THE GROUP OF HOMOMORPHISMS OF ABELIAN TORSION GROUPS

Let G and A be abelian torsion groups. In[5], R. S. Pierce develops a complete set of invariants for Hom(G, A). To compute these invariants he introduces, and uses extensively, the group of small homomorphisms of G into A. Also, using some of Pierce's methods, Fuchs characterizes this group in [1]. Our purpose in this paper is to characterize Hom(G, A) in what seems to be a more natural manner than either of the treatments just mentioned.


M.W. LEGG numbers.
In this paper we will make use of methods introduced in [3] to cal- culate these invariants for Hom(G,A).
The notation used here will mainly conform to that of Fuchs [I].In this paper group means additive abelian group.The maximal divisible subgroup of the group G will be denoted dG If dG 0, G is called reduced.We will use G to p denote the p-primary component of G and tG to represent the torsion subgroup of G.
If Q is the group of rationals and Z is the group of integers, then (Q/Z)p is denoted Z(p).Z(n) denotes the cyclic group of order n.

PRELIMINARIES.
A group G is called algebraically compact (Maranda 4) if for every group X, any homomorphlsm from a pure subgroup of X to G can be extended to a homomorphlsm of X to G.
We now recall two definitions and the main result from the paper by Legg and Walker that was mentioned above.
Let G be a group, p a prime, and n a non-negatlve integer.Let G (n) denote P the dimension of (pnG)Ep/(pn+IG)p as a vector space over the integers modulo p.That is, Gp(n) is the n-th Ulm invarlant of G (with respect to p).If G is a group and p is a prime, then G0( p shall denote the dimension of G/(pG+tG) as a vector space over the inte{ers modulo p.We call GO( p the p-th torsion free number of G. The Ulm invariants and the torsion free numbers are a complete set of invariants for algebraically compact .groups.That is, two algebraically compact groups A and B are isomorphic if and only if (i) dA -dB (ii) A 0(p) B 0(p) for all primes p, and (iii) Ap(n) Bp(n) for all primes p and all non-negative integers n.
We now proceed to calculate Horn (G,A)(n) and Hom0(G,A)(p) for the algebrlcally P compact Hom(G,A).
It will suffice, as we shall show later, to confine our calculations to p- primary groups.So, for the time being, all groups will be understood to be of this type.
Our calculation of the Ulm invariants of Hom(G,A) requires that we determine the adjusted component of Hom (G,A).This result is interesting in itself.

THE ADJUSTED COMPONENT OF HOMG A).
The following definition and results are due to Harrison [2].These results were subsequently obtained by Legg and Walker [3] using quite elementary methods.
A reduced algebraically compact group is called adjusted if it has no torsion free direct summands.A reduced algebraically compact group A is the direct sum of an adjusted algebraically compact group and a torson free algebraically compact group.
The adjusted component of A is unique.Further, G is the adjusted part of A if and only if G/tA d(A/tA). (For an arbitrary G, the adjusted component of the reduced part of G is called the adjusted component of G).
The adjusted component of Hom(G,A) will be denoted AdJ(G,A) and its elements will be called adjusted homomorphisms.THEOREM 3.1.AdJ(G,A) is the set of all e Hom(G,A) such that for each integer m > 0 there is an integer n >-0 such that (i) (pnG) c__ p A and (il) (pnG)[pm] c__ Ker .
PROOF.Let S(G,A) be the subgup of Hom(G,A) consisting' f those homomorphisms that satisfy conditions (i) and (ii) of the theorem.Let e AdJ(G,A) and m >_ 0 be an integer.
By the remarks preceding this theorem, we will be done if we show that M.W. LEGG S(G,A)/tHom(G,A) is divisible.
Since Hom(G,A) is q-divislble for primes q # p and since Hom(G,A)/S(G,A) is clearly q-torsion free, it follows that S(G,A)/tHom(G,A) is q-dlvlsible.We show now that it is p-divisible.Let e S(G,A) and choose n > 0 such that pn+ 1A Gn+1 / An+I by 8 l(pn+Ix) (pnx).It is easy to check that 81 is well- n+l defined.Let D be the divisible hull of An+I.Since p Gn+1 is essential in n+l.
Gn+l, we can extend 81 to a homomorphism 82 from Gn+I to D. Also p an+I is essential in An+I so that 82(Gn+I) c__ An+I.
It remains to show that 8 e S(G,A).It is routine that pn e S(G,A); thus pn+ 18 e S(G,A).We must show that Hom(G,A)/S(G,A) is p-torsion free Suppose y e Hom(G,A) and py e S(G,A).Let m > 0 be an integer.Choose n > 0 such that py (pnG) pm+n+iA and (pnG)[ pro+l] c Ker py.Then 7 (Pn+IG) py(pnG) _c pm+n+iA_c pm+nA.Also (pn+IG)[ pm] _c Ker The proof is complete.
We next investigate the functorial behavior of Adj (G,A).The proof of our first lemma depends on the following result [3]: If A is pure in B, then there exists a group C, containing B, with C/A divisible and A pure in C.
LEMMA 3.2.If A is pure in B and b AdJ(A,X), then can be extended to an adjusted homomorphism of B into X.
PROOF.It is immediate that the restriction of an adjusted homomorphism is adjusted.Hence, by the remark preceding this lemma, we can assume that B/A is Suppose also (pkA)[pm] c__ Ker , and assume k > n.Write b Pk + ak' wlth O(bk) o(b k + A).Then a k Alpm].We have that pn(b n pk-nb k) a k an.Since n A is pure in B, there is an a A so that p a a Thus a "-'(pnA)tpmj.
Hence _(a n) (a k) so that is well-deflned.It is routine to check that is a homomorphism.
Finally, we show that is adjusted.Let m > 0 be an integer.Then there is an integer n > 0 so that (pnA)[pm] c__ Ker and (pnA) c__ p X. Let b e (pnB) [pm]    so that, by purity, there is an a E A such that p a a ,i.e. a E It follows that (b) (a s) pnx.Thus (pnB) c__ pm+nx.
We have shown that Adj (B,X).Since extends , the lemma is proved.
If Adj (X,Y) and a Hom(A,X), then it is routine to show that AdJ (A,Y).
We must show that is adjusted.Let m >_ 0 be an integer.Since is adjusted, there is an integer n >_ 0 such that (pnB)[pm] c_ Ker and (pnB)c_ pmx.
Clearly (pn(B/A)) =_ pmX.Thus is adjusted.Hence is adjusted, which shows that Ker * c Im q*.We have shown exactness at AdJ (B,X).
The fact that * is onto is a consequence of Lemma 3.2.
Hence (*) is pure exact.Since AdJ(B/A,X) is algebraically compact, (*) is split exact.The theorem is proved.
Two important results follow immediately from Theorem 3.3.
(n) for all (n) Hom(B,A)p COROLLARY 3 5 If B is basic in G, then Hom(G,A)p integers n >_ 0.
PROOF.This follows from Theorem 3.4 and the observation that the Ulm invariants of an algebraically compact group are the same as the Ulm Invariants of its adjusted part.
We are now ready to calculate the Ulm invarlants for Hom(G,A).i=l Horn ,A) --" i_l i-l)

P P pn
Observe that for m > n, (pn(A[pm]))[p] (pnA) [p] and for m n, (A[pm]) 0. Thus p where r(pnA) denotes the rank of pnA.Hence we have proved the following theorem.G (n) r(pnA) i--n+l G (i) A (n) P P P In Theorem 4. i, as in certain subsequent theorems, we prefer not to compute the dimensions of the Z(p)-spaces involved; such computations are easy, but the resulting expressions are clumsy.

THE TORSION FREE NUMBER OF HOM(G,A).
The computation of Hom(G,A)0(p) will be done in three steps by the following LEMMA 5.1.Let G and A be divisible.Then Hom(G,A)0(p) dim r.
LEMMA 5.2.Let G be reduced and A be divisible.Then Hom(G,A)o(p) dim n l Z(p).fr(G) r(A) PROOF.If fr(G) is finite, it is zero and the result is clear.Hence we can assume that fr(G) is infinite.By Fuchs [I] we can write G G I G2, where G I is bounded and r(G 2) fr(G 2) fr(G).Hence, since Hom(G,A)0(p) Hom(G2,A)0(p), we can assume that fr(G) r(G).Let B be basic in G such that r(G/B) r(G).
Thus, since Hom(G/B,A)0 (p) is infinite, we have Hom(G,A)0 (p) Hom(G/B,A)0(p).This yields Hom(G,A)0(p) dim Z Z(p) on application of Lemma 5.1.fr (G) r (A) The final rank of a basic subgroup of a p-group G is an invariant of G and is called the critical number of G.We shall denote it cr(G) (cf.Szele [63).If is finite it is zero and B is bounded.Thus G is bounded; hence so is Hom(G,A).
It follows that Hom(G,A)0(p 0. Similarily, if 8 is finite the result follows. Hence we may take both = and 8 to be infinite.By remarks similar to those at the beginning of the previous proof, we can assue that r(B) fr(B) and r(C) fr(C).
Let B E B i and C E C i in the usual notation.i=l Let {bi'lJ3 e Ji} and {cljlj e K i} be bases for B i and C i respectively.We note that, for each n, IJil e and IKil 8.It will be clear from the proof that we i=n i=n {Kij can assume that IKil > 1 for each i.Choose a subsequence }i=l of {K}'i=l such that {IKil} j=l is non-decreaslng and E IK i s denote IK i by 8j.j j= j j Next choose a subsequence {Ji.j=l of {Ji i=l such that{ l}j= 1 is non-decreasing, such that IJi. e and so that the common order of the generators indexed by j=l 3 J i.3 is greater than the order of the generators indexed by Kij let Jij j.
Let J. ' for almost all J. Now if is an epimorphism from G to B and 6,6' are two InequSvalentmaps, then 6 # 6'.Suppose 6 6' py + T where y,T e Hom(G,A) and pnT 0. Then n n+l p 6-6') p According to the construction, there is a generator, say c. i k n of order greater than p as a component of some element in the image of 6-6'.
Thus there is an x e G such that Cik is a component of (6 6')(x).Since Clk has n+l height zero, pn( 6-6')(x) has height exactly n.But pn(6 6')(x) p 7(x) which gives a contradiction.Hence no such y and T exist.We have shown that if and 6' are inequivalent maps from B to C, then 6 and 6' determine different cosets in Hom(G,A)/(pHom(G,A) + tHom(G,A)).Hom(G,A) -Hom(G I,AI) Hom(G l,dA) Hom(dG,dA).
Fom the preceding three lemmas we obtain Hom(G,A)o (p) dim cr(G I) cr(A I) Z(p) + dim H 7.
Since Hom(G,A) is q-divisible for all primes q # p, Hom(G,A)0(p) and each Hom(G,A)q(n) are zero for such q.Thus we can summarize our results in the following theroem.
Finally, it is easy to see that Theorem 5.5 allows us to calculate the Ulm invariants and the torsion free numbers for Hom(X,Y), where X and Y are arbitrary torsion groups Let t X denote the p-primary component of X.Then Hom(X,Y) = H Hom(tqX,tqY), P q (tpX, tpg (n).
where q ranges over the primes.Thus for each n Hom(X,Y) (n)
Hence the invarlants calculated in Theorem 5.5 characterize Hom(X,Y).
pro] implies that y py(pnx) 0. Thus y E S(G,A), as we n+l B wished to show.Hence p S(G,A) implies that S(G,A)/tHom(G,A) is divisible.
that (pnA)[pm] c__Ker .Since A is pure in B and B/A is divisible, there exist b B and a A such that b Pn + an and o(b n) o(b n + A).Note that b + A pn + A. Thus, since 0 Hence aA[pm]o(b n) --o(b n + A) the fact that p 0 implies that pm n n Now define :B / X by (b) (an).

A(
n) o(b n + A).Note that a A[p k] c A[pm].We have pn(bl-b n) a so that n by purity, there is an a ) 0;i.e.we have shown that (pnB)[pm] 5_ Ker .k Now let b pnB, say b P'I'-with o(b) p Choose s > n so that pk] .a + as with o (b s) o (b s + A) Then pnb I pSbs s

4 .
THE ULM INVARIANTS OF HOM<G,A).Let G and A be p-groups with B basic in G.

THEOREM 4 .
1.If G and A are p-groups, then Hom(G,A)p(n)

LEMMA 5 . 3 .
Let G and A be reduced.Then Hom(G,A)0(p) dim H Z Z(p).cr(G) cr(A) PROOF.Let B be basic in G and C basic in A. Let cr(G) a and cr(A) B.

3 3 It
follows that the cardinality of the set of equivalence classes is B.ej J-I J and it is routine to show that B ej B e j=l 3 Thus 8 <_ IHom(G,A)/(pHom(G,A) + tHom(G,A)) Hm(G,A) O(p)-The result follows.THEOREM 5.4.If G and A are p-groups, then Hom(G,A)0(p) dim l Z(p) + dim H l Z(p).cr(G) cr(A) fr(G) r(dA) PROOF.Let G G ldG and A =A ldA.Then Complete invariants for Hom(G,A) are Hom(G,A)o(p) dim H l Z(p) + dim H Z(p) cr(G) cr(A) fr(G) r(dA) index B and K i index C