We will now prove the structure theorem for finitely generated abelian
groups, since it will be crucial for much of what we will do later.
denote the ring of integers, and
for each positive integer let
denote the ring of integers
modulo , which is a cyclic abelian group of order under
if there exists
such that every element of
can be obtained from the
We will prove the theorem as follows. We first remark that any
subgroup of a finitely generated free abelian group is finitely
generated. Then we see that finitely generated abelian groups can be
presented as quotients of finite rank free abelian groups, and such a
presentation can be reinterpreted in terms of matrices over the
integers. Next we describe how to use row and column operations over
the integers to show that every matrix over the integers is equivalent
to one in a canonical diagonal form, called the Smith normal form. We
obtain a proof of the theorem by reinterpreting in
terms of groups.
Suppose is a free abelian group of finite rank , and is a
subgroup of . Then is a free abelian group generated by at
The key reason that this is true is that is a finitely generated
module over the principal ideal domain
. We will give a complete
proof of a beautiful generalization of this result in the context of
Noetherian rings next time, but will not prove this proposition here.
Suppose is a finitely generated abelian group. Then there are
finitely generated free abelian groups and such that
be generators for
be the map that sends the
a surjective homomorphism, and by Proposition 3.0.4
is a finitely generated free abelian group.
This proves the corollary.
Suppose is a nonzero finitely generated abelian group. By the
corollary, there are free abelian groups and such that
. Choosing a basis for , we obtain an
, for some positive integer . By
, for some integer
, and the inclusion map
induces a map
. This homomorphism is left
multiplication by the matrix whose columns are the
images of the generators of in
. The of
this homomorphism is the quotient of
by the image of , and the
cokernel is isomorphic to . By augmenting with zero columns on
the right we obtain a square matrix with the same
cokernel. The following proposition implies that we may choose bases
such that the matrix is diagonal, and then the structure of
the cokernel of will be easy to understand.
We will see in the proof of Theorem 3.0.3 that
is uniquely determined by .
will be a product of matrices that define elementary
row operations and
will be a product corresponding to elementary
column operations. The elementary operations are:
- Add an integer multiple of one row to another (or a multiple
of one column to another).
- Interchange two rows or two columns.
- Multiply a row by .
Each of these operations is given by left or right multiplying by an
with integer entries, where
is the result of
applying the given operation to the identity matrix, and
invertible because each operation can be reversed using another row or
column operation over the integers.
To see that the proposition must be true, assume and perform
the following steps (compare [Art91, pg. 459]):
- By permuting rows and columns, move a nonzero entry of with
smallest absolute value to the upper left corner of . Now attempt
to make all other entries in the first row and column 0 by adding
multiples of row or column 1 to other rows (see step 2 below). If an
operation produces a nonzero entry in the matrix with absolute value
smaller than , start the process over by permuting rows and
columns to move that entry to the upper left corner of . Since the
integers are a decreasing sequence of positive integers, we
will not have to move an entry to the upper left corner infinitely
- Suppose is a nonzero entry in the first column, with
. Using the division algorithm, write
. Now add times the first row to the
th row. If , then go to step 1 (so that an entry with
absolute value at most is the upper left corner). Since we will
only perform step 1 finitely many times, we may assume .
Repeating this procedure we set all entries in the first column
(except ) to 0. A similar process using column operations
sets each entry in the first row (except ) to 0.
- We may now assume that is the only nonzero entry in the
first row and column. If some entry of is not divisible
by , add the column of containing to the first
column, thus producing an entry in the first column that is nonzero.
When we perform step 2, the remainder will be greater than 0.
Permuting rows and columns results in a smaller . Since
can only shrink finitely many times, eventually we will get
to a point where every is divisible by . If
is negative, multiple the first row by .
After performing the above operations, the first row and column
are zero except for
which is positive and divides
all other entries of
. We repeat the above steps for the
by deleting the first row and column.
The upper left entry of the resulting matrix will be divisible by
, since every entry of
is. Repeating the argument
inductively proves the proposition.
and the matrix
is equivalent to
Note that the determinants match, up to sign.
is a finitely generated abelian group, which we may assume
is nonzero. As in the paragraph before Proposition 3.0.6
we use Corollary 3.0.5
as a the cokernel
. By Proposition 3.0.6
there are isomorphisms
is a diagonal matrix with entries
is isomorphic to the cokernel of the diagonal matrix
as claimed. The
are determined by
smallest positive integer
requires at most
generators (we see from the representation (3.0.1
a product that
has this property and that no smaller positive