Filed under: Algebra, Math | Tags: abelian groups, algebra, commutative, group theory, Math

Wow, it’s been a long time since I’ve written anything on this blog. I’m taking algebraic topology and an algebraic number theory course this semester, and I started reading through Atiyah and MacDonald’s *Commutative Algebra* over the winter. So I thought I’d continue with a little algebra. The algebra we’ve done thus far has been highly noncommutative, for the most part — we investigated groups like free groups, symmetric groups, matrix groups, and dihedral groups in which the order of operations mattered. As you might expect, with abelian groups, the theory becomes much simpler, and the subject called “commutative algebra” is just the study of abelian groups with extra structure — something like a scalar multiplication, as in the case of vector spaces, or some other operation. But first, we need to understand abelian groups.

When talking about abelian groups specifically, we usually write them additively: the group operation applied to and is , and then we can build expressions like . The proof I give below is due to J. S. Milne, who in turn says it’s similar to Kronecker’s original proof. Of course, I’ve added more detail in places where I thought it was necessary, and taken it out where I thought it wasn’t. There are other, more common proofs, typically using matrices, but I find them unwieldy and inelegant.

For arbitrary groups, we had constructed this nice presentation by generators and relations that expressed the group as a quotient of a free group by another free group. A group is **finitely generated** if the first free group can be chosen to have a finite basis, and **finitely presented** if both free groups can. What does the generators-and-relations presentation of an abelian group look like? We have to require that any two of the generators commute (check for yourself that this makes the whole group commutative), so the relations must generate every element of the form , for generators.

(Aside: this is called the **commutator** of and , written , and the subgroup of a group generated by its commutators is called the **commutator subgroup**, . So for any group, we can get an abelian quotient group . This is called the **abelianization** of , written .)

If we’re given that a group is abelian, this implies a certain redundancy in its generators-and-relations presentation. So maybe there’s a better way of forming such a presentation. A good solution is to consider **free abelian groups** — the abelianizations of free groups. Elements of such a group are of the form , where are integers and are generators. As you might expect, for any set , there is a free abelian group with it as basis, and bijections of the bases give isomorphisms of the groups. In fact, the free abelian group with basis is just (recall that this is the “direct sum” over of , meaning the group of -tuples of integers only finitely many of which is nonzero). The only *finitely generated* free abelian groups, then, up to isomorphism, are .

For vector spaces, we had these nice things called “bases.” We required a basis to span the vector space, so every vector was a linear combination , where were real numbers. This generalizes directly to abelian groups — just require to be integers instead. We also required “linear independence” — if , then all . Here we have to be more careful, since in , for example, we have for any . So most abelian groups won’t have linearly independent sets. What we *can* require is that if , then all . We define a **basis** for an abelian group to be a spanning set that satisfies this requirement.

Unlike vector spaces, it isn’t immediately clear that abelian groups have bases. If an abelian group *does* have a basis, then we can write it as a direct sum of the groups generated by each element of the basis, , and conversely. This is a pretty nice way to express a group. To prove that bases exist, we argue by induction on the number of generators of . Since is finitely generated, there must be some minimum number of generators (and this is necessarily the cardinality of a basis, if it exists). If this is , then we’re done: . If it’s , then we will prove that . Then the second summand is generated by generators, so by induction, we can find a basis for it with generators and write it as a direct sum, and the proof will be done.

Consider all the -element generating sets for and choose the one where has the smallest possible order. If is not the direct sum , then we have, for some , , where . Let be the gcd of the , and let . Then has order . If we can show that is the first element in some -element generating set, we have a contradiction.

So suppose that are a generating set for and are relatively prime integers (their gcd is 1). We will prove the existence of a that generate with . First observe that by replacing with if necessary, we can assume all the are nonnegative. Now we use induction on the sum of the . If this is , then for some , so can be the relabelled. If it is more, then we must have at least two nonzero — reorder so that they are and , with . But now is *still* a generating set, are *still* relatively prime, and their sum is strictly less than the original sum of the . So by induction, there is a generating set with . Clearly this is the generating set we want.

Now go back to our original situation and find the generating set . We now have , so is the first member of a -element generating set and has order , which is smaller than the order of , which is a contradiction since is minimal. So, in fact, . By induction, .

But each either has infinite order or order . So is respectively , or . And so, reordering, we have:

**Classification Theorem for Finitely Generated Abelian Groups.***Every finitely generated abelian group is a direct sum of cyclic groups, that is, of the form . The first summands are the***torsion subgroup**, and the last one is the**free subgroup**.

Ordinarily I’d be glad to stop here, but the theorem is usually stated with a little more detail, so I’ll go on. If you’re interested in finite group theory, this says that every *finite* abelian group is a direct sum of *finite* cyclic groups , and so the question becomes how to canonically write the . There are actually different ways to do this: in , for relatively prime, the element has order , and the group has elements, so generates the whole group, and we have .

So a cyclic group of order with prime factorization is isomorphic to . In particular, a finitely generated abelian group can be written as a direct sum of a free subgroup and a finite set of cyclic groups *of prime power order*. There is only one way to do this, as can be seen by counting elements: if there are elements of order , then occurs in of the factors, and so on. We call the the **elementary divisors** of .

Alternatively, starting with an elementary divisor decomposition, we could let where there is one copy of each *distinct* prime from among the invariant factors, with its highest exponent. Then we could let be the product of all the distinct prime powers remaining, again with their highest exponents, and so on. If the elementary divisors are then we have . We have , because at each stage we’re collecting cyclic groups of relatively prime order. The also have the property that divides for each . We call them the **invariant factors** — since the are unique, so are the . Both of these decompositions are occasionally useful, especially for finite abelian groups.

We haven’t forgotten the on the end! is the **rank** of the group, something like its dimension, and it is also invariant of the group. If you know some more advanced commutative algebra, it’s easy to see this by tensoring with (or any field of characteristic not occurring in the ), but then this page probably isn’t for you. It’s a nice exercise to come up with a simpler, group-theoretic proof that is unique.

I’ll end with a word on non-finitely-generated abelian groups. There are very many of these, and at a certain stage it becomes hard to tell the difference. For example, and are both non-finitely-generated, and includes as a subgroup, but what exactly is the difference? This is the stage when we have to appeal to analysis, with notions like limits: we can get, say, by defining it as a solution of the polynomial with rational coefficients, but we can’t do this for or . If we want to keep talking on algebraic lines, we have to consider different ideas of “finite generation”: for example, finite-dimensional vector spaces are infinitely-generated abelian groups, so we introduce the vector space “picture” to find a way to make them finite. Now that we have a good picture of abelian groups, I’ll start generalizing substantially and show a couple more pictures of structures, leading the way to more advanced algebra.

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[…] I sort of breeze through a couple of really awesome and really important concepts. Last time, we classified abelian groups — now we’ll see what happens if we require additional structure on the groups. In […]

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