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18 Divisible Groups 18 Divisible groups Every group splits as a direct sum G = D © R where D is divisible and R is reduced. However, the divisible subgroup is natural whereas the reduced subgroup is not. [Recall that all groups are abelian in this chapter.] Definition 18.1. A group G is called divisible if for every x 2 G and every positive integer n there is a y 2 G so that ny = x, i.e., every element of G is divisible by every positive integer. For example the additive group of rational numbers Q and real numbers R and any vector space over Q or R are divisible but Z is not divisible. [In fact there are no nontrivial finitely generated divisible groups.] Another example of a divisible group is Z=p1. This is usually defined as a direct limit: Z=p1 = lim Z=pk ! where Z=pk maps to Z=pk+1 by multiplication by p. This is the quotient of the direct sum ©Z=pk by the relation that n 2 Z=pk is identified with pjn 2 Z=pj+k. Another way to say this is that Z=p1 is generated by elements x0; x1; x2; x3; ¢ ¢ ¢ which are related by xk = pxk+1 and x0 = 0. Since xk has order pk it is divisible by any number relatively prime to p and is divisible j by any power of p since xk = p xj+k. Yet another description of Z=p1 is the multiplicative group of all p-power roots of unity. These are complex numbers of the form e2¼in=pk where n is an integer. An isomorphism Á with the additive version above is given by 2¼i=pk Á(xk) = e : Divisible group satisfy properties similar to those of torsion groups: Proposition 18.2. 1. Every product of divisible groups is divisible. 2. Every direct sum of divisible groups is divisible. 3. Every quotient of a divisible group is divisible. Q Proof. Let x = (x®) 2 G® where each G® is divisible. Then each coor- dinate x® 2 G® is divisible by any given n so there exists y® 2 G® so that ny® = x®. Then n(y®) = (ny®) = x. This proves (1). To prove (2) note that if almost all coordinates x® are zero we can choose the corresponding y® to be zero making y 2 ©G®. To prove that the image f(G) of a divisible group G is divisible note that for any y = f(x) and n > 0 there is a z 2 G so that nz = x so y = f(nz) = nf(z) is divisible in f(G). 1 Definition 18.3. For any group G let dG be the subgroup of G generated by all divisible subgroups of G. If dG = 0 then G is called reduced. As a consequence of the second and third statements in Proposition 18.2 we get the following. Corollary 18.4. dG is a divisible subgroup of G and any homomorphism G ! H sends dG into dH, i.e., d is a functor. The main property of divisible groups is that they are injective. To prove this we need the following lemma. Lemma 18.5 (Pasting lemma). If A; B are subgroups of C and f : A ! G, g : B ! G are homomorphisms which agree on A \ B (i.e., fjA \ B = gjA \ B), then there is a unique extension f + g of f and g to A + B given by (f + g)(a + b) = f(a) + g(b): Proof. There is a short exact sequence j 0 ! A \ B ¡!∆ A © B ¡! A + B ! 0 where ∆; j are given by ∆(x) = (x; ¡x) and j(a; b) = a + b. The condition on f; g is equivalent to saying that ∆(A \ B) lies in the kernel of f © g : A © B ! G: Consequently, we get an induced map on the quotient: f +g : A+B ! G. Theorem 18.6. A group G is divisible if and only if it satisfies the following “injectivity” condition: Any homomorphism f : A ! G from any group A into G extends to any group B which contains A, i.e., there exists a homo- morphism f : B ! G so that fjA = f. Proof. We use Zorn’s Lemma. Consider the partially ordered set P of all pairs (C; g) where C is a subgroup of B containing A and g : C ! G is an extension of f. Let (C; g) · (D; h) if C · D and g = hjC. The set P is nonempty since it contains (A; f). Also any tower (C®; g®) in P has an upper bound ([C®; [g®) 2 P . Consequently, by Zorn’s Lemma, P has a maximal element, say (C; g). We claim that C = B and g is the desired extension of f to B. To see this suppose C < B. Then there exists an x 2 B so that x2 = C. There are two cases. Either x + C has finite order in B=C or it has infinite order. In the second case, hC; xi = C © hxi so g : C ! G can be extended to g © 0 : C © hxi ! G contradicting the maximality of (C; g). In the first case, let n be the order of x+C in B=C, i.e., n > 0 is smallest positive integer so that nx 2 C. Since G is divisible there is a z 2 G so that nz = g(nx). By the Pasting lemma, g : C ! G can be extended to the 2 homomorphism g + h : C + hxi ! G where h : hxi ! G is given by h(x) = z. This contradicts the maximality of (C; g). Therefore, divisible groups are injective. Conversely, if G is injective then it is obviously divisible: given any x 2 G and n > 0 let f : nZ ! G be given by f(n) = x. If G is injective this extends to a homomorphism f : Z ! G. But then nf(1) = f(n) = x. Corollary 18.7. Any divisible subgroup of G splits, i.e., if D · G is divisible then D has a complement H so that G = D © H. Proof. Since D is divisible, it is injective. Thus the identity mapping D ! D extends to a homomorphism r : G ! D. Then H = ker r is a complement for D in G. Corollary 18.8. Every group G splits as a direct sum G = dG © R where dG is the divisible subgroup of G and R is a (noncanonical) reduced subgroup of G. Proof. By Corollary 18.7, dG has a complement H. We just need to show that H is reduced. But dH · H · G is a divisible subgroup of G and is therefore contained in dG. So dH · H \ dG = 0 and H is reduced. Lemma 18.9. The torsion subgroup tD of any divisible group D is divisible, D=tD is a vector space over Q and D »= tD © D=tD. Proof. Any x 2 tD is divisible by n > 0 in D. So there is a y 2 D s.t. ny = x. But then mny = mx = 0 for some m so y is torsion and lies in tD. Since tD is torsion it has a complement isomorphic to D=tD. Since G = D=tD is a quotient of a divisible group it is also divisible. Since G is also torsion-free it is uniquely divisible (if x 2 G and y; y0 are two elements of G so that ny = x = ny0 then y ¡ y0 is n-torsion so must be zero.) a Consequently, there is an action of Q on G given by taking b x to be the unique element y 2 G so that by = ax. Thus G is a vector space over Q. Theorem 18.10 (Classification of divisible groups). Divisible groups are direct sums of the groups Q and Z=p1. Furthermore, two divisible groups G; H are isomorphic iff 1. dimQ(G=tG) = dimQ(H=tH) and 2. dimGF (p) G[p] = dimGF (p) H[p] for all primes p. Proof. Given any divisible group D, D = tD©D=tD. By the lemma, D=tD is a direct sum of many copies of the divisible group Q. The torsion subgroup tD is a direct sum of primary subgroups: tD = ©tDp. Since each tDp is 3 a quotient of D, it is also divisible. Thus it remains to show that every p-primary divisible group P is a direct sum of copies of Z=p1. Let fb®g®2I be a basis for the Z=p-vector space P [p]. Construct a direct 1 1 sum V = ©I Z=p of copies of Z=p one for every element of the index set I. Then V [p] is a direct sum of copies of Z=p, one for each element of I so it has a Z=p-basis fc®g®2I indexed by I. Thus there is a (unique) isomorphism of Z=p-vector spaces Á : P [p] ! V [p] sending each b® to c®. Since V is injective, this extends to a homomorphism Á : P ! V . Since Á is a monomorphism on P [p] it is a monomorphism. [The kernel K of Á must be zero since K[p] = K \ P [p] = 0.] The image is thus an injective subgroup of V which must also split: V = Á(P ) © H. But the image of Á contains V [p] so H[p] = H \ V [p] = 0 implies H = 0 and Á is an isomorphism P »= V . The theorem follows. 4.
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