R-Hopfian and L-co-Hopfian Abelian Groups (with an Appendix by A.L.S. Corner on Near Automorphisms of an Abelian Group)

Abstract

The notions of Hopfian and co-Hopfian groups are well known in both non-commutative and Abelian group theory. In this work we begin a systematic investigation of natural generalizations of these concepts and, in the case of Abelian p-groups, give a complete characterization of the generalizations in terms of the original concepts. The final section of the paper contains an unpublished result of A.L.S. Corner on near automorphisms which has been useful in a number of contexts.

In memoriam Rüdiger Göbel

Appendix: Near Automorphisms of an Abelian Group

1.1 A.L.S. Corner, Late of Worcester College, Oxford

1.1.1 Introductory Remarks

The notion of a near automorphism was discussed by Corner in this paper dating from sometime around the early 1960s. It should be noted that it is a different concept from that used nowadays in the theory of torsion-free Abelian groups of finite rank. The note was discovered among Corner’s papers after his death. The hand-written material has been transcribed by Brendan Goldsmith and some comments (in italics) have been added. Please note that in this Appendix, mappings are always written on the right.

1.1.2 The Main Result

Let G be an Abelian group and ι the identity map on G.

Definition

An endomorphism ɛ of G is called a near automorphism if q(ɛ − ı) = 0 for some integer q ≥ 1.

We characterize those groups whose near automorphisms are all automorphisms and those whose monomorphic or epimorphic near automorphisms are all automorphisms.

Theorem A.12

Every near automorphism of G is an automorphism if, and only if, G has no nonzero bounded pure subgroups.

Theorem A.13

The following are equivalent:

• (M) Every monic near automorphism of G is an automorphism

• (E) Every epic near automorphism of G is an automorphism

• (B) G has no bounded pure subgroup of infinite rank.

If q is a positive integer, we call an endomorphism ɛ of G a q-map if q(ɛ −α) = 0 for some automorphism α of G. We say that a group G is q-bounded if qG = 0. It is clear that Theorems A.12 and A.13 are contained in the more precise:

Theorem A.14

Every q-map of G is an automorphism if, and only if, G has no nonzero q-bounded pure subgroup.

Theorem A.15

The following are equivalent:

(M _q ):

Every monic q-map of G is an automorphism

(E _q ):

Every epic q-map of G is an automorphism

(B _q ):

G has no q-bounded pure subgroup of infinite rank.

The proofs of Theorems  A.14 , A.15 are based on two lemmas which are largely computational in nature. In the proof of Lemma A.16 below, Corner used the unexplained term E(x) in relation to the element x of a p-group G; clearly this was intended to mean the exponent, i.e., the least integer n such that p ⁿ x = 0. In modern notation this is often denoted by either e(x) or O(x).

Lemma A.16

If G has no nonzero q-bounded pure subgroup and if ϕ is an endomorphism of G such that qϕ = 0, then ϕ ⁿ = 0 for some integer n ≥ 1.

Proof

• (i) If G is torsion-free, then qϕ = 0 implies that ϕ = 0; so we may take n = 1.

• (ii) Suppose that G is a p-group and let p ^k be the highest power of p dividing q. Since multiplication by qp ^−k affects an automorphism of G, therefore p ^k ϕ = 0; and it is clear that G has no p ^k-bounded pure subgroup. We prove that ϕ ^k+1 = 0. If k = 0 there is nothing to prove; so we suppose that k ≥ 1.

Note first that if x ∈ G and E(x) = 1, then h _G (x) ≥ k. For if h _G (x) = l < k, then x = p ^l y for some y ∈ G, and it is clear that 〈y〉 is a pure subgroup of order p ^l+1, a factor of p ^k, contrary to our hypothesis.

Let $\mathscr{P}(n)$ denote the proposition: x ∈ G, E(x) = n ≤ k ⇒ xϕ ⁿ = 0. We prove $\mathscr{P}(n)$ by induction on n. Since $\mathscr{P}(0)$ is trivial, we may suppose that 1 ≤ n ≤ k and that $\mathscr{P}(r)$ is true for r < n. If x ∈ G and E(x) = n, then E(p ⁿ⁻¹ x) = 1, so h _G (p ⁿ⁻¹ x) ≥ k and therefore p ⁿ⁻¹ x = p ^k z for some z ∈ G. So p ⁿ⁻¹(xϕ) = z(p ^k ϕ) = 0, whence E(xϕ) ≤ n − 1 and so (xϕ)ϕ ⁿ⁻¹ = 0, i.e., xϕ ⁿ = 0.

Since for each x ∈ G we have p ^k(xϕ) = 0, so that E(xϕ) ≤ k, therefore it follows that xϕ ^k+1 = (xϕ)ϕ ^k = 0. Thus ϕ ^k+1 = 0.

• (iii) If G is mixed, write q = ∏ _p p ^k(p) and set n = 1 + max _p k(p). For each prime p, the p-component T _p of the torsion subgroup T of G is mapped into itself by ϕ, so that ϕ induces an endomorphism of T _p . Since this endomorphism of T _p is annihilated by p ^k(p), ϕ ⁿ vanishes on T _p by (ii). Consequently ϕ ⁿ vanishes on T. But the endomorphism of the torsion-free group G∕T induced by ϕ is annihilated by q and so vanishes by (i). Thus ϕ ⁿ vanishes on T and induces the zero endomorphism of G∕T; so ϕ ⁿ = 0 by the Five Lemma.¹

□

Lemma A.17

Let ɛ be an endomorphism of G with q(ɛ − ı) = 0, let A be a maximal q-bounded pure subgroup of G, and B a direct complement of the direct summand A: G = A ⊕ B. Then

$$\displaystyle{ G = A +\mathrm{ Im}\varepsilon \ \ \mathrm{and}\ \ B \cap \mathop{\mathrm{Ker}}\nolimits \varepsilon = 0. }$$

Moreover, there exist endomorphisms λ, μ, λ′, μ′ of G such that

$$\displaystyle\begin{array}{rcl} & \lambda = \imath _{A}\lambda \imath _{A} = (\imath _{A} +\mu \imath _{B})\varepsilon,& {}\\ & \lambda ' = \imath _{A}\lambda '\imath _{A} =\varepsilon (\imath _{A} + \imath _{B}\mu '),& {}\\ \end{array}$$

where ı _A , ı _B are the projections of G onto A, B corresponding to the direct decomposition G = A ⊕ B.

Proof

Write ɛ = ı + ϕ so that qϕ = 0. Then ı _B ϕı _B may be regarded as an endomorphism of B. Since q(ı _B ϕı _B ) = 0 and B has no nonzero q-bounded pure subgroup, it follows from Lemma A.16 that

$$\displaystyle{ (\imath _{B}\phi \imath _{B})^{n-1} = 0\ \ \mathrm{for\ some\ integer}\ n \geq 2. }$$

(1)

Write θ = ı _B − (ı _B ϕı _B ) + (ı _B ϕı _B )² − … + (−ı _B ϕı _B )ⁿ⁻²; since ı _B ɛı _B = ı _B + ı _B ϕı _B , we have, in view of (1)

$$\displaystyle{ \theta \imath _{B}\varepsilon \imath _{B} = \imath _{B}\varepsilon \imath _{B}\theta = \imath _{B}. }$$

(2)

The first claims now follow easily: since ı _B = ı − ı _A , for any x ∈ G we have x = xı _A + xı _B = (x − xθı _B ɛ)ı _A + (xθı _B )ɛ ∈ A + Gɛ. And if $y \in B \cap \mathop{\mathrm{Ker}}\nolimits \varepsilon$, then y = yı _B = yı _B ɛı _B θ = yɛı _B θ = 0.

Since ı _B ı _B = ı _B , we may write (1) in either of the forms ı _B (ϕı _B )ⁿ⁻¹ = 0 or (ı _B ϕ)ⁿ⁻¹ ı _B = 0. Pre- and post-multiplying by ϕ, these become

$$\displaystyle{ (\phi \imath _{B})^{n}\phi = 0\quad \quad \quad \mathrm{and}\quad \quad \quad \phi (\imath _{ B}\phi )^{n} = 0. }$$

(3)

Substituting ϕı _B = ɛı _B + ı _A − ı in the first of these, we find that

$$\displaystyle\begin{array}{rcl} 0& =& (-\phi \imath _{B})^{n}\phi \imath _{ A} = [\imath - (\varepsilon \imath _{B} + \imath _{A})]^{n}\phi \imath _{ A} = (\sum \limits _{r=0}^{n}(-)^{r}\binom{n}{r}(\varepsilon \imath _{ B} + \imath _{A})^{r})\phi \imath _{ A} {}\\ & =& (\varepsilon -\imath )\imath _{A} + (\varepsilon \imath _{B} + \imath _{A})(\sum \limits _{r=1}^{n}(-)^{r}\binom{n}{r}(\varepsilon \imath _{ B} + \imath _{A})^{r-1})\phi \imath _{ A}, {}\\ \end{array}$$

whence

$[\imath _{A} - \imath _{A}\sum \limits _{r=1}^{n}(-)^{r}\binom{n}{r}(\varepsilon \imath _{b} + \imath _{A})^{r-1}\phi \imath _{A}] =\varepsilon [\imath _{a} + \imath _{B}\sum \limits _{r=1}^{n}(-)^{r}\binom{n}{r}(\varepsilon \imath _{b} + \imath _{A})^{r-1}\phi \imath _{A}]$.

Taking λ′ to be the left-hand side, and μ′ to be the summation on the right, we see that λ′ = ɛ(ı _A + ı _B μ′); and it is clear that λ′ = ı _A λ′ı _A . The proof of the corresponding statement for (ı _A + ı _B μ)ɛ is similar. □

With Lemmas  A.16 , A.17 established, it is now easy to give the desired proof of Theorem  A.14 .

Proof of Theorem A.14

( ⇐ ) Let G be a group with no nonzero q-bounded pure subgroup, and let ɛ be a q-map of G, so that q(ɛ −α) = 0 for some automorphism α. Since q(ɛα ⁻¹ − ı) = 0, and ɛ is an automorphism, monomorphism or epimorphism if, and only if, ɛα ⁻¹ is one, it is enough to consider the case α = ı. Then q(ɛ − ı) = 0. In Lemma A.17 we may take A = 0, B = G. Then we have $G =\mathrm{ Im}\varepsilon,\ \mathop{\mathrm{Ker}}\nolimits \varepsilon = 0$; which proves that ɛ is an automorphism of G.

( ⇒ ) Let G be a group with a q-bounded pure subgroup A > 0. Then A is a direct summand of G, so it has a direct complement B (say). If ı _B is the corresponding projection of G onto B, then ı _B is not an automorphism of G; but it is a q-map because q(ı _B − ı) = qı _A = 0. □

The proof of Theorem  A.15 proceeds by showing firstly that (B _q ) implies both (M _q ) and (E _q ); the reverse implications are established using a counter-positive argument.

Proof of Theorem A.15

• (1) Suppose first that G satisfies (B _q ), i.e., that G has no q-bounded pure subgroup of infinite rank, and let ɛ be a q-map of G which is either (i) a monomorphism or (ii) an epimorphism. We prove that in either case ɛ is an automorphism of G. As in the case of Theorem A.14 we may suppose that q(ɛ − ı) = 0. Take A, B as in Lemma A.17, and let λ, λ′, μ, μ′ be endomorphisms as given in that lemma. Note that A being a q-bounded pure subgroup of G is of finite rank, and so is in fact a finite group.

  • (i) If $x \in A \cap \mathop{\mathrm{Ker}}\nolimits \lambda$, then 0 = xλ = (x + xμı _B )ɛ, whence x + xμı _B = 0 because ɛ is a monomorphism, so x = −xμı _B ∈ A ∩ B, i.e., x = 0. Now it follows from the properties of λ given in Lemma A.17 that λ may be regarded as an endomorphism of A; what we have just proved shows that, so regarded, λ is a monomorphism. Since A is finite, it follows that λ maps A onto itself. Consequently, given any x ∈ A, there exists y ∈ A such that x = yλ, i.e., x = y(ı _A + μı _B )ɛ ∈ Gɛ. So A ≤ Gɛ; whence G = Gɛ. Thus the monomorphism ɛ is also an epimorphism; so it is an automorphism, as required.

  • (ii) If x ∈ A, then, because ɛ is given to be an epimorphism, there exists y ∈ G such that x = yɛ; but x = xı _A , so from the properties of λ′ we find that x = yɛı _A = yλ′ − yɛı _B μ′ = yλ′ − xı _B μ′; and since $x \in A =\mathop{ \mathrm{Ker}}\nolimits \imath _{B}$, it follows that x = yλ′. Thus λ′, regarded as an endomorphism of the finite group A, is epic, and therefore monic; so $A \cap \mathop{\mathrm{Ker}}\nolimits \lambda ' = 0$. Now, if $x \in \mathop{\mathrm{Ker}}\nolimits \varepsilon$, we have from the properties of λ′ that xı _A λ′ = xɛ(ı _A + ı _B μ′) = 0, so xı _A = 0 and therefore $x = x\imath _{B} \in B \cap \mathop{\mathrm{Ker}}\nolimits \varepsilon$; whence x = 0. We conclude that the epimorphism ɛ is also a monomorphism, and so an automorphism.

  • (2) Suppose that G does not satisfy (B _q ), so that G admits a q-bounded pure subgroup A (say) of infinite rank. Now A is a direct sum of cyclic groups of orders dividing q; passing to a direct summand of A, if necessary, we may suppose that A is a direct sum of countably many isomorphic cyclic subgroups, say $A =\bigoplus \limits _{ i=1}^{\infty }\langle e_{i}\rangle$. Let B be a direct complement of A in G. Now it is clear that A admits monomorphic q-maps which are not epic, and epimorphic q-maps which are not monic; e.g., the endomorphisms defined by e _i ↦ e _i+1 (i ≥ 1) and by e ₁ ↦ 0,  e _i ↦ e _i−1 (i ≥ 2). If we extend such an endomorphism of A to the whole of G by requiring it to coincide with the identity on B, then the resulting endomorphism of G is clearly a monomorphic or epimorphic q-map, but not an automorphism. □