measurable.
In addition, by [2, 3], Proposition 94.2, Q[i] will be slender.
Again, since for i≁i′, Hom(Q[i′],Q[i])=0, a standard argument using [2, 3], Corollary 94.5, shows that there is a natural isomorphism EH≅∏[i]∈ℐEQ[i].
Therefore, if σ∈EH, then we can write σ as (σ[i])[i]∈ℐ, and the image of σ can be expressed as ∏[i]∈ℐQ[i]σ[i].
So if Gσ is pure in G, each Q[i]σ[i] will be pure in Q[i].
But by Lemma 3.2, each Q[i]σ[i] will be a summand of Q[i].
This, in turn, implies that Gσ is a summand of G, as required.
∎
Suppose G is a group with the PEIS-property.
If H⊆G is both a pure subgroup and an endomorphic image, then it is actually a summand of G.
And by Lemma 2.1, this implies that H also has the PEIS-property.
On the other hand, simple examples using Theorem 3.5 show that we must assume both of these two conditions to conclude that H also satisfies the PEIS-property.
Example 3.6.
There is a finite rank torsion-free group G with the PEIS-property with a pure subgroup P that does not have the PEIS-property and a torsion-free epimorphic image E that does not have the PEIS-property.
Proof.
Let p, q be distinct primes.
Using the notation of [2, 3], Section 88, let
G=〈p-∞𝐞1,p-∞𝐞2,q-∞𝐞3〉⊆ℚ3.
Since these two types are incomparable, it follows that G is a completely decomposable group with the PEIS-property.
On the other hand,
P:=〈p-∞𝐞1,𝐞2+𝐞3〉
will be a pure subgroup of G that does not have the PEIS-property.
In addition, mapping 𝐞1↦𝐟1 and 𝐞2,𝐞3↦𝐟2 gives a surjection
G→E:=〈p-∞𝐟1,p-∞𝐟2,q-∞𝐟2〉⊆ℚ2.
Clearly, E does not have the PEIS-property.
∎
The following observation shows that going from “finite rank completely decomposable” to “finite rank almost completely decomposable” does not yield anything new.
Proposition 3.7.
Suppose G is a finite rank torsion-free group that is almost completely decomposable.
If G has the PEIS-property, then it is completely decomposable.
Proof.
Suppose C=Q1⊕⋯⊕Qk is a subgroup of G such that G/C is finite and each Qi has rank 1.
The projection C→Q1 extends to a homomorphism π:G→ℚQ1 such that Gπ/Q1 is finite.
Therefore, if Q1* is the purification of Q1 in G, then Q1⊆Q1*⊆Gπ are all isomorphic.
So, after possibly composing π with multiplication by some rational number, we can conclude that Q1* is a pure-endomorphic image of G.
Therefore, G=Q1*⊕G′, where G′ will also be almost completely decomposable with the PEIS-property.
It follows by induction that G′, and hence G, is completely decomposable.
∎
Using Theorem 3.5, it is straightforward to construct a finite rank torsion-free completely decomposable group G with the PEIS-property and a group H that is quasi-equal to G but not completely decomposable (so it will be almost completely decomposable; for example, G=〈2-∞𝐞1,3-∞𝐞2〉, and H=G+〈15(𝐞1+𝐞2)〉).
Thus, even though G has the PEIS-property and G is quasi-equal to H, by Proposition 3.7, H fails to have the PEIS-property.
The following observation clearly generalizes the fact that a finite rank torsion-free homogeneous group has the PEIS-property.
Proposition 3.8.
Suppose A is a torsion-free group such that R:=EA is a PID for which any ideal is of the form (n) for some n∈Z.
Then
G=Ak=A⊕A⊕⋯⊕A
will have the PEIS-property.
Proof.
We can think of any endomorphism σ∈EG as being a matrix S=[σi,j] (for 1≤i,j≤k), where each σi,j∈R.
It follows that there are k by k unimodular (i.e., invertible) matrices P, Q over R such that PSQ is diagonal with entries that are in ℤ.
If Gσ is pure in G, it follows that each of the diagonal entries of PSQ are 0 or 1.
Using the R-module bases implied by these matrices, Gσ corresponds to the R-span of the standard basis elements corresponding to where the 1’s appear.
This is evidently a summand of G.
∎
Some of our examples, and results such as Proposition 3.8, seem to imply that the PEIS-property is essentially ring-theoretic, especially in the torsion-free case.
In other words, if G and H are torsion-free groups with isomorphic endomorphism rings, then G has the PEIS-property if and only if the same holds for H.
The following example, however, shows even quasi-equal torsion-free finite rank groups with the same endomorphism ring can differ with respect to the PEIS-property.
Example 3.9.
There are quasi-equal torsion-free groups G and H of rank 4 such that EG=EH, for which G does not have the PEIS-property, but H does.
Proof.
Suppose 𝒬={p∈𝒫:p>7} and ℚτ:=〈1p:p∈𝒬〉⊆ℚ.
If V=ℚ𝐞1⊕ℚ𝐞2, W=ℚ𝐟1⊕ℚ𝐟2, then let G⊆V⊕W be defined by
G=〈3-∞𝐞1,5-∞𝐞2,7-∞(𝐞1+𝐞2),3-∞𝐟1,5-∞𝐟2,7-∞(𝐟1+𝐟2),1p2𝐞1+1p𝐟1,1p2𝐞2+1p𝐟2:p∈𝒬〉.
Let K:=G∩V so that K is pure in G.
It can be seen that
G(τ)=〈3-∞𝐞1,5-∞𝐞2,7-∞(𝐞1+𝐞2)〉=K(τ)
so that this subgroup is fully invariant in G.
In addition, we will have
K=K(τ)+(ℚτ𝐞1⊕ℚτ𝐞2).
It easily follows that EK(τ)=EG(τ)≅ℤ, i.e., every such endomorphism is multiplication by an integer.
And if γ:K→G is a homomorphism, then γ restricts to an endomorphism K(τ)→G(τ)(=K(τ)).
Since K(τ) is full-rank in K, it follows that γ must equal nξ, where n∈ℤ and ξ:K→G is the inclusion map, i.e., Hom(K,G)=ℤξ.
If 𝐟1′, 𝐟2′ are the images of 𝐟1 and 𝐟2 in G/K, it follows that
G/K=〈3-∞𝐟1′,5-∞𝐟2′,7-∞(𝐟1′+𝐟2′)〉+(ℚτ𝐟1′⊕ℚτ𝐟2′)≅K⊆G,
where this composition, which we denote by μ:G/K→G, is defined by 𝐟1′μ=𝐞1 and 𝐟2′μ=𝐞2.
Therefore, if γ:(G/K)(≅K)→G, then γ=mμ for some m∈ℤ, i.e., Hom((G/K),G)=ℤμ.
Suppose now that ϕ∈EG is defined by 𝐞1ϕ=𝐞2ϕ=0 and 𝐟1ϕ=𝐞1, 𝐟2ϕ=𝐞2, i.e., it is the natural composition G→G/K→𝜇G.
Let R⊆EG be the subring generated by 1G and ϕ, i.e., those endomorphisms of the form n+mϕ, where n,m∈ℤ.
There is an exact sequence
0→(ℤμ=)Hom(G/K,G)→EG→Hom(K,G)(=ℤξ).
Since R⊆EG contains Hom(G/K,G) and maps onto Hom(K,G), we must have R=EG.
Put another way, EG can be identified with right multiplication of ordered pairs of vectors in V⊕W by integer-valued matrices of the form
[n0mn],
where ϕ corresponds to the matrix with m=1, n=0.
In particular, if σ=n+mϕ∈EG is an idempotent, then σ2=σ readily implies that n=0,1 and m=0.
In other words, G is indecomposable.
Therefore, K=Gϕ is a pure endomorphic image that is not a summand of G, so that G does not have the PEIS-property.
We now let H=G+〈12𝐞1〉, so that G and H are quasi-equal.
We first claim that EH=EG.
Using the above notation, it is easy to see that Hϕ=K⊆H so that ϕ∈EH.
So, for any m,n∈ℤ, n+mϕ∈EH, i.e., EG⊆EH.
Suppose now that σ∈EH.
Since H/G≅ℤ2, it follows that
G(2σ)⊆2Hσ⊆2H⊆G
so that 2σ∈EG.
In other words, 2σ=n+mϕ for some m,n∈ℤ.
Therefore,
2(𝐟2σ)=(𝐟2)2σ=(𝐟2)(n+mϕ)=m𝐞2+n𝐟2
so that
(m/2)𝐞2+(n/2)𝐟2=𝐟2σ∈H.
From the above description of H, it follows that m and n are even.
Therefore, σ=(n/2)+(m/2)ϕ∈EG, as desired.
Suppose now that σ=n+mϕ∈EH=EG has a pure image, Hσ⊆H; we want to show that Hσ must be a summand of H.
If n≠0, it follows that σ is injective (the matrix for the corresponding vector space endomorphism on ℚ4 will have determinant n4≠0).
In that case, Hσ will have rank 4.
So if it is pure in H, then H=Hσ, which is certainly a summand.
So assume n=0.
If m=0, as well, then clearly 0=Hσ is a summand.
Finally, if n=0 and m≠0, then we have Hσ=mHϕ=mK.
It follows that 12𝐞1∈H∖K⊆H∖Hσ, but 2m(12𝐞1)=m𝐞1∈Hσ.
In other words, Hσ will not be pure in H, and this contradiction completes the argument.
∎
In [4], a group G was said to be pure-image simple if the only pure endomorphic images Gσ⊆G are 0 and G.
In [4, Lemma 8 (3)], it was asserted that if G is torsion-free and homogeneous of rank at most 3, then G is pure-image simple if and only if it is strongly indecomposable.
The proof of this assertion was somewhat opaque (at least to this reader).
In the remainder of this work, we aim to clarify this proof and to relate it to the PEIS-property.
In particular, we replace the condition “strongly indecomposable” by the more straightforward “indecomposable”. (In fact, for torsion-free homogeneous groups of rank at most 3, the two notions can be shown to be equivalent.)
Suppose G is a torsion-free finite rank group and A, B are quasi-equal subgroups of G; i.e., there are n,m∈ℤ such that mA⊆B and nB⊆A.
If p is relatively prime to mn, it is well known that (G/A)p≅(G/B)p, and that, even for the finite number of primes that divide mn, the maximal divisible parts of (G/A)p and (G/B)p will be isomorphic.
So G/A and G/B really only differ by finite summands.
If τ is a type, we will say a torsion-free group A of finite rank is τ-completely decomposable if it is completely decomposable and homogeneous of type τ.
If G is any finite rank torsion-free group, then F⊆G will be said to be τ-full-rank if F is τ-completely decomposable and full-rank in G (i.e., G/F is torsion).
It is easy to check that G has a τ-full-rank subgroup if and only if G(τ)=G, and that this condition is equivalent to τ≤γ, where γ is the inner type of G (see [1, Proposition 1.7]).
In addition, if F is a τ-full-rank subgroup of G, then a subgroup F′⊆G will also be τ-full-rank if and only if F and F′ are quasi-equal (see [2, 3], Proposition 98.1).
We also remind ourselves of the following consequence of a result of Baer.
Lemma 3.10 ([2, 3], Proposition 86.5).
Suppose τ is a type and G is a torsion-free group with G(τ)=G.
If A is a pure subgroup of G such that G/A is τ-completely decomposable, then A is a summand of G.
If G is a group and A is a pure subgroup of G, we will say a subgroup K⊆G is a kernel for A if G/K≅A; in other words, there is a σ∈EG with image A and kernel K.
This brings us to the following interesting result.
Theorem 3.11.
Suppose G is a finite rank torsion-free group with inner type τ and A⊆G is a pure subgroup.
Then A has a τ-completely decomposable kernel if and only if A has a τ-completely decomposable complementary summand, i.e., G=A⊕B, where B is τ-completely decomposable.
Proof.
Certainly, if G=A⊕B as above, then B will be a τ-completely decomposable kernel for A.
So assume K is a τ-completely decomposable kernel for A; we want to produce a τ-completely decomposable complementary summand of A.
Setting B:=G/A, we have pure-exact sequences
0→K→G→A→0 and 0→A→G→B→0.
If we can show that B is also τ-completely decomposable, then the second sequence will split by Lemma 3.10, completing the proof.
Let F⊆G be such that K⊕F is a τ-full-rank subgroup of G.
It follows that F¯:=(K⊕F)/K≅F can be viewed as a τ-full-rank subgroup of A.
There is an isomorphism
(†)G/(K⊕F)≅(G/K)/(K⊕F/K)=(G/K)/F¯≅A/F¯.
We can extend F¯, which is a τ-full-rank subgroup for A, to a τ-full-rank subgroup for G of the form F¯⊕E; in particular, (A⊕E)/A:=E¯ is τ-full-rank in B.
This leads to an exact sequence
(‡)0→A/F¯→G/(F¯⊕E)→B/E¯→0.
Since F¯⊕E and K⊕F are both τ-full-rank in G, they are quasi-equal.
In particular, there is a set 𝒬 containing all but finitely many primes, such that, for all p∈𝒬,
(G/(F¯⊕E))p≅(G/(K⊕F))p,
which, by (†), is isomorphic to (A/F¯)p.
So, since all these p-groups have finite rank, and finite rank p-groups do not have isomorphic proper subgroups, for all of those primes p∈𝒬, (‡) implies we must have (B/E¯)p=0.
And even if p is a prime not in 𝒬,
(G/(F¯⊕E))p and (G/(K⊕F))p≅(A/F¯)p
have isomorphic maximal divisible subgroups.
Therefore, (B/E¯)p will be finite for these p∉𝒬.
It follows that B/E¯ will be finite, i.e., B and E¯ are quasi-equal.
So B, like E¯, is τ-completely decomposable, finishing the proof.
∎
Corollary 3.12.
Suppose G is a torsion-free homogeneous group of rank n<ω.
If A is a pure endomorphic image of G of rank n-1, then A is a summand of G.
Proof.
Let τ be the type of G.
Now A has a kernel K of rank 1.
Trivially, K is τ-completely decomposable.
The result, therefore, follows from Theorem 3.11.∎
Corollary 3.13.
Suppose G is a homogeneous torsion-free group of rank at most 3.
Then G is pure-image simple if and only if it is indecomposable.
Proof.
Certainly, if G is pure-image simple, then it must be indecomposable.
Conversely, suppose G fails to be pure-image simple.
Suppose A is a non-trivial pure endomorphic image of G of rank 1 or 3-1=2.
If A has rank 1, then it has type τ and a proper splitting of G follows from Lemma 3.10.
And if A has rank 2, then G has rank 3 and a proper splitting follows from Corollary 3.12.
∎
Corollary 3.14.
Suppose G is a homogeneous torsion-free group of rank at most 3.
Then G has the PEIS-property if and only if either it is completely decomposable or pure-image simple.
Proof.
Clearly, if G is completely decomposable or pure-image simple, then it has the PEIS-property.
So suppose G is homogeneous and not pure-image simple; we want to show that it is completely decomposable.
By Corollary 3.13, there is a decomposition G=A⊕B, where A has rank 1.
But any pure rank-1 subgroup of B will be isomorphic to A, and so an endomorphic image of G.
So if G has the PEIS-property, B will decompose into rank-1 summands, so that G is completely decomposable.
∎
On the other hand, it is not difficult to construct a homogeneous torsion-free group A of rank 2 such that EA is equal to the integers localized at a collection of primes (see, for example, [1, Example 2.4]).
It follows from Proposition 3.8 that G=A⊕A has rank 4 and satisfies that PEIS-property, but it is neither completely decomposable nor pure-image simple.
