Gracious Living

The Classification Theorem for Finitely Generated Abelian Groups
January 29, 2011, 16:23
Filed under: Algebra, Math | Tags: , , , ,

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 $a$ and $b$ is $a+b$, and then we can build expressions like $3a+2b$.  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 $aba^{-1}b^{-1}$, for $a,b$ generators.

(Aside: this is called the commutator of $a$ and $b$, written $[a,b]$, and the subgroup of a group $G$ generated by its commutators is called the commutator subgroup, $[G,G]$.  So for any group, we can get an abelian quotient group $G/[G,G]$.  This is called the abelianization of $G$, written $G^{ab}$.)

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 $a_ix_i+\dotsb+a_nx_n$, where $a_1,\dotsc,a_n$ are integers and $x_1,\dotsc,x_n$ are generators.  As you might expect, for any set $B$, 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 $B$ is just $\bigoplus_B \mathbb{Z}$ (recall that this is the “direct sum” over $B$ of $\mathbb{Z}$, meaning the group of $B$-tuples of integers only finitely many of which is nonzero).  The only finitely generated free abelian groups, then, up to isomorphism, are $\bigoplus_n\mathbb{Z}=\mathbb{Z}^n$.

For vector spaces, we had these nice things called “bases.”  We required a basis $x_1,\dotsc,x_n$ to span the vector space, so every vector was a linear combination $a_1x_1+\dotsb+a_nx_n$, where $a_1,\dotsc,a_n$ were real numbers.  This generalizes directly to abelian groups — just require $a_1,\dotsc,a_n$ to be integers instead.  We also required “linear independence” — if $a_1x_1+\dotsb+a_nx_n=0$, then all $a_i=0$.  Here we have to be more careful, since in $\mathbb{Z}/3\mathbb{Z}$, for example, we have $3x=0$ for any $x$.  So most abelian groups won’t have linearly independent sets.  What we can require is that if $a_1x_1+\dotsb+a_nx_n=0$, then all $a_ix_i=0$.  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, $G\cong\langle x_1\rangle\oplus\dotsb\oplus\langle x_n\rangle$, 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 $G$.  Since $G$ 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 $1$, then we’re done: $G\cong\langle x_1\rangle$.  If it’s $k$, then we will prove that $G\cong\langle x_1\rangle\oplus\langle x_2,\dotsc,x_k\rangle$.  Then the second summand is generated by $k-1$ generators, so by induction, we can find a basis for it with $k-1$ generators and write it as a direct sum, and the proof will be done.

Consider all the $k$-element generating sets for $G$ and choose the one where $x_1$ has the smallest possible order.  If $G$ is not the direct sum $\langle x_1\rangle\oplus\langle x_2,\dotsc,x_k\rangle$, then we have, for some $m_i$, $m_1x_1+\dotsb+m_kx_k=0$, where $m_1x_1\ne 0$.  Let $d$ be the gcd of the $m_i$, and let $c_i=m_i/d$.  Then $y_1=c_1x_1+\dotsb+c_kx_k$ has order $d\le m_i<\mathrm{order}(x_1)$.  If we can show that $y_1$ is the first element in some $k$-element generating set, we have a contradiction.

So suppose that $x_1,\dotsc,x_k$ are a generating set for $G$ and $c_1,\dotsc,c_k$ are relatively prime integers (their gcd is 1).  We will prove the existence of a $y_1,\dotsc,y_k$ that generate $G$ with $y_1=c_1x_1+\dotsb+c_kx_k$.  First observe that by replacing $x_i$ with $-x_i$ if necessary, we can assume all the $c_i$ are nonnegative.  Now we use induction on the sum of the $c_i$.  If this is $1$, then $y_1=x_i$ for some $i$, so $y_i$ can be the $x_i$ relabelled.  If it is more, then we must have at least two nonzero $c_i$ — reorder so that they are $c_1$ and $c_2$, with $c_1\ge c_2$.  But now $\{x_1,x_1+x_2,x_3,\dotsc,x_k\}$ is still a generating set, $\{c_1-c_2,c_2,c_3\dotsc,c_k\}$ are still relatively prime, and their sum is strictly less than the original sum of the $c_i$.  So by induction, there is a generating set $y_1,\dotsc,y_k$ with $y_1=(c_1-c_2)x_1+c_2(x_1+x_2)+\dotsb+c_kx_k=c_1x_1+\dotsb+c_kx_k$.  Clearly this is the generating set we want.

Now go back to our original situation and find the generating set $\{y_i\}$.  We now have $dy_1=m_1x_1+\dotsb+m_kx_k=0$, so $y_1$ is the first member of a $k$-element generating set and has order $d$, which is smaller than the order of $x_1$, which is a contradiction since $x_1$ is minimal.  So, in fact, $G\cong\langle x_1\rangle\oplus\langle x_2,\dotsc,x_k\rangle$.  By induction, $G\cong\langle x_1\rangle\oplus\langle x_2\rangle\oplus\dotsb\oplus \langle x_k\rangle$.

But each $x_i$ either has infinite order or order $n_i>0$.  So $\langle x_i\rangle$ is respectively $\mathbb{Z}$, or $\mathbb{Z}/n_i\mathbb{Z}$.  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 $\mathbb{Z}/n_1\mathbb{Z}\oplus\dotsb\oplus\mathbb{Z}/n_k\mathbb{Z}\oplus\mathbb{Z}^r$.  The first $k$ 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 $\mathbb{Z}/n_i\mathbb{Z}$, and so the question becomes how to canonically write the $\mathbb{Z}/n_i\mathbb{Z}$.  There are actually different ways to do this: in $\mathbb{Z}/n\mathbb{Z}\oplus\mathbb{Z}/m\mathbb{Z}$, for $n,m$ relatively prime, the element $(1,1)$ has order $mn$, and the group has $mn$ elements, so $(1,1)$ generates the whole group, and we have $\mathbb{Z}/n\mathbb{Z}\oplus \mathbb{Z}/m\mathbb{Z}\cong\mathbb{z}/mn\mathbb{Z}$.

So a cyclic group of order $n$ with prime factorization $p_1^{e_1}\dotsm p_k^{e_k}$ is isomorphic to $\mathbb{Z}/p_1^{e_1}\mathbb{Z}\oplus\dotsb\oplus\mathbb{Z}/p_k^{e_k}\mathbb{Z}$.  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 $p^a$ elements of order $p$, then $p$ occurs in $a$ of the factors, and so on.  We call the $p_i^{e_i}$ the elementary divisors of $G$.

Alternatively, starting with an elementary divisor decomposition, we could let $n_1=\prod p_i^{e_i}$ where there is one copy of each distinct prime from among the invariant factors, with its highest exponent.  Then we could let $n_2$ be the product of all the distinct prime powers remaining, again with their highest exponents, and so on.  If the elementary divisors are $4, 64, 8, 27, 25, 5,$ then we have $n_1=64\cdot 27\cdot 25,n_2=8\cdot 5, n_3=4$.  We have $\mathbb{Z}/p_1^{e_1}\mathbb{Z}\oplus\dotsb\oplus\mathbb{Z}/ p_k^{e_k}\mathbb{Z}\cong \mathbb{Z}/n_1\mathbb{Z}\oplus\dotsb\oplus\mathbb{Z}/n_j\mathbb{Z}$, because at each stage we’re collecting cyclic groups of relatively prime order.  The $n_i$ also have the property that $n_{i+1}$ divides $n_i$ for each $i$.  We call them the invariant factors — since the $p_i^{e_i}$ are unique, so are the $n_i$.  Both of these decompositions are occasionally useful, especially for finite abelian groups.

We haven’t forgotten the $\mathbb{Z}^r$ on the end!  $r$ 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 $\mathbb{Q}$ (or any field of characteristic not occurring in the $p_i$), but then this page probably isn’t for you.  It’s a nice exercise to come up with a simpler, group-theoretic proof that $r$ 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, $\mathbb{Q}$ and $\mathbb{R}$ are both non-finitely-generated, and $\mathbb{R}$ includes $\mathbb{Q}$ 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, $\sqrt{2}$ by defining it as a solution of the polynomial $x^2-2$ with rational coefficients, but we can’t do this for $e$ or $\pi$.  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.