Class Note for MATH 511A at UA
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Date Created: 02/06/15
THE FUNDAMENTAL THEOREM OF FINITELY GENERATED ABELIAN GROUPS In this handout we give a proof of the fundamental theorem of nitely generated abelian groups Theorem 1 Every nitely generated abelian group is the product af nitely many cyclic groups As a starting point we will need the following lemma Lemma 2 fA is a subgroup of Z then A is nitely generated Proof If we regard Z as a subset of Q then we can regard elements of A as elements of Q and in this context it makes sense to speak of a subset of A as being linearly independent or not Let v1 i i vk be any maximal linearly independent subset of A note that k S n since Q is ndimensionall Let B be the subgroup of A generated by v1 i vkl What we want to do is prove that B has nite index in A to see that this suf ces observe that if A z B is nite and wl i i wg are coset representatives for B in A then A is generated by v1 i vk w1l wgl Proving that A z B is nite is a little bit trickier than one might hopel Here7s one argument not necessarily the shortest The key is to nd an integer N such that N A C B Extend the linearly independent set v1 i i vk to a basis v1 i 1 of Q clearing the denominators of vk1 i i 1 if necessary we may assume that vk1l vn also lie in an By linear algebra any element of Q will be a rational linear combination of v1 i 1 in a unique manner If ei E Z is a standard basis vector we write 6139 alivi 39 39 39 anivn with each aji E Let N be a common denominator for all the aji l S ij S Then Nei is an integer linear combination of v1 i i 1 for all i it follows that Nu is an integer linear combination of v1 i i 1 for any w E an Now suppose that w 6 Al By the previous paragraph there is a unique expres sion 01 Nwc1v1c vn and in this expression we have ci 6 Z for all if Since v1 i i vk are linearly indepen dent but w v1 i i vk are linearly dependent we see that w is a linear combination of v1luvk alone The same must be true of Nu and since the linear com bination Oil is unique we conclude that ck1ulcn are all equal to zero and Nu c1111 ckvk with ci 6 Z for all i This proves that Nu E B for all wEAsothatNACBl We are almost done Since B C A we have N B C N A and we have the inclusions N B C N A C B and in particular N A z N B S B N Since B is generated by k elements we have B N B nite in fact it is at most Nk and therefore N A z N B is nite Note that the multiplicationbyN map A gt gt N A is an 2 THE FUNDAMENTAL THEOREM OF FINITELY GENERATED ABELIAN GROUPS isomorphism because A contains no elements of nite order and this map carries B to N B Hence AB E N AN B and in particular A z B N A N We have already seen that the latter is nite and therefore A z B is D Now we return to the main theoremi Proof of Theorem 1 Suppose that A is a nitely generated abelian group and let v1iuvn be generators of A There is a surjection go Z A A sending the ith standard basis vector if Z to vi By Lemma 2 the kernel B kergo is nitely generated Let 1211 1 1127 be generators of B There is a map f Zm A Z sending the ith standard basis vector in Zm to 12 and imf B It follows that A coker We are now reduced to the statement if G is a free abelian group on a basis of size n and f Zm A G is a homomorphism then cokerf is a product of cyclic groupsi Recall that once we have chosen a basis of Zm and a basis of G then the map f is given by an nbym matrix as follows Let 611 1 i em be our basis of Zm and 111 In our basis of G if f j Eilmijzi then M That is the jth column of M gives the image of 61 Suppose we replace M with another matrix M obtained by performing any one of the following three column operations on M 1 interchange any two columns 2 replace a column with its negative 3 add an integer multiple of one column to a different column If we form the map f Z7 A G corresponding to this new matrix M observe that imf imfi For instance if we add 5 times the jth column to the kth column then fek Cf j fek and the subspace of G generated by fel i i fem is the same as the subspace of G generated by feli i Cf j feki i femi Since imf imf we also have cokerf cokerf and therefore we can perform column operations on M without changing the isomorphism type of the cokerneli Next we want to prove the same statement for row operations which is a bit more complicated One checks that any of the following operations yield another basis of G 1 interchange z and zj for some ij 2 replace z with 71139 for some i 3 replace z with z 7 czj for somej i and c E Z Suppose we interchange z and I in terms of this new basis the matrix of the map f becomes the matrix M obtained by swapping rows ij of Mr The other operations are similar note that if we replace 1 with z 7 or the effect is to add 5 times row j from row Since the map f is unchanged the cokernel is unchanged Thus we may perform the following row operations on M without changing the cokernel of f 1 interchange any two rows 2 replace a row with its negative 3 add an integer multiple of one row to a different rowi THE FUNDAMENTAL THEOREM OF FINITELY GENERATED ABELIAN GROUPS 3 We have shown that if we begin with a map f Z A G given by the matrix M then we can perform any combination of row and column operations on M without changing the cokerneli Suppose M has a nonzero entry in either the rst row or the rst column then it is possible through swaps and sign changes to put a positive integer in the 11 entryi Let Mi be a matrix that can be obtained from M by these operations that has the smallest possible positive integer in the l lentryi Let all denote this entry I claim that an divides all other entries in the rst row and column Suppose eg that a is an entry in the rst row the division algorithm gives a qau T with 0 S T lt uni Subtracting 4 times the rst column from the column containing a puts an T in the rst row and swapping this column with the rst column puts T in the l lentryi By choice of Mi we have T 0 Now subtract the appropriate multiple of the rst row from other rows and the appropriate multiple of the rst column from other columns so that all other entries in the rst row and column become zeroi At the end of this procedure we obtain a matrix M1 in which the 11entry is a nonnegative integer mm and all other entries in the rst row and column are zero Note that an 0 if and only if the entire rst row and column were zero to begin with Repeating this procedure with the second rowcolumn and so forth we can use these row and column operation to transform M into a matrix M in which the only nonzero entries are along the diagonal Denote these diagonal elements by ail for l S i S miniji Let aii 0 for minij lt i S maxiji We are now reduced to considering maps f Z7 A G where G is a free abelian group on the basis 11 i i In and in which f given by the matrix M send ei gt gt ailzl for l S i S mi The cokernel of G is evidently isomorphic to ZauZ gtlt gtlt ZannZ and we are done D Corollary 3 fA is a subgToup of Z then A is a fTee abelian gToup 0f Tank m foT some m S n Contrast this with the behavior of free nonabelian groups in which a subgroup of a free group may be free of larger ranki PToof By Lemma 2 the group A is nitely generated The group A contains no elements of nite order since the same is true by Z and so by Theorem 1 the group A is free abelian of rank m for some mi Since the generators of A C Z are linearly independent in Z hence in Q we get m S n D
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