The structure theorem for finitely generated abelian groups asserts that any finitely generated abelian group is a direct sum of cyclic groups, some finite and some infinite. The orders of the finite cyclic groups can be arranged so that each order divides the next, but this extra requirement can be easily deduced once we know that the group is a direct sum of cyclic groups. There is a similar statement for for finitely generated modules over principal domains. The usual proof involves reducing matrices to some canonical form using elementary transformations and takes a few pages. There is a good exposition of the standard method in Chapter 3, of N.Jocobson's "Basic Algebra", Volume 1 (W.H. Freeman and Company, 1985). I taught this method for several years. Recently, writing some expository stuff, I noticed a simple argument which may be useful for teachers. Here it is:

Let A be a finitely generated (f.g.) free abelian group with basis {x(1),..., x(n)}. The fundamental theorem is equivalent to the following statement:

If B is a subgroup of A, we can find a basis {y(1),..., y(n)} of A, integers d(1),...d(n), which are greater than or equal to zero and some of which may be zero such that the non-zero elements of d(1)y(1),...,d(n)y(n) form a basis for B.

We proceed to the proof. We call An element x of A a 'basis element' if it is part of a basis of A. Given any b in B, we can write it as dx+ c, where c is a combination of the complementary basis elements. Note that if d=0 for all b in B, then B is contained in the subgroup generated by the complementary basis elements and we are through by induction since this number is (n-1). The number d depends not only on b,d but also on the basis and we may get different values of d for the same b and x. Choose all positive d (for every d, -d appears for -b and thus there are both positive and negative d) as b varies over B and x varies over the basis elements of A. This set will have a minimal element. Suppose this is achieved for some x and the minimum achieved is d(x). This means that firstly we can write for some b in B, b=d(x)x+c, with respect to some basis containing x. Further if we take any basis, and if y is one of the basis elements, then the y-coefficient of b for any b in B is either zero or a multiple of d(x).

We next claim that we can choose x so that b=d(x)x. First, we have with the above choice b=d(x)x+d(2)x(2)+...+d(n)x(n) where x(2),...,x(n)are the rest of the elements of the basis in the above choice. Let d be the greatest common divisor of d(x),d(2),...,d(n). We can write b=d[c(x)x+c(2)x(2)+...+c(n)x(n)]and the greatest common divisor of c(x), c(2),...,c(n) is one. In this case y=c(x)x+c(2)x(2)+...+c(n)x(n) is also a basis element of A and we have b=dy. Since d divides d(x), this is a contradiction unless d=d(x). Renaming y as x, we may assume that for our minimum d(x), b=d(x)x. Now fix a basis containing x, say {x,x(2),...,x(n)}. Now any c in B is of the form c=kd(x)x+c' for c' a linear combination of x(2),...,x(n). If we call C the free abelian group generated by x(2),...,x(n), then A is a direct sum the subgroup Zx generated x and C. Here k is an integer which may be zero. In any case kd(x)x is in B, and since c is also in B, we see that c' is in B. Thus we have expressed any element c of B as a sum of two elements of B, one of which is in Zx and the other in C. This gives a direct sum decomposition of B compatible with the decomposition in to Zx and C of A. We have the intersection of B and Zx is the subgroup generated by d(x)x. By induction, there is a basis {y(2),...,y(n)} of C and integers d(2),...,d(n) such that the non-zero elements of d(2)y(2),...d(n)y(n) form a basis for the intersection of B and C. Thus the basis {x,y(2),...,y(n)} and the numbers d(x),d(2),...,d(n) satisfy the conclusions of the theorem.

Remark 1. We used the fact that y=c(x)x+c(2)x(2)+...+c(n)x(n) when the greatest common divisor of c(x), c(2),...,c(n) is one. This is easily proved when the basis has two elements and this used again in induction.

Remark 2. To deduce the stronger version of the fundamental theorem, use the fact that if d=mn with m,n coprime, then the cyclic group of order d is isomorphic to the direct sum of cyclic groups of orders m and n.

Remark 3. For principal ideal domains take d(x)to be one of the d with minimal number factors in the first paragraph of the argument. But we have not checked this.

P.S. I have not yet learnt how to upload files. A more formal version may appear later.

P.P.S. (30th July) So far there are two responses. The first with in an hour from Mladen Bestvina said that it seemed good to him. The second from John Groves, a day later, said that Remark 1 can be used to see that d(x) divides d(2) and thus one can see more quickly (without Remark 2) that one can find the orders each dividing the next in the stronger statement.

P.P.P.S. (6th August)

Similar argument apears in Bourbaki:: Algebre

Chapitre 7, Modules sur les anneaux principaux, Hermann, Paris 1964.

See pages 82-84. So, we now have a complete argument!

## Tuesday, July 28, 2009

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