The main theorem of this lecture is:
Theorem 10.2.1 (Classification of finitely generated abelian groups). Every finitely generated abelian group is isomorphic to a product of cyclic groups: where and . Moreover, using the Chinese Remainder Theorem ( when ), we can arrange each to be a prime power. In this form, the decomposition is unique up to reordering.
The proof reduces to “linear algebra over ,” following Artin §14.4–14.7. The strategy has two ingredients: first, express any finitely generated -module as for some ; second, diagonalize by changes of basis over .
10.2.1 Step 1: Presenting a Module as a Quotient
Proposition 10.2.1 (Presentation of finitely generated -modules). If is a finitely generated -module, then there exist and such that Equivalently, there is an exact sequence .
This says: quotient by a -submodule of is the same as quotient of an abelian group by a subgroup. We need a lemma:
Lemma 10.2.1 (Submodules of are free). Any submodule of is finitely generated and free of rank .
Proof. By induction on . For : subgroups of are or for some , both free of rank .
Assume the result for , and let be a submodule. The projection sending restricts to a homomorphism .
The image is a submodule of , hence free and finitely generated by induction. The kernel is a subgroup of , hence free of rank 0 or 1.
If and are both finitely generated (and free), then so is . Let be generators of (a basis) and generators of . For any , write , so , meaning . Thus , so generates . That this set is a basis (i.e., free) follows from the directness of . ∎
Proof of the proposition. Let be finitely generated with generators . Define by . This is surjective. The kernel is a submodule, hence finitely generated by the lemma. Let be generators of . Define by . Then , giving the exact sequence and thus . ∎
The key difference from linear algebra over a field: we cannot divide freely in . The substitute is the Euclidean algorithm, which governs what “row and column operations” are available.
Definition 10.2.1 (Divisibility). The divisibility of a nonzero element is the largest such that for some . Equivalently, . An element is primitive if its divisibility is 1.
Lemma 10.2.2 (Primitive elements and bases). An element of can be part of a basis if and only if it is primitive. More generally, with divisibility satisfies where is a basis element.
Proof. Elements of a basis are primitive: linear independence prevents for .
Conversely, let be primitive. Without loss of generality, and . For each , write by Euclidean division. Change basis to . In this basis, with .
Repeating this process (which strictly decreases the minimum nonzero coefficient at each step) terminates in finitely many steps with for some basis vector . Since is primitive, . ∎
Theorem 10.2.2 (Smith Normal Form). For any , there exist bases of and of , and positive integers (where is the rank of ), such that In matrix form: Moreover, the construction gives of all entries of .
Proof. By induction on .
If , the statement is trivial for any .
Case : is determined by . Let . By the lemma, where can be extended to a basis of . Take .
Inductive step: Assume . Let , and choose with . Since has divisibility , write with primitive, and extend to a basis of . Extend to a basis of (using the lemma, since is primitive: if then , contradicting minimality unless ).
In the bases and , the matrix of has the form where is the restriction to composed with projection to .
For , write with . Change basis of to . The matrix becomes If any , then , contradicting the minimality of . So for all , and the matrix is By induction on , we can diagonalize by further basis changes in and , giving the desired diagonal form. ∎
Note. The Smith Normal Form is the integer analogue of row reduction. Over a field, Gaussian elimination diagonalizes any matrix (in fact reduces it to have 1’s and 0’s on the diagonal). Over , the Euclidean algorithm plays the role of division, and the diagonal entries need not be 1. The construction actually forces (the invariant factor divisibility chain), though we do not prove this here.
10.2.3 Step 3: Completing the Classification
Proof of the classification theorem. By Proposition 1 (presentation as quotient), any finitely generated abelian group satisfies for some . By the Smith Normal Form, after choosing appropriate bases, is spanned by for some positive integers and . Therefore The factors with are trivial and can be dropped. Relabeling the remaining as and setting gives the decomposition By the Chinese Remainder Theorem, each with decomposes as . Rearranging gives the prime power form. ∎
Example 10.2.1 (Finitely generated abelian groups of small order).
- Order 4: or . These are not isomorphic (the first has an element of order 4, the second does not).
- Order 6: (only one, since ).
- Order 8: , , or .
- Order 12: , or .
Note. The number is the free rank (or Betti number) of , and the integers (in prime power form) are the elementary divisors. These invariants completely determine up to isomorphism. The free rank counts the number of “copies of ,” while the elementary divisors encode the torsion (elements of finite order).
Exercise 10.2.1 (Modules and finitely generated abelian groups exercises).
- Classify all abelian groups of order 36 up to isomorphism.
- Let be the linear map with matrix . Compute the Smith Normal Form of and determine the group .
- Prove that a finitely generated abelian group is free (i.e., isomorphic to for some ) if and only if it is torsion-free (i.e., the only element of finite order is 0).
- Let be a finitely generated abelian group with for a prime . Show that where and . How many such groups are there (as a function of the partition structure of )?
- Show that . (Hint: use the presentation and the right-exactness of the tensor product.)
Show solution
(1) We have . For the 2-part: or . For the 3-part: or . Taking all combinations: So there are exactly 4 abelian groups of order 36.
(2) Start with . The gcd of all entries is 1, but let us apply row/column operations over .
: . Swap rows: . : . : . : . : .
Smith Normal Form: . So .
Check: , and . Consistent.
(3) One direction: if , then is torsion-free (only has finite order).
Conversely, if is finitely generated and torsion-free, write by the classification theorem. Each contributes a nonzero element of finite order (namely , which has order ). Since is torsion-free, , so .
(4) Since , the classification gives with and , i.e., . The number of such groups equals the number of partitions of . For example, gives partitions , so 5 groups.
(5) From the short exact sequence , tensor with . Right-exactness gives: The map has image , which is the cyclic subgroup of order , i.e., . The quotient therefore has order :