Kalyan Mukherjea

My friend Kalyan Mukherjea is a talented musician and mathematician. He always had poor eye sight and that was one of the reasons he learnt sarod in his childhood. Now he is totally blind and partly paralyzed but carries on an enviable active lfe partly due to a software developed by T.V. Raman. Here is an article about T.V. Raman and software: For the Blind, Technology Does What a Guide Dog Can’t.
Kalyan has already written one book and now he is completing another. He also teaches some friends music. It seems that my observation about abelian groups has just come in time. He writes: "Is there some kind of telepathic link which works across space? I am in the process of finishing off my notes on Algebraic Topology;
this is just a way of keeping myself occupied. I am writing a preliminary chapter on prerequisites and was having trouble writing up the structure theorem for finitely generated abelian groups. In fact, I had asked K to come over and give me an oral account of the best exposition that he knows of. But he has been rather preoccupied. Your note is a gift from heaven! I hope that I'll be able to write up your note in LaTeX and include it in my book."
Here is another exposition that I wrote for Kalyan:

Kalyan,
I too started writing a book on algebraic topology (about 60 pages) and it came out as a part of it since I did not have too many algebra books at home. I am glad it will be useful to you. It is much simpler than what I wrote there. I was talking of internal direct sums but it is unnecessary depending on what definition of free abelian groups you take. Here it is again.
One observation needed is that if you have a free abelian group and a linear combination of the basis elements where the coefficients are co prime, that combination is also a basis element. Now start with a free abelian group A of rank n, a subgroup B, and look at all basis elements x of A and expressions for elements b of B. Choose the positive non zero coefficients of x. As x and b vary, choose the minimum of such, call this d. So we have a b=dx+ c, where c is written in terms of some other bsis elements. Now we can change x so that b=dx. To see this, if not, take out the common factor of d and the coefficients in the expression of c, so that we can write b=ky. Now k divides d and y is written in terms so that the coefficients are co prime. So y is also a basis element. So k=d. Now in the new basis if we write any other element f of B, the y coefficient of f must be a multiple of d (zero multiple included). Since this part is already in B, the remaining expression must be also in B. But the rest of the basis consists of (n-1) elements. So for the intersection of B with this part (by induction), there is basis such that for the interesection some mltiples form a basis. For the first bit we have dy. A little more thought shows that d must divide the other integers.
This is easier than my blog post.
Swarup