First, we need a result rectifying the theorem that a subgroup of a f.g. free abelian group is free.

In a way, it is a generalisation of a well-known property of a subspace of a vectorspace : we can change the basis of to, say, so that will be generated by As we cannot, generally speaking, invert our integer coefficients in the case of abelian groups, the ‘s will get extra multipliers. More precisely, the following, known as Smith Normal Form Theorem for abelian groups, holds.

Let be a subgroup of a f.g. free abelian group Then in there exists a free generating set such that is freely generated by with positive integers and dividing for

We know that is free of So is freely generated by , and there is an integer matrix such that where are free generators for We apply to the Smith Normal Form algorithm (in this reference one can ignore the PID etc bits, just assuming that matrix entries are integers), something akin to Gauss elimination, although without division.

More precisely, given a free abelian group and an (ordered) basis for it, we are allowed to (check this!)

- multiply basis elements by -1;
- swap two basis elements;
- add to a basis element another one, multiplied by an integer

without “breaking” the basis property (the resulting set of elements will remain a basis for the group.)

These are the operations we will perform on and on Respectively, they affect, in the way just described, rows and columns of Thus, after applying Smith Normal Form algorithm to we obtain a diagonal matrix and bases and of and respectively, so that (We denoted here by the all-0 matrix of size .) This completes the proof.

Our work will bear fruit here. We will prove the following **Classification theorem for finitely generated abelian groups.** One more piece of notation: with direct summands on the right-hand side.

Let be an abelian group generated by Then Moreover, for and for some are allowed to be 1.

Consider the homomorphism given by By the 1st Isomorphism Theorem one has Changing, if necessary, the generating set in we can assume by Smith Normal Form Theorem that is freely generated by with and freely generating

Now consider the homomorphism given by It can happen that then the corresponding first direct summands must be dropped – they correspond to trivial subgroups. Then and so This completes the proof (we will have ).

As a 1st example, consider an abelian group where the generators satisfy just one irredundant relation The latter just means that the kernel of the homomorphism is Then, according to the procedure outlined in the proof, we are to compute the Smith normal form of the matrix Applying the column operation of adding the times the 1st column to the 2nd column, we obtain the normal form Thus we see that

It is instructive to see what happens to the basis of and to the relation as this basis change is carried out. Note that the column operation we used is given by the matrix (To see this, perform this operation on the matrix and read off from the 2nd and 3rd rows of the result.) So the new basis is equal to and The new relation will be written as so in the new basis

As an exercise, do the corresponding computation with in the place of in the definition of Note that you will need more than one column transformation in this case!

While Smith Normal Form for a given group is unique, often (e.g. for purposes of ring theory) we are more interested in decomposing into a direct sum of as many cyclic subgroups as possible.

We can do a bit more work to arrive at a direct sum decomposition that is, in fact, unique in the latter sense, up to permuting summands. Here we just note that the number that counts -summands, does not depend upon a particular decomposition. To see that, we introduce the notion of *torsion subgroup* of

- Let be an abelian group generated by that satisfy the relations Determine using the Theorem above and the Smith normal form computation.
- Let and i.e. and relatively prime. Use the Theorem above and the Smith normal form computation to show that (As usual, GCD=Greatest Common Divisor.)
- More generally, show, using the method from the previos problem, that where
- Show that and integer matrix satisfying can be transformed into the identity matrix by the elementary row operations. What about elementary column operations?
- Find the matrices such that for an matrix the elementary row operations on can be described as computing the products That is,
- that corresponds to the multiplication of the -th row by -1;
- that corresponds to swapping of the -th and the -th rows;
- that corresponds to adding to the -th row the -th row multiplied by ;

- Show that the operation of swapping the rows in the Smith normal form computation is redundant, i.e. the same can be accomplished using the other two operations only.
- Let be an integer matrix and its Smith normal form. Investigate what happens if we consider the “block” matrix and apply the row operations on the topmost rows and the column operations on the leftmost columns to obtain a matrix

Find out how to use the matrix product of to obtain - Investigate how and from the previous problem change the generating sets of the corresponding free abelian groups;
- check that

(As the definition of does not depend on a particular decomposition as in the theorem, and we see that is invariant, as claimed.)

10 February, 2009 at 16:33 |

maybe you can teach us more about how each elementary operation on Q change the basis for B and H?

10 February, 2009 at 17:52 |

this is in fact something that ought to be done in a linear algebra course… but I’ll do what I can.

8 March, 2009 at 21:45 |

Is the Smith Normal Form unique? From your last paragraph before exercises, it seems that it is not unique.

9 March, 2009 at 9:21 |

The Smith Normal Form (SNF) is unique (can’t you prove it?), but sometimes hides the important information about the cyclic structure of the group, i.e. SNF for is rather than I’ll fix the text to explain more on what I wanted to say here.