13 More on free abelian groups Recall. G is a free abelian group if G ∼= Z i I for some set I. ∈ 13.1 Definition. Let G be an abelian group. A set B G is a basis of G if ⊆ B generates G • if for some x1,...xk B and n1,...,nk Z we have • ∈ ∈ n x + + n x =0 1 1 ··· k k then n = = n =0. 1 ··· k 13.2 Theorem. An abelian group G has a basis iff G is a free abelian group. Proof. If B is a basis of G then the map f : Z G, f((nx)x B)= nxx → ∈ x B x B ∈ ∈ is an isomorphism (check!). Conversely, if we have an isomorphism f : Z G → i I ∈ then for j I take δj i I Z where δj =(ni)i I such that ∈ ∈ ∈ ∈ 1 if i = j ni = 0 otherwise Then B = f(δj) i I is a basis of G (check!). { } ∈ 42 13.3 Proposition. If G is a free abelian group then any two basis of G have the same cardinality. Notation. If X is a set then X = cardinality of X. | | 13.4 Lemma. If Gi i I is a family of abelian groups and Hi is a subgroup of ∈ G for i I then { } i ∈ G1 / Hi ∼= (Gi/Hi) i I i I i I ∈ ∈ ∈ Proof. Exercise. Proof of Proposition 13.3. Let B, B be two bases of G.Wewanttoshow: B = B . | | | | We have isomorphisms Z ∼= G ∼= Z x B y B ∈ ∈ Case 1. B,B -finitesets, B = m , B = n. | | | | Take 2G := 2a a G .ThisisasubgroupG.SinceG ∼= x B Z,using Lemma 13.4 {we obtain| ∈ } ∈ G/2G ∼= Z/2Z x B ∈ Similarly, since G = x B Z we have ∼ ∈ G/2G ∼= Z/2Z y B ∈ This gives: 2m = Z/2Z = G/2G = Z/2Z =2n | | | | | | x B y B ∈ ∈ so m = n. 43 Case 2. B -finiteset,B -infiniteset. As before this would give: Z/2Z ∼= G/2G ∼= Z/2Z x B x B ∈ ∈ This is however impossible since /2 is a finite group and /2 x B Z Z y B Z Z is an infinite group. ∈ ∈ Case 3. B,B -infinitesets. Check: if B is an infinite set then B = Z | | | | x B ∈ It follows that we have B = Z = G = Z = B | | | | | | | | | | x B y B ∈ ∈ 13.5 Definition. If G is a free abelian group then the rank of G is the cardinality of a basis of G. Note. By (13.3)rankofG does not depend on the choice of basis of G. 13.6 Theorem. Let G be a free abelian group of a finite rank n and let H be asubgroupofG.ThenH is a free abelian group and rank H rank G ≤ Note. This theorem is true also if G is a free abelian group of an infinite rank. 13.7 Lemma. If f : G H is an epimorphism of abelian groups and H is free abelian group then → G = H Ker(f) ∼ ⊕ 44 Proof. Recall (from homework): if we have homomorphisms f : G H, g: H G → → such that fg =idH then G ∼= H Ker(f).Iffollowsthatweonlyneedto construct the homomorphism g. ⊕ Take a basis B of H.Sincef is onto, for every x B there is ax G such that f(a )=x.Letg : H G be the unique homomorphism∈ satisfying∈ x → g(x)=ax for all x B.Thenfg(x)=x for all x B and so fg =id . ∈ ∈ H Proof of Theorem 13.6. We can assume that G = Zn. We want to show: if H Zn then H is a free abelian group and rank H n. ⊆ ≤ Induction with respect to n: If n =1then H = kZ for some k 0 so H = 0 or H = Z. ≥ { } ∼ Next, assume that for some n every subgroup of Zn is a free abelian group of rank n,andletH Zn+1.Takethehomomorphism ≤ ⊆ n+1 f : Z Z,f(m1,...,mn+1)=mn+1 → We have: n Ker(f)= (m1,...,mn, 0) mi Z = Z { | ∈ } ∼ We have an epimorphism: f : H Im(f) |H → Since Im(f H ) Z,thusIm(f H ) is a free abelian group and so by Lemma 13.7 we have | ⊆ | G = Im(f ) Ker(f ) ∼ |H ⊕ |H We also have: Ker(f )=Ker(f) H |H ∩ 45 It follows that that Ker(f ) is a subgroup of Ker(f),andsinceKer(f) is a free |H abelian group of rank n by the inductive assumption we get that Ker(f H ) is a free abelian group of rank n.Therefore | ≤ H = Im(f ) Ker(f ) ∼ |H ⊕ |H free abelian free abelian rank 1 rank n ≤ ≤ and so H is a free abelian group of rank n +1. ≤ 46 14 Finitely generated abelian groups Goal. Describe all isomorphism types of finitely generated abelian groups. 14.1 Proposition. If G is an abelian group generated by n elements then G ∼= F/H where F is a free abelian group of rank n and H is some subgroup of F . Proof. We have G = a1,...an for some a1,...,an G.Let x1,...,xn be abasisofF .Wehaveahomomorphism ∈ { } f : F G, f(x )=a → i i Take H =Ker(f).Sincef is an epimorphism by the First Isomorphism Theorem (6.1)wehave G ∼= F/H Recall. If rank F = n and H F then H is a free abelian group of rank n. ⊆ ≤ 14.2 Theorem. Let F be a free abelian group of rank n and let H be a subgroup of F There exists a basis x1,...,xn of F and integers d1,...,dr > 0 such that { } d d for i =1,...,r • i| i+1 d x ,...,d x is a basis of H. • { 1 1 r r} 14.3 Theorem. If G is a finitely generated abelian group then k G = (Z/d1Z) ... (Z/drZ) Z ∼ ⊕ ⊕ ⊕ for some k 0 and d ,...,d > 0 such that d d for i =1,...,r. ≥ 1 r i| i+1 47 Proof. By (14.1)wehave G ∼= F/H for some free abelian group F of finite rank and H F .Letrank F = n.By ⊆ Theorem 14.2 there is a basis x1,...,xn of F such that d1x1,...,drxr is abasisofH for some d ,...,d{ > 0, d d} . { } 1 r i| i+1 We have an isomorphism f : F Z Z → ⊕···⊕ n times where f(x1)=(1, 0,...,0),f(x2)=(0 , 1,...,0),...,f(xn)=(0,...,0, 1) Notice that f(H)=d1Z ... drZ 0 ... 0 ⊕ ⊕ ⊕{ }⊕ ⊕{ } .Thisgives G = (Z Z)/(d1Z ... drZ 0 ... 0 ) ∼ ⊕···⊕ ⊕ ⊕ ⊕{ }⊕ ⊕{ } Using (13.4)weobtain k G = (Z/d1Z) ... (Z/drZ) Z ∼ ⊕ ⊕ ⊕ where k = n r. − Proof of Theorem 14.2. Let y1,...,yn be any basis of F and let h1,...,hm H be any set generating H.Wehave{ } { }⊆ hi = ai1y1 + ai2y2 + ainyn for some aij Z.Considerthematrix ∈ a11 ... a1n . A = . am1 ... amn Note: columns of A correspond to basis elements of F and rows of A correspond to generators of H. Consider the following operations on matrices: 48 1) interchange of two rows 2) multiplication of a row by ( 1) − 3) addition of a multiple of one row to another row. These operations are called elementary row operations for matrices of integers. Elementary column operations are defined analogously. Notice that: application of an elementary row operation to the matrix A corresponds to • replacing of the set of generators of H by another set of generators of H; application of an elementary column operation to the matrix A corresponds • to passing to a new basis of F and rewriting the generators of H in terms of this new basis. Key step. Starting with any matrix of integers A and applying a sequence of elementary row and column operations we can obtain a matrix of the form D 0 B = 00 where 0’s denote zero matrices (of appropriate dimensions) and D is a square diagonal matrix d1 0 ... 0 0 d2 ... 0 D = . . .. 00... dr for some d ,...,d > 0 such thatd d for all i. 1 r i| i+1 Note. The matrix B is called the Smith normal form of the matrix A. Let x ,...,x be the basis of F corresponding to columns of the matrix B.In { 1 n} terms of this basis a set of generators of H is given by d1x1,...,drxr, 0,...,0 . It follows that d x ,...d x is a basis of H. { } { 1 1 r r} 49 Key step. How to compute the Smith normal form of a matrix A. Step 1. Produce a matrix of the form a11 0 ... 0 0 A1 = . A 1 0 This can be done as follows. (1a) By interchanging rows and columns if necessary we can make sure that a =0. Also, by multiplying the first row by ( 1) we can get a > 0. 11 − 11 (1b) By adding multiples of the first row to the other rows (and multiples of the first column to the other columns) we can make all other entries in the first row and the first column positive and smaller than a11. (1c) If all these other entries of the first row and column are 0 we are done. If some entry is non-zero then by replacing rows (or columns) we can move that entry to the (1, 1)-position. Then we go back to (1b). (1d) After a finite number of iterations we get a matrix of the form of A1. Step 2. Given a matrix A1 as above make sure that the entry a11 divides all entries of A1 .