
If the only (associative) ring with additive group G is the zeroring, then G is said to be an (associative) nil group, denoted by (A) nil group.
The two ring properties π considered in this paper are: 1. every two-sided ideal is principal, denoted by PI, 2. every two-sided ideal is finitely generated, denoted by N. In (ii) a complete characterization of the torsion (A )SPI groups will be given. It will be shown that there are no mixed (A)SPI groups. Some results concerning torsion free (A)SPI groups will be obtained. In (iii), the torsion, and mixed SN groups will be completely characterized. Some results concerning torsion free SN groups will be given.
(ii) If X is a nonempty subset of a group or ring, (X) denotes the additive subgroup generated by X, and (X) denotes the ideal generated by X.
If G = Giφ G 2 is a group, π Gί is the natural projection of G on G n for i = 1,2. 
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Proof. Suppose that H is not A nil. Let S be an associative ring on H which is not the zeroring on if, and let T be the zeroring on K. The ring direct sum R = S φ T is an associative ring on G, which is not the zeroring on G. T is an ideal in R, and hence T = (x). Clearly K = (x). Therefore K is not A nil. The above argument, interchanging the roles of H and K, yields that H is cyclic.
COROLLARY. LetG=H@K, Hέ 0,K^0 be an SPI-group. Then H and K are cyclic. Proof. It suffices to negate that H and K are both A nil. Suppose this is so. Let R be a ring on G which is not the zeroring on G.
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Proof. (1) Φ (2): Nontrivial cyclic groups are clearly SPI. Suppose that G = (xθφ(x 2 ) with | x f | = p a prime, ί = 1,2. Let R be a ring on G which is not the zeroring on G, and let / be a proper ideal in R. Then | /1 = 0 or p, and hence / is generated by a single element. It therefore suffices to show that JR is generated by a single element. We may assume that R^(x 1 ) , and that R^(x 2 ). Hence (Xi) = (x.) for i = 1,2. This implies that
Put / = (xi + x 2 ). Let r,s be integers such that rk^ sp = I. Then rx 1 (x 1 + x 2 )= rfciXi = (1-5p)xi = Xi. Hence XiE/, and so (xi + x 2 )-Xj = x 2 E /. Therefore / = JR. (2) 
Hence G admits a ring structure isomorphic to the ring direct sum Z n 0 Z m . (3) φ (1): [3, Satz 1, and Zusatz] . Hence G is cyclic. By Lemma 1 and [1, Theorem 120.3] we may assume G = (xi)0(x 2 ) with \x t \ = n t ,i = 1,2. If (n u n 2 ) = 1, G is cyclic; otherwise, let p be a prime divisor of (n u n 2 ). Then G = (yi)0(y 2 )0/ί, with |y, | = p"\ i = 1,2, and 1 g m, m 2 .
Since (yi) Proof. Let G be a mixed ASP/-group. G is decomposable, [1, Corollary 27.4 ], so by Lemma 1, G = H0K, Hέ 0, X^ 0, with H and K either both cyclic or both A nil.
(1) Suppose that H and K are both A nil. There are no mixed A nil groups [4, hilfssatz 9] so we may assume that H is a torsion group, and that K is torsion free. Let R be an associative ring on G, such that R is not a zeroring. Clearly H is an ideal in R, and so H = (h). Let m = I ft |, so mft = 0. There are no nontrivial, bounded A nil-groups [3, Satz 1 and Zusatz], a contradiction. (2) Suppose that H = (x), and K = (e) with | JC | = n < <», and | e | = oo. Let 1? be the ring on G with multiplication induced by JC 2 = xe = ex = 0, e 2 = ne. Clearly, R is an associative ring on G, and 1? is not a zeroring. Hence i? = {sx + te), s,tEZ.
Every y E R is of the form y = m y 5X + (m y + u y n)te, m y , w y E Z. In particular, (m e + u e n)t = 1. Hence ί = ± 1. Therefore m* + u x n = 0, so that n | m*. However x = m x sx = 0, a contradiction. THEOREM 
Let Gbe a torsion free, ASPI-group. Then G is either indecomposable, or the direct sum of Wo A nil-groups.
Proof. By Lemma 1, it suffices to negate that G = (*i)0(*2), Xi ^ 0, i = 1,2. Suppose this is so. Let R be the ring on G with multiplication induced by r xi + 35 xi = ±1, so that r xi = ±l(mod3). However, r xl + 3ί xl = 0, so that r X] = 0(mod3), a contradiction.
COROLLARY.
Lei G be a torsion free SPI-group. Then G is indecomposable.
Proof. Theorem 3, and the Corollary to Lemma 1. The above Corollary yields that if G is a torsion free SP/-group then either G -Q, the group of rational numbers, or G is reduced, [1, Theorem 21.3] . (1) G is bounded. (2) φ (3): Obvious. (3) φ (1): Let i? be a Pi-ring on G which is not the zeroring on G. Then R = (x). Let n = \x |. Clearly, nG = 0.
COROLLARY. Lei G be a mixed group. If G is an (A)PI-group, then T(G) (the torsion part of G) is bounded, and T/T(G) is an (A)PI-group. Conversely, if T(G) is bounded, and if there exists a unital (A)PI-ring on G/T(G), then G is (A)PI.
Proof. Let G be an (A)PI-group, and let R be an (A)PI-ring on G which is not a zeroring.
Since T(G) is an ideal in R, T(G) = (x), and nT(G) = 0, n = \x |. Clearly R/T(G) is an (A)PI-ring with identity on G/T(G).

Suppose that T{G) is bounded, and that there exists an (A)P/-ring with identity on G/T(G). Then G -T(G)®G/T(G),
[1, Theorem 100.1]. There exists an API-ring i^ with identity on T(G), [1, Lemma 122.3] . Let R 2 be a unital (A)PJ-ring on G/T(G). Let R be the ring direct sum JR = R 1 Q)R 2 , with e, the identity of R x , i = 1,2. Let / be an ideal in R. Then / = (/ Π i?i)0(/ Π i? 2 ). Since /ΠJ?, is an ideal in R h lΠR ι =(x ι ),i = l, 2. Clearly, {x λ + x 2 ) C /. However JC, = e, (xi -f -JC 2 ) G (*i + JC 2 ) for / = 1,2. Hence / = (jti + x 2 ).
Additional information concerning PJ-groups and the classification of 7r-groups for other ring properties π may be found in [2] . LEMMA 
If a group G is finitely generated, then G is SN.
Proof Obvious. LEMMA 
Let a group G = HφK,H^0,K^0, be SN. Then either G is finitely generated, or H and K are both nil.
Proof Suppose that H is not nil. Let S be a nonzeroring on H, and let T be the zeroring on K. The ring direct sum R = 5 © T is a ring on G which is not the zeroring, with ideal T. Let t x , , t n be a finite set of generators for T. Then K = (t x , , t n ). This implies that K is not nil. The same argument, interchanging the roles of H and K, yields that H is finitely generated. Hence G is finitely generated. THEOREM 
Let G be a non torsion free group. G is SN if and only if G is finitely generated.
Proof By Lemma 2, it suffices to show that if G is non torsion free and SN, then G is finitely generated.
(1) Suppose that G is a torsion group. If G is indecomposable then G is cyclic [1, Corollary 27.4 and Theorem 120.3] . We may assume, by Lemma 3, that G = H@K, H^ 0, K^ 0, with H and K both nil. This implies that G is nil [1, Theorem 120.3] . A contradiction.
(2) Suppose that G is a mixed group. Then G is decomposable [1, Corollary 27.3] . By Lemma 3, it suffices to negate that G = HφK,H^0,K^0, with H and K both nil. Suppose that this is so. By [1, Theorem 120.3] we may assume that H is a torsion group and that K is torsion free. However, by [1, Proposition 126.2] H is bounded and hence not nil, a contradiction.
COROLLARY. Let G be an SN group. Then T{G) and G/T(G) are SN.
Proof. If G is torsion free then the statement is trivial. Otherwise, G is finitely generated by Theorem 5, and so T(G) and G/T(G) are SN by Lemma 2.
A torsion-free 5JV-group need not be finitely generated; e.g. Q the group of rational numbers. However, we have the following: THEOREM 6. Let G be an SN~group. Then G is either indecomposable, or finitely generated.
Proof. By Lemma 3, it suffices to negate that G = KφK, #y 0, K^ 0, with H and K both nil. Suppose this is so. Let R be a ring on G which is not a zeroring. Then R = (χ l9 -,x n ), n a positive integer. Put x, = h> + k h hi E H, fe, 6K,lg/gn.
(1) Suppose that R 2 CK. Let /ι6H. Since hER,h = ΣΓ=i r^h, + fc I )+x,r,EZ,lSi^n,xER 2 . However R 2 C K. Hence /ι = Σ"=i Φi, and fί is finitely generated. This contradicts the fact that H is nil.
(2) Suppose that R 2 £K. For g h g 2 €G define g λ x g 2 = flΓ/ί (gig2) Then S = (G, x) is a ring on G which is not a zeroring satisfying S 2 CH. The above argument yields that K is finitely generated, contradicting the fact that K is nil.

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